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Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

• A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
• Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
• Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
• Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
• A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
• Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
• A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
• A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

• A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
• A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
• A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
• As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
• Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
• A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
• A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
• A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
• Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
• Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
• Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
• Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
• On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
• Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Real World Algebra by Edward Zaccaro

Algebra is often taught abstractly with little or no emphasis on what algebra is or how it can be used to solve real problems. Just as English can be translated into other languages, word problems can be "translated" into the math language of algebra and easily solved. Real World Algebra explains this process in an easy to understand format using cartoons and drawings. This makes self-learning easy for both the student and any teacher who never did quite understand algebra. Includes chapters on algebra and money, algebra and geometry, algebra and physics, algebra and levers and many more. Designed for children in grades 4-9 with higher math ability and interest but could be used by older students and adults as well. Contains 22 chapters with instruction and problems at three levels of difficulty.

One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

Multiplication and division

Converting units of length (distance)

Rearranging the subject of a formula

Substitution into formulae

This topic is relevant for:

Speed Distance Time

Speed Distance Time Triangle

Here we will learn about the speed distance time triangle including how they relate to each other, how to calculate each one and how to solve problems involving them.

There are also speed distance time triangle worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is speed distance time?

Speed distance time is the formula used to explain the relationship between speed, distance and time. That is speed = distance ÷ time . Or to put it another way distance divided by speed will give you the time. Provided you know two of the inputs you can work out the third.

For example if a car travels for 2 hours and covers 120 miles we can work out speed as 120 ÷ 2 = 60 miles per hour.

The units of the the distance and time tell you the units for the speed.

What is the speed distance time triangle?

The speed distance time triangle is a way to describe the relationship between speed, distance and time as shown by the formula below.

\textbf{Speed } \bf{=} \textbf{ distance } \bf{\div} \textbf{ time}

“Speed equals distance divided by time”

Let’s look at an example to calculate speed.

If a car travels 66km in 1.5 hours then we can use this formula to calculate the speed.

This formula can also be rearranged to calculate distance or calculate time given the other two measures. An easy way to remember the formula and the different rearrangements is to use this speed distance time triangle. 

From this triangle we can work out how to calculate each measure: We can ‘cover up’ what we are trying to find and the formula triangle tells us what calculation to do.

Let’s look at an example to calculate time.

How long does it take for a car to travel 34 miles at a speed of 68 miles per hour?

Let’s look at an example to calculate distance.

What distance does a bike cover if it travels at a speed of 7 metres per second for 50 seconds?

What is the speed distance time formula?

The speed distance time formula is just another way of referring to the speed distance time triangle or calculation you can use to determine speed, time or distance.

• speed = distance ÷ time
• time = distance ÷ speed
• distance = speed x time

Time problem

We can solve problems involving time by remembering the formula for speed , distance and time . 

Calculate the time if a car travels at 15 miles at a speed of 36 mph.

Time = distance ÷ speed

Time = 15 ÷ 36 = 0.42 hours

0.42 ✕ 60 = 25.2 minutes

A train travels 42km between two stops at an average speed of 36 km/h. 

If the train departs at 4 pm, when does the train arrive?

Time = 42 ÷ 36 = 1.17 hours

1.17 ✕ 60= 70 minutes = 1 hour 10 minutes.

The average speed of a scooter is 18 km/h and the average speed of a cycle is 10 km/h. 

When both have travelled 99 km what is the difference in the time taken?

Time A = 99 ÷ 18 = 5.5 hours

Time B= 99 ÷ 10 = 9.9 hours

Difference in time = 9.9 – 5.5 = 4.4 hours

4.4 hours = 4 hours and 24 minutes

Units of speed, distance and time

• The speed of an object is the magnitude of its velocity. We measure speed most commonly in metres per second (m/s), miles per hour (mph) and kilometres per hour (km/hr).

The average speed of a small plane is 124mph.

The average walking speed of a person is 1.4m/s.

• We measure the distance an object has travelled most commonly in millimetres (mm), centimetres (cm), metres (m) and kilometres (km).

The distance from London to Birmingham is 162.54km.

• We measure time taken in milliseconds, seconds, minutes, hours, days, weeks, months and years. 

The time taken for the Earth to orbit the sun is 1 year or 365 days. We don’t measure this in smaller units like minutes of hours.

A short bus journey however, would be measured in minutes.

Speed, distance and time are proportional.

If we know two of the measurements we can find the other.

A car drives 150 miles in 3 hours.

Calculate the average speed, in mph, of the car.

Distance = 150 miles

Time = 3 hours

Speed = 150÷ 3= 50mph

Speed, distance, time and units of measure

It is very important to be aware of the units being used when calculating speed, distance and time.

• Examples of units of distance: mm, \ cm, \ m, \ km, \ miles
• Examples of units of time: seconds (sec), minutes, (mins) hours (hrs), days
• Examples of units of speed: metres per second (m/s), miles per hour (mph)

Note that speed is a compound measure and therefore involves two units; a combination of a distance in relation to a time.

When you use the speed distance time formula you must check that each measure  is in the appropriate unit before you carry out the calculation. Sometimes you will need to convert a measure into different units. Here are some useful conversions to remember.

Units of length

Units of time

1 minute = 60 seconds

1 hour = 60 minutes

1 day = 24 hours

Let’s look at an example.

What distance does a bike cover if it travels at a speed of 5 metres per second for 3 minutes?

Note here that the speed involves seconds, but the time given is in minutes. So before using the formula you must change 3 minutes into seconds.

3 minutes = 3\times 60 =180 seconds

Note also that sometimes you may need to convert an answer into different units at the end of a calculation.

Constant speed / average speed

For the GCSE course you will be asked to calculate either a constant speed or an average speed . Both of these can be calculated using the same formula as shown above.

However, this terminology is used because in real life speed varies throughout a journey. You should also be familiar with the terms acceleration (getting faster) and deceleration (getting slower).

Constant speed

A part of a journey where the speed stays the same.

Average speed

A journey might involve a variety of different constant speeds and some acceleration and deceleration. We can use the formula for speed to calculate the average speed over the course of the whole journey.

Average Speed Formula

Average speed is the total distance travelled by an object divided by the total time taken. To do this we can use the formula

Average speed =\frac{Total\, distance}{Total\, time}

If we are calculating an average speed in mph or km/h, we will need to ensure we have decimalised the time before we divide.

How to calculate speed distance time

In order to calculate speed, distance or time:

Write down the values of the measures you know with the units.

Check that the units are compatible with each other, converting them if necessary.

Substitute the values into the selected formula and carry out the resulting calculation.

Write your final answer with the required units.

Explain how to calculate speed distance time

Speed distance time triangle worksheet

Get your free speed distance time triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on   compound measures

Speed distance time  is part of our series of lessons to support revision on  compound measures . You may find it helpful to start with the main compound measures lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Compound measures
• Mass density volume
• Pressure force area
• Formula for speed
• Average speed formula

Speed distance time triangle examples

Example 1: calculating average speed.

Calculate the average speed of a car which travels 68 miles in 2 hours.

Speed: unknown

Distance: 68 miles

Time: 2 hours

2 Write down the formula you need to use from the speed, distance, time triangle.

3 Check that the units are compatible with each other, converting them if necessary.

The distance is in miles and the time is in hours. These units are compatible to give the speed in miles per hour.

4 Substitute the values into the formula and carry out the resulting calculation.

5 Write your final answer with the required units.

Example 2: calculating time

A golden eagle can fly at a speed of 55 kilometres per hour. Calculate the time taken for a golden eagle to fly 66 \ km, giving your answer in hours.

Speed: 55 \ km/hour

Distance: 66 \ km

Time: unknown

Write down the formula  you need to use from the speed, distance, time  triangle.

T=\frac{D}{S}

Time = distance \div speed

Speed is in km per hour and the distance is in km , so these are compatible to give an answer for time in hours.

Example 3: calculating distance

Calculate the distance covered by a train travelling at a constant speed of 112 miles per hour for 4 hours.

Speed: 112 \ mph

Distance: unknown

Time: 4 hours

D= S \times T

Distance = speed \times time

Speed is in miles per hour. The time is in hours. These units are compatible to find the distance in miles.

Example 4: calculating speed with unit conversion

A car travels for 1 hour and 45 minutes, covering a distance of 63 miles. Calculate the average speed of the car giving your answer in miles per hour (mph) .

Distance: 63 miles

Time: 1 hour and 45 minutes

S = \frac{D}{T}

Speed = distance \div time

The distance is in miles . The time is in hours and minutes. To calculate the speed in miles per hour , the time needs to be converted into hours only.

1 hour 45 minutes = 1\frac{3}{4} hours = 1.75 hours

Example 5: calculating time with unit conversion

A small plane can travel at an average speed of 120 miles per hour. Calculate the time taken for this plane to fly 80 miles giving your answer in minutes.

Speed: 120 \ mph

Distance: 80 \ miles

T = \frac{D}{S}

Speed is in miles per hour and the distance is in miles . These units are compatible to find the time in hours.

\frac{2}{3} hours in minutes

\frac{2}{3} \times 60 = 40

Example 6: calculating distance with unit conversion

A train travels at a constant speed of 96 miles per hour for 135 minutes. Calculate the distance covered giving your answer in miles.

Speed: 96 \ mph

Time: 135 minutes

D = S \times T

The speed is in miles per hour , but the time is in minutes. To make these compatible the time needs changing into hours and then the calculation will give the distance in miles .

135 minutes

135 \div 60 = \frac{9}{4} = 2\frac{1}{4} = 2.25

Common misconceptions

• Incorrectly rearranging the formula Speed = distance \div time

Make sure you rearrange the formula correctly. One of the simplest ways of doing this is to use the formula triangle. In the triangle you cover up the measure you want to find out and then the triangle shows you what calculation to do with the other two measures.

• Using incompatible units in a calculation

When using the speed distance time formula you must ensure that the units of the measures are compatible. For example, if a car travels at 80 \ km per hour for 30 minutes and you are asked to calculate the distance, a common error is to substitute the values straight into the formula and do the following calculation. Distance = speed \times time = 80 \times 30 = 2400 \ km The correct way is to notice that the speed uses hours but the time given is in minutes. Therefore you must change 30 minutes into 0.5 hours and substitute these compatible values into the formula and do the following calculation. Distance = speed \times time = 80 \times 0.5 = 40 \ km

Practice speed distance time triangle questions

1. A car drives 120 miles in 3 hours. Calculate its average speed.

2. A cyclist travels 100 miles at an average speed of 20 \ mph. Calculate how long the journey takes.

3. An eagle flies for 30 minutes at a speed of 66 \ km per hour. Calculate the total distance the bird has flown.

30 minutes = 0.5 hours

4. Calculate the average speed of a lorry travelling 54 miles in 90 minutes. Give your answer in miles per hour (mph).

Firstly convert 90 minutes to hours. 90 minutes = 1.5 hours

5. Calculate the time taken for a plane to fly 90 miles at an average speed of 120 \ mph. Give your answer in minutes.

180 minutes

Convert 0.75 hours to minutes

6. A helicopter flies 18 \ km in 20 minutes. Calculate its average speed in km/h .

Firstly convert 20 minutes to hours. 20 minutes is a third of an hour or \frac{1}{3} hours. \begin{aligned} &Speed = distance \div time \\\\ &Speed =18 \div \frac{1}{3} \\\\ &Speed = 54 \\\\ &54 \ km/h \end{aligned}

Speed distance time triangle GCSE questions

1. A commercial aircraft travels from its origin to its destination in a time of 2 hours and 15 minutes. The journey is 1462.5 \ km.

What is the average speed of the plane in km/hour?

2 hours 15 minutes = 2\frac{15}{60} = 2\frac{1}{4} = 2.25

2. John travelled 30 \ km in 90 minutes.

Nadine travelled 52.5 \ km in 2.5 hours.

Who had the greater average speed?

You must show your working.

90 minutes = 1.5 hours

John = 30 \div 1.5 = 20 \ km/h

Nadine = 52.5 \div 2.5 = 21 \ km/h

Nadine has the greater average speed.

3. The distance from Birmingham to Rugby is 40 miles.

Omar drives from Rugby to Birmingham at 60 \ mph.

Ayushi drives from Rugby to Birmingham at 50 \ mph.

How much longer was Ayushi’s journey compared to Omar’s journey? Give your answer in minutes.

For calculating time in hours for Omar or Ayushi.

For converting hours into minutes for Omar or Ayushi.

For correct final answer of 8 minutes.

Learning checklist

You have now learned how to:

• Use compound units such as speed
• Solve simple kinematic problem involving distance and speed
• Change freely between related standard units (e.g. time, length) and compound units (e.g. speed) in numerical contexts
• Work with compound units in a numerical context

The next lessons are

• Best buy maths
• Scale maths

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Solving Problems Involving Distance, Rate, and Time

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In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by d in math problems .

The rate is the speed at which an object or person travels. It is usually denoted by  r  in equations . Time is the measured or measurable period during which an action, process, or condition exists or continues. In distance, rate, and time problems, time is measured as the fraction in which a particular distance is traveled. Time is usually denoted by t in equations. 

Solving for Distance, Rate, or Time

When you are solving problems for distance, rate, and time, you will find it helpful to use diagrams or charts to organize the information and help you solve the problem. You will also apply the formula that solves distance, rate, and time, which is  distance = rate x tim e. It is abbreviated as:

There are many examples where you might use this formula in real life. For example, if you know the time and rate a person is traveling on a train, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled on a plane, you could quickly figure the distance she traveled simply by reconfiguring the formula.

Distance, Rate, and Time Example

You'll usually encounter a distance, rate, and time question as a word problem in mathematics. Once you read the problem, simply plug the numbers into the formula.

For example, suppose a train leaves Deb's house and travels at 50 mph. Two hours later, another train leaves from Deb's house on the track beside or parallel to the first train but it travels at 100 mph. How far away from Deb's house will the faster train pass the other train?

To solve the problem, remember that d represents the distance in miles from Deb's house and t  represents the time that the slower train has been traveling. You may wish to draw a diagram to show what is happening. Organize the information you have in a chart format if you haven't solved these types of problems before. Remember the formula:

distance = rate x time

When identifying the parts of the word problem, distance is typically given in units of miles, meters, kilometers, or inches. Time is in units of seconds, minutes, hours, or years. Rate is distance per time, so its units could be mph, meters per second, or inches per year.

Now you can solve the system of equations:

50t = 100(t - 2) (Multiply both values inside the parentheses by 100.) 50t = 100t - 200 200 = 50t (Divide 200 by 50 to solve for t.) t = 4

Substitute t = 4 into train No. 1

d = 50t = 50(4) = 200

Now you can write your statement. "The faster train will pass the slower train 200 miles from Deb's house."

Sample Problems

Try solving similar problems. Remember to use the formula that supports what you're looking for—distance, rate, or time.

d = rt (multiply) r = d/t (divide) t = d/r (divide)

Practice Question 1

A train left Chicago and traveled toward Dallas. Five hours later another train left for Dallas traveling at 40 mph with a goal of catching up with the first train bound for Dallas. The second train finally caught up with the first train after traveling for three hours. How fast was the train that left first going?

Remember to use a diagram to arrange your information. Then write two equations to solve your problem. Start with the second train, since you know the time and rate it traveled:

Second train t x r = d 3 x 40 = 120 miles First train t x r = d 8 hours x r = 120 miles Divide each side by 8 hours to solve for r. 8 hours/8 hours x r = 120 miles/8 hours r = 15 mph

Practice Question 2

One train left the station and traveled toward its destination at 65 mph. Later, another train left the station traveling in the opposite direction of the first train at 75 mph. After the first train had traveled for 14 hours, it was 1,960 miles apart from the second train. How long did the second train travel? First, consider what you know:

First train r = 65 mph, t = 14 hours, d = 65 x 14 miles Second train r = 75 mph, t = x hours, d = 75x miles

Then use the d = rt formula as follows:

d (of train 1) + d (of train 2) = 1,960 miles 75x + 910 = 1,960 75x = 1,050 x = 14 hours (the time the second train traveled)

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Distance, Time and Speed Word Problems | GMAT GRE Maths

Before you get into distance, time and speed word problems, check out how you can add a little power to your resume by getting our mini-MBA certificate.  

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time   and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems.   Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Time = Distance / speed = 20/4 = 5 hours.   Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Speed = Distance/time = 15/2 = 7.5 miles per hour.   Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

Distance covered = 4*40 = 160 miles

Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph   Now, take a look at the following example:

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

For these kinds of questions, a table like this might make it easier to solve.

  Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9            →        9d = 4(d+7.5)   →        9d=4d+30        →        d = 6.

So, he covered a distance of 6 miles in 1.5 hours.   Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

  s*t = (1/3)s*(t+30)      →        t = t/3 + 10      →        t = 15.

So the actual time taken to cover the distance is 15 minutes.

Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

Solved Questions on Trains

Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

So the two trains meet after 4 hours.   Example 2. A train leaves from a station and moves at a certain speed. After 2 hours, another train leaves from the same station and moves in the same direction at a speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed of the first train?

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.  

Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles.   Example 2. With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind?

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:  

  4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.  

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus.   Example 2. A plane covered a distance of 630 miles in 6 hours. For the first part of the trip, the average speed was 100 mph and for the second part of the trip, the average speed was 110 mph. what is the time it flew at each speed?

Our table looks like this.

Assuming the distance covered in the 1 st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

So the plane flew the first part of the journey in 3 hours and the second part in 3 hours.   Example 2. Two persons are walking towards each other on a walking path that is 20 miles long. One is walking at the rate of 3 mph and the other at 4 mph. After how much time will they meet each other?

  Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.  

Practice Questions for you to solve

Problem 1: Click here

A boat covers a certain distance in 2 hours, while it comes back in 3 hours. If the speed of the stream is 4 kmph, what is the boat’s speed in still water?

A) 30 kmph B) 20 kmph C) 15 kmph D) 40 kmph

Answer 1: Click here

Explanation

Let the speed of the boat be ‘s’ kmph.

Then, 2(s+4) = 3(s-4) → s = 20

Problem 2: Click here

A cyclist travels for 3 hours, travelling for the first half of the journey at 12 mph and the second half at 15 mph. Find the total distance he covered.

A) 30 miles B) 35 miles C) 40 miles D) 180 miles

Answer 2: Click here

Since it is mentioned, that the first ‘half’ of the journey is covered in 12 mph and the second in 15, the equation looks like

(d/2)/12 + (d/2)/15 = 3

Solving this gives d = 40 miles

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17 thoughts on “Distance, Time and Speed Word Problems | GMAT GRE Maths”

Meera walked to school at a speed of 3 miles per hour. Once she reached the school, she realized that she forgot to bring her books, so rushed back home at a speed of 6 miles per hour. She then walked back to school at a speed of 4 miles per hour. All the times, she walked in the same route. please explain above problem

When she walks faster the time she takes to reach her home and school is lower. There is nothing wrong with the statement. They never mentioned how long she took every time.

a man covers a distance on a toy train.if the train moved 4km/hr faster,it would take 30 min. less. if it moved 2km/hr slower, it would have taken 20 min. more .find the distance.

Let the speed be x. and time be y. A.T.Q, (x+4)(y-1/2)=d and (x-2)(y+1/3)=d. Equate these two and get the answer

Could you explain how ? you have two equations and there are 3 variables.

The 3rd equation is d=xy. Now, you have 3 equations with 3 unknowns. The variables x and y represent the usual speed and usual time to travel distance d.

Speed comes out to be 20 km/hr and the time taken is 3 hrs. The distance traveled is 60 km.

(s + 4) (t – 1/2)= st 1…new equotion = -1/2s + 4t = 2

(s – 2) (t + 1/3)= st 2…new equotion = 1/3s – 2t = 2/3

Multiply all by 6 1… -3s + 24t = 12 2… 2s – 12t = 4 Next, use elimination t= 3 Find s: -3s + 24t = 12 -3s + 24(3) = 12 -3s = -60 s= 20

st or distance = 3 x 20 = 60 km/h

It’s probably the average speed that we are looking for here. Ave. Speed= total distance/ total time. Since it’s harder to look for one variable since both are absent, you can use, 3d/ d( V2V3 + V1V3 + V1V2/ V1V2V3)

2 girls meenu and priya start at the same time to ride from madurai to manamadurai, 60 km away.meenu travels 4kmph slower than priya. priya reaches manamadurai and at turns back meeting meenu 12km from manamaduai. find meenu’s speed?

Hi, when the two girls meet, they have taken equal time to travel their respective distance. So, we just need to equate their time equations

Distance travelled by Meenu = 60 -12 = 48 Distance travelled by Priya = 60 + 12 = 72 Let ‘s’ be the speed of Meenu

Time taken by Meenu => t1 = 48/s Time taken by Priya => t2 = 72/(s+4)

t1=t2 Thus, 48/s = 72/(s+4) => 24s = 192 => s = 8Km/hr

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 KMS away from A at the same time. On the way, however the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Let speed of the CAR BE x kmph.

Then, speed of the train = 3/2(x) .’. 75/x – 75/(3/2)x= 125/(10*60) — subtracting the times travelled by two them hence trains wastage time

therefore x= 120 kmph

A cyclist completes a distance of 60 km at the same speed throughout. She travels 10 km in one hour. She stops every 20 km for one hour to have a break. What are the two variables involved in this situation?

For the answer, not variables: 60km divided by 10km/h=6 hours 60 divided by 20= 3 hours 3 hours+6 hours= 9 hours Answer: 9 hours

Let the length of the train to prod past a point be the intrinsic distance (D) of the train and its speed be S. Its speed, S in passing the electric pole of negligible length is = D/12. The length of the platform added to the intrinsic length of the train. So, the total distance = D + 200. The time = 20 secs. The Speed, S = (D + 200)/20 At constant speed, D/12 = (D + 200)/20 Cross-multiplying, 20D = 12D + 200*12 20D – 12D = 200*12; 8D = 200*12 D = 200*12/8 = 300m. 4th Aug, 2018

Can anyone solve this? Nathan and Philip agree to meet up at the park at 5:00 pm. Nathan lives 300 m due north of the park, and Philip lives 500 m due west of the park. Philip leaves his house at 4:54 pm and walks towards the park at a pace of 1.5 m/s, but Nathan loses track of time and doesn’t leave until 4:59 pm. Trying to avoid being too late, he jogs towards the park at 2.5 m/s. At what rate is the distance between the two friends changing 30 seconds after Nathan has departed?

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Speed, Distance, and Time

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Classical mechanics.

Hardcore training for the aspiring physicist.

A common set of physics problems ask students to determine either the speed, distance, or travel time of something given the other two variables. These problems are interesting since they describe very basic situations that occur regularly for many people. For example, a problem might say: "Find the distance a car has traveled in fifteen minutes if it travels at a constant speed of \(75 \text {km/hr}\)." Often in these problems, we work with an average velocity or speed, which simplifies the laws of motion used to calculate the desired quantity. Let's see how that works.

Application and Extensions

As long as the speed is constant or average, the relationship between speed , distance , and time is expressed in this equation

\[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}},\]

which can also be rearranged as

\[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}\]

\[\mbox{Distance} = \mbox{Speed} \times \mbox{Time}.\]

Speed, distance, and time problems ask to solve for one of the three variables given certain information. In these problems, objects are moving at either constant speeds or average speeds.

Most problems will give values for two variables and ask for the third.

Bernie boards a train at 1:00 PM and gets off at 5:00 PM. During this trip, the train traveled 360 kilometers. What was the train's average speed in kilometers per hour? In this problem, the total time is 4 hours and the total distance is \(360\text{ km},\) which we can plug into the equation: \[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}}= \frac{360~\mbox{km}}{4~\mbox{h}} = 90~\mbox{km/h}. \ _\square \]

When working with these problems, always pay attention to the units for speed, distance, and time. Converting units may be necessary to obtaining a correct answer.

A horse is trotting along at a constant speed of 8 miles per hour. How many miles will it travel in 45 minutes? The equation for calculating distance is \[\mbox{Distance} = \mbox{Speed} \times \mbox{Time},\] but we won't arrive at the correct answer if we just multiply 8 and 45 together, as the answer would be in units of \(\mbox{miles} \times \mbox{minute} / \mbox{hour}\). To fix this, we incorporate a unit conversion: \[\mbox{Distance} = \frac{8~\mbox{miles}}{~\mbox{hour}} \times 45~\mbox{minutes} \times \frac{1~\mbox{hour}}{60~\mbox{minutes}} = 6~\mbox{miles}. \ _\square \] Alternatively, we can convert the speed to units of miles per minute and calculate for distance: \[\mbox{Distance} = \frac{2}{15}~\frac{\mbox{miles}}{\mbox{minute}} \times 45~\mbox{minutes} = 6~\mbox{miles},\] or we can convert time to units of hours before calculating: \[\mbox{Distance} = 8~\frac{\mbox{miles}}{\mbox{hour}} \times \frac{3}{4}~\mbox{hours} = 6~\mbox{miles}.\] Any of these methods will give the correct units and answer. \(_\square\)

In more involved problems, it is convenient to use variables such as \(v\), \(d\), and \(t\) for speed, distance, and velocity, respectively.

Alice, Bob, Carly, and Dave are in a flying race!

Alice's plane is twice as fast as Bob's plane. When Alice finishes the race, the distance between her and Carly is \(D.\) When Bob finishes the race, the distance between him and Dave is \(D.\)

If Bob's plane is three times as fast as Carly's plane, then how many times faster is Alice's plane than Dave's plane?

Albert and Danny are running in a long-distance race. Albert runs at 6 miles per hour while Danny runs at 5 miles per hour. You may assume they run at a constant speed throughout the race. When Danny reaches the 25 mile mark, Albert is exactly 40 minutes away from finishing. What is the race's distance in miles? \[\] Let's begin by calculating how long it takes for Danny to run 25 miles: \[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}= \frac{25~\mbox{miles}}{5~\mbox{miles/hour}}= 5~\mbox{hours}.\] So, it will take Albert \(5~\mbox{hours} + 40~\mbox{minutes}\), or \(\frac{17}{3}~\mbox{hours}\), to finish the race. Now we can calculate the race's distance: \[\begin{align} \mbox{Distance} &= \mbox{Speed} \times \mbox{Time} \\ &= (6~\mbox{miles/hour}) \times \left(\frac{17}{3}~\mbox{hours}\right) \\ &= 34~\mbox{miles}.\ _\square \end{align}\]

A cheetah spots a gazelle \(300\text{ m}\) away and sprints towards it at \(100\text{ km/h}.\) At the same time, the gazelle runs away from the cheetah at \(80\text{ km/h}.\) How many seconds does it take for the cheetah to catch the gazelle? \[\] Let's set up equations representing the distance the cheetah travels and the distance the gazelle travels. If we set distance \(d\) equal to \(0\) as the cheetah's starting point, we have \[\begin{align} d_\text{cheetah} &= 100t \\ d_\text{gazelle} &= 0.3 + 80t. \end{align}\] Note that time \(t\) here is in units of hours, and \(300\text{ m}\) was converted to \(0.3\text{ km}.\) The cheetah catches the gazelle when \[\begin{align} d_\text{cheetah} &=d_\text{gazelle} \\ 100t &= 0.3 + 80t \\ 20t &= 0.3 \\ t &= 0.015~\mbox{hours}. \end{align}\] Converting that answer to seconds, we find that the cheetah catches the gazelle in \(54~\mbox{seconds}\). \(_\square\)

Two friends are crossing a hundred meter railroad bridge when they suddenly hear a train whistle. Desperate, each friend starts running, one towards the train and one away from the train. The one that ran towards the train gets to safety just before the train passes, and so does the one that ran in the same direction as the train.

If the train is five times faster than each friend, then what is the train-to-friends distance when the train whistled (in meters)?

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"Distance" Word Problems

Explanation More Examples

What is a "distance" word problem?

"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate", "velocity", or "speed"), or else you are told to regard to object as moving at some average speed.

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Distance Word Problems

Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt , where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time.

Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using mis-matched units, and you have to catch this and convert to the correct units.

In case you're wondering, this type of exercise requires that the rate be fixed and steady (that is, unchanging) for the d = rt formula to work. The only way you can deal with a speed that might be changing over time is to take the average speed over the time or distance in question. Working directly with changing speeds will be something you'll encounter in calculus, as it requires calculus-based (or more advanced) methods.

What is the difference between a fixed speed and an average speed?

A fixed-speed exercise is one in which the car, say, is always going exactly sixty miles an hour; in three hours, the car, on cruise-control, will have gone 180 miles. An average-speed exercise is one in which the car, say, averaged forty miles an hour, but this average includes the different speeds related to stop lights, highways, and back roads; in three hours the car went 120 miles, though the car's speed was not constant. Most of the exercises you'll see will be fixed-speed exercises, but obviously they're not very "real world". It's a simplification they do in order to make the situation feasible using only algebraic methods.

What is an example of a "distance" word problem?

• A 555 -mile, 5 -hour plane trip was flown at two speeds. For the first part of the trip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed?

There is a method for setting up and solving these exercises that I first encountered well after I'd actually been doing them while taking a class as an undergraduate. But, as soon as I was introduced to the method, I switched over, because it is *so* way easier.

First I set up a grid, with the columns being labelled with the variables from the "distance" formula, and the rows being labelled with the "parts" involved:

In the first part, the plane covered some distance. I don't know how much, so I'll need a variable to stand for this unknown value. I'll use the variable they give me in the distance equation:

They gave me the speed, or rate, for this part, so I'll add this to my table:

The plane flew for some amount of time during this first part, but I don't know how long that was. So I need a variable to stand for this unknown value; I'll use the variable from the distance equation:

For the second part, the plane travelled the rest of the total distance. I don't know the exact distance that was flown during this second part, but I do know that it was "however much was left of the 555 miles, after the first d miles were flown in the first part. "How much was left after [some amount] was taken out" is expressed with subtraction: I take the amount that has been taken care of already, and subtract this from whatever was the total. Adding this to my table, I get:

They've given me the speed, or rate, for the second part, and I can use the same "How much is left?" construction for whatever was the time for this second part. So now my table looks like this:

For the "total" row, I add down (or take info from the exercise statement):

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Why did I not add down in the "rate" column? Because I cannot add rates! In this exercise, adding the rates would have said that the average rate for the entire trip was 105 + 115 = 220 miles per hour. But obviously this makes no sense.

The genius of this table-based method of set-up is that I can now create equations from the rows and columns. In this exercise, there is more than one way to proceed. I'll work with the "distance" equation to create expressions for the distances covered in each part.

Multiplying across, the first row tells me that the distance covered in the first part of the flight was:

1st part distance: 105 t

Again multiplying across, the second row tells me that the distance covered in the second part of the flight was:

2nd part distance: 115(5 − t )

I can add these two partial-distance expressions, and set them equal to the known total distance:

105 t + 115(5 − t ) = 555

This is an equation in one variable, which I can solve:

105 t + 115(5 − t ) = 555 105 t + 575 − 115 t = 555 575 − 10 t = 555 20 = 10 t 2 = t

Looking back at my table, I see the I had defined t to be the time that the plane spent in the air on the first part of its journey. Looking back at the original exercise, I see that they want to know the times that the plane spent at each of the two speeds.

I now have the time for the first part of the flight; the time was two hours. The exercise said that the entire trip was five hours, so the second part must have taken three hours (found by subtracting the first-part time from the total time). They haven't asked for the partial distances, so I now have all the information I need; no further computations are necessary. My answer is:

first part: 2 hours second part: 3 hours

When I was setting up my equation, I mentioned that there was more than one way to proceed. What was the other way? I could have used the table to create an expression for each of the two partial times, added, set the result equal to the given total, and solved for the variable d . Since the distance equation is d  =  rt , then the expressions for the partial times would be created by solving the equation for t  = . My work would have looked like this:

first part: d /105

second part: (555 − d )/115

adding: d /105 + (555 − d )/115 = 5 23 d + 11,655 − 21 d = 12,075 2 d = 420 d = 210

Looking back at my table, I would have seen that this gives me the distance covered in the first half of the flight. Looking back at the exercise, I would have seeing that they are wanting times, not distances. So I would have back-solved for the time for the first part, and then done the subtraction to find the time for the second part. My work would have had more steps, but my answer would have been the same.

There are three things that I hope you take from the above example:

• Using a table or grid to keep track of what you're doing can be incredibly helpful.
• It is important to clearly define your variables, so you know (by the end) what you'd meant (back in the beginning), so you can apply your results correctly.
• You should always check the original exercise, so you can be sure that you're answering the question that they'd actually asked.

(My value for the distance, found above, is correct, but was not what they'd asked for.) But even more important to understand is this:

NEVER TRY TO ADD RATES! Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph? Of course not.

Can I even average the rates? If I drove at 20 mph for one hour, and then drove 60 mph for two hours, then I would have travelled 140 miles in three hours, or a little under 47 mph. But 47 is not the average of 20 and 60 .

As you can see, the actual math involved in solving this type of exercise is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed. Try your hand at solving, and click on the links to get pop-ups from which to check your equations and solutions.

• An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices.

I will start in the usual way, by setting up my table:

I have labelled my rows so it's clear how they relate to the exercise. Now I need to fill in the rows. As before, I don't know the distance or the time for the part where the executive was driving, so I'll use variables for these unknowns, along with the given rate.

For the flying portion of the trip, I'll use the "how much is left" construction, along with the given rate, to fill in my second row. I'll also fill in the totals.

The first row gives me the equation d  = 30 t . The second row is messier, giving me the equation:

150 − d = 60(3 − t )

There are various ways I can go from here; I think I'll solve this second equation for the variable d , and then set the results equal to each other.

150 − d = 60(3 − t ) 150 − 60(3 − t ) = d

Setting equal these two expressions for d , I get:

30 t = 150 − 60(3 − t )

Solve for t ; interpret the value; state the final answer.

• A car and a bus set out at 2 p.m. from the same point, headed in the same direction. The average speed of the car is 30 mph slower than twice the speed of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of the car.

Both vehicles travelled for the same amount of time.

The car's values are expressed in terms of the bus' values, so I'll use variables for the bus' unknowns, and then define the car in terms of the bus' variables. This gives me:

(As it turns out, I won't need the "total" row this time.) The car's row gives me:

d + 20 = 2(2 r − 30)

This is not terribly helpful. The second row gives me:

I'll use the second equation to simplify the first equation by substituting " 2 r " from the second equation in for the " d " in the first equation. Then I'll solve the equation for the value of " r ". Finally, I'll need to interpret this value within the context of the exercise, and then I'll state the final answer.

(Remember that the expression for the car's speed, from the table, was 2 r  − 30 , so all you need to do is find the numerical value of this expression. Just evaluate; don't try to solve — again — for the value of r .)

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Distance, speed, time

In this lesson we will look at three physical quantities: distance , speed , and time .

We have already studied distance in the units of measurement lesson . Simply put, distance is the length from one point to another. (Example: the distance from home to school is 2 kilometers.)

When dealing with long distances, they will mostly be measured in meters and kilometers. Distance is denoted by the Latin letter S . You can also denote by another letter, but the letter S is generally accepted.

Speed is the distance traveled by a body in a unit of time. A unit of time is one hour, one minute, or one second.

Suppose that two students decided to compete and run from the yard to the playground. The distance from the yard to the playground is 100 meters. The first pupil runs in 25 seconds. The second one ran in 50 seconds. Who is the fastest student?

The one who ran the greater distance in 1 second is the fastest. He is said to have more speed. In this case, the speed of the students is the distance they run in 1 second.

To find the speed, you have to divide the distance by the time of movement.  Let's find the speed of the first schoolboy. To do this, divide 100 meters by the time of movement of the first schoolboy, that is, by 25 seconds:

100 m : 25 s = 4

If distance is given in meters and travel time in seconds, speed is measured in meters per second (m/s) . If distance is given in kilometers and travel time in hours, speed is measured in kilometers per hour (km/h) . 

We have distance in meters and time in seconds. So the speed is measured in meters per second (m/s).

100 m : 25 s = 4 (m/s)

So, the speed of the first student is 4 meters per second (m/s).

Now let's find the speed of the second pupil. To do this, divide the distance by the time of movement of the second student, i.e., by 50 seconds:

100 m : 50 s = 2 (m/s)

So the speed of the second student is 2 meters per second (m/s).

The speed of the first student is 4 (m/s) The speed of the second student is 2 (m/s)

4 (m/s) > 2 (m/s)

The speed of the first student is faster. So he got to the playground faster. Speed is denoted by the Latin letter - v .

Sometimes there is a situation where you want to know how long it takes the object to cover a particular distance (travel that distance).

For example, it is 1000 meters from the house to the sports club. We need to get there on a bicycle. Our speed will be 500 meters per minute (500 m/min). How long will it take us to get to the athletic section?

If we will travel 500 meters in one minute, how many such minutes (with five hundred meters each) will be in 1000 meters?

Obviously, we have to divide 1,000 meters by the distance we travel in one minute. That is, 500 meters. The result will be the time in which we will reach a sports club:

1000 : 500 = 2 (min)

The time of movement is denoted by the small Latin letter - t .

The relationship of speed, time, distance

Speed is usually denoted by the small Latin letter v ,

time of movement by the small letter - t ,

the distance traveled by the small letter - s .

Speed, time and distance are related to each other.

If you know the speed and time of movement, you can find the distance. It is equal to speed multiplied by time:

For example, we left the house and went to the store. It took us 10 minutes to get to the store. Our speed was 50 meters per minute. If we know our speed and time, we can find the distance.

If we walked 50 meters in one minute, how many of these 50 meters will we walk in 10 minutes? Obviously, by multiplying 50 meters by 10, we will determine the distance from the house to the store:

v = 50 (m/min)

t = 10 minutes

s = v × t = 50 × 10 = 500 (meters to the store)

If time and distance are known, you can find the speed:

For example, the distance from home to the school is 900 meters. It took the student 10 minutes to reach the school. What was his speed?

The speed of a schoolboy is the distance he travels in one minute. If he traveled 900 meters in 10 minutes, what distance did he travel in one minute?

To answer this one, you have to divide the distance by the time of the schoolboy's movement:

s = 900 meters

v = s : t = 900 : 10 = 90 (m/min)

If you know the speed and distance, you can find the time:

For example, we have to walk 500 meters from our house to the sport club. Our speed will be 100 meters per minute (100 m/min). How long will it take us to reach it?

If we walk 100 meters in one minute, how many such minutes with 100 meters are in 500 meters?

To answer this question we need to divide 500 meters by the distance we will walk in one minute, that is, by 100. Then we will get the time in which we will reach the sports section:

s = 500 meters

v = 100 (m/minute)

t = s : v = 500 : 100 = 5 (minutes to the sports section)

Video lesson

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Time, Speed and Distance : Question Types

• Type 2. Speed and distance formula: Average Speed = Total Distance Travelled/Total Time Taken Illustration 4: Suraj drives first 120 km in 2 hrs and next 180 km in next 4 hrs. What is his average speed for the entire trip in km per hour? Sol: Total Distance travelled = 120 + 180 = 300 km. Total Time taken = 2 + 4 = 6 hrs. Average Speed =Total Distance Travelled/Total Time Taken = 300/6 = 50 km/hr. Illustration 5: A train covered first 120 km at a speed of 20 km an hour and then covered the remaining 180 km at a speed of 45 km an hour. Find its average speed. Sol: Total distance = 120 + 180 = 300 km. Time taken for the first 120 km = 120/20 → 6 hrs. Time taken for the next 180 km = 180/45 → 4 hrs. Total time taken = 6 + 4 = 10 hrs. Average Speed =Total Distance Travelled/Total Time Taken = 300/10 = 30 km/hr. Illustration 6: A cyclist travels at 10 km/hr for 2 hours and then at 13 km/hr for 1 hour. Find his average speed. Sol: Distance travelled in first 2 hours =10 × 2 = 20 km. Distance travelled in next 1 hour =13 × 1 = 13 km. Total Distance travelled = 20 + 13 = 33 km. Total time taken = 2 + 1 = 3 hrs. Average Speed =Total Distance Travelled/Total Time Taken =33/3 = 11 km/hr.
• Type 3. Average Speed: When the time travelled is the same: The average speed of traveling at two different speeds for the same time span is just the simple average of these two speeds. Illustration 7: A motorist travels one hour at an average speed of 45 kmph and the next hour at an average speed of 65 kmph. Then what is his average speed? Sol: (45 + 65) ÷ 2 = 55 kmph. The total distance traveled by the motorist in these two hours = 65 + 45 = 110 km and he has taken two hours. Therefore, his average speed = 55 kmph.
• Type 4. Average Speed: When the distance travelled is the same: However, the above simple average rule does not work when the time span of each of the different speeds is different and only the distance is the same. In this case, one should take the simple average of the inverses of the two speeds and then again inverse the speed. Illustration 8: On my way from the office to the Pimpri class, I drive at 30 kmph and on the return journey I drive at 45 kmph. What is my average speed of travel? Sol: 37.5 kmph is incorrect as the time traveled is different in both the cases and only the distances are same. Let the distance between the office and Pimpri class be x km. ∴Time taken on my onward journey = x/30 hours and time taken on my return journey = x/45 . ∴The total time taken for my onward and return journey = x/30 + x/45 = 5x/90 hours. The total distance traveled both ways = 2x km ∴ Average speed = 2x/(5x/90) = 36 kmph.
• Theory & Concepts
• Solved Examples
• Practice Problems: Level 01
• Question Types
• When two objects are moving in the same direction, then their relative speed is the difference between the two speeds.
• When it crosses a stationary man / lamp post / sign post / pole - in all these cases the object which the train crosses is stationary - and the distance traveled is the length of the train.
• When it crosses a platform / bridge - in these cases, the object which the train crosses is stationary - and the distance traveled is the length of the train + length of the object.
• When it crosses another train which is moving at a particular speed in the same / opposite direction - in these cases, the other train is also moving and the relative speed between them is taken depending upon the direction of the other train - and the distance is the sum of the lengths of both the trains.
• When it crosses a car / bicycle / a mobile man - in these cases again the relative speed between the train and the object is taken depending upon the direction of the movement of the other object relative to the train - and the distance traveled is the length of the train.

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Problems on Calculating Speed

Here we will learn to solve different types of problems on calculating speed.

We know, the speed of a moving body is the distance traveled by it in unit time.             

Formula to find out speed = distance/time

Word problems on calculating speed:

1.  A man walks 20 km in 4 hours. Find his speed.

Solution:            

Distance covered = 20 km

Time taken = 4 hours

We know, speed = distance/time            

                       = 20/4 km/hr

Therefore, speed = 5 km/hr

2. A car covers a distance of 450 m in 1 minute whereas a train covers 69 km in 45 minutes. Find the ratio of their speeds.

Speed of car = Distance covered/Time taken = 450/60 m/sec = 15/2

                                                            = 15/2 × 18/5 km/hr

                                                            = 27 km/hr

Distance covered by train = 69 km

Time taken = 45 min = 45/60 hr = 3/4 hr

Therefore, speed of trains = 69/(3/4) km/hr

                                    = 69/1 × 4/3 km/hr

                                    = 92 km/hr

Therefore, ratio of their speed i.e., speed of car/speed of train = 27/92 = 27 : 92

3. Kate travels a distance of 9 km from her house to the school by auto-rickshaw at 18 km/hr and returns on rickshaw at 15 km/hr. Find the average speed for the whole journey.

Time taken by Kate to reach school = distance/speed = 9/18 hr = 1/2 hr

Time taken by Kate to reach house to school = 9/15 = 3/5 hr

Total time of journey = (1/2 + 3/5) hr

Total time of journey = (5 + 6)/10 = 11/10 hr

Total distance covered = (9 + 9) km = 18 km

Therefore, average speed for the whole journey = distance/speed = 18/(11/10) km/hr

= 18/1 × 10/11 = (18 × 10)/(1 × 11) km/hr

                      = 180/11 km/hr

                      = 16.3 km/hr (approximately)

Speed of Train

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• Speed Distance Time Word Problems with Solutions

Introduction

Speed, distance, and time are the three main pillars behind mathematics and physics. Whenever you are presented with a question related to any kind of transportation, you should immediately relate to these concepts. The train speed, the relation between the time taken by two cars at different speeds or even a simple question about a person walking from one place to another can be answered using simple formulas.

What is Speed?

When you think of speed, you must think of how fast? Speed is all about calculating the rate at which something can be accomplished.

A Device Used to Measure Speed Known as a ‘Speedometer’

What is Distance?

Distance in this concept refers to the distance travelled. It means how far? It can be measured in meters, for small distances and kilometres, for larger distances.

Distance is the Measurement From One Place to Another

What is Time?

Time is the answer to the question, how long? The best way to measure time is using the ‘world clock.’ In ancient times, time was measured with the help of the sun and the moon, and other celestial objects.

Time as Shown by a Clock

Relation Between Speed Distance and Time

There is a very simple mathematical relation between speed, distance and time.

$\text{speed=}\frac{\text{distance}}{\text{time}}$

$\text{time=}\frac{\text{distance}}{\text{speed}}$

$\text{distance=speed }\!\!\times\!\!\text{ time}$

These relations can be memorised using the triangle method. See the following image to understand the use of this method.

Explaining the Triangle Method

Speed Distance Time Word Problems With Solutions

Q1. A train is travelling at a speed of 160 km/hour. It takes 15 hours to cover the distance from city A to city B. Find the distance between the two cities.

Speed= 160 km/hr

Time= 15 hours

Using the formula, $\text{distance=speed }\!\!\times\!\!\text{ time}$

Distance= $160\times 15=2400km$

Answer: The distance between city A and city B is 2400km

Q2. A car travelling from city A to city C completes the journey in 3 hours, whereas a person travelling on a bike completes the journey in 5 hours. What is the speed of the man on the bike if the car is travelling at 45 kmph?

Find the distance using the information on the car

Speed= 45km/hour

Time= 3 hours

$\therefore \text{distance}=45\times 3=135km$

Using the distance, we can find the speed of the bike

Formula used, $\text{speed=}\frac{\text{distance}}{\text{time}}$

Time taken by bike is 5 hours

$\therefore \text{speed}=\frac{135}{5}=27$

Answer: The bike is moving at a speed of 27 kmph.

Q3. A person travels at a speed of 15 kmph from point A to point B, which are 25 km from each other. Another person is travelling from point C to point B, 44 km from each other. Both people start their journey at the same time; the second person arrives at the point 2 hours after the first. What is the speed at which the second person was travelling?

Speed of person 1 = 15km/hour

Distance travelled by person 1= 25 km/hour

Time taken by first-person = $\frac{\text{distance}}{\text{speed}}=\frac{25}{15}=\frac{5}{3}$hours

Note: While solving the questions, keep the answers in fractions until you reach the final answer to make calculations easier.

Time taken by second person =$\frac{5}{3}+2=\frac{11}{3}$hours

Distance travelled by second person= 44 km

$\therefore \text{speed}=\frac{\text{distance}}{\text{time}}=\frac{44}{\frac{11}{3}}=\frac{44\times 3}{11}=4\times 3=12$kmph

Answer: The second person was travelling at a speed of 12kmph.

Q4. Raju is travelling from one station to another 560km away in a train which is moving at a speed of 125kmph. Monu is travelling to the same destination at a speed of 80kmph. Monu starts his journey 3 hours after Raju, from 120 km closer to the destination. How long will Raju have to wait at the train station for Monu to arrive?

Distance travelled by Raju = 560km

Speed of Raju= 125km/hour

∴ Time taken by Raju to reach the station = $\frac{\text{distance}}{\text{speed}}=\frac{560}{125}=\frac{112}{25}$hours

Distance travelled by Monu = $560-120=440$km

The speed at which Monu travelled = 80kmph

Time taken by Monu to complete journey = $\frac{440}{80}=\frac{22}{4}=\frac{11}{2}$hours

Total time taken by Monu = $\frac{11}{2}+3=\frac{17}{2}$hours

Therefore, the time Raju had to wait = difference between the two times taken

$\therefore \frac{17}{2}-\frac{112}{25}=\frac{17\times 25}{50}-\frac{112\times 2}{50}=\frac{201}{50}=4.02$hours

Answer: Raju had to wait for 4.02 hours at the train station

Convert 4.02 hours to hours and minutes

$\therefore 0.02=\frac{2}{100}\times 60=1.2$minutes

$\therefore 0.2=\frac{2}{10}\times 60=12$seconds

Therefore, to be precise, Raju had to wait at the train station for 4 hours, 1 minute and 12 seconds.

FAQs on Speed Distance Time Word Problems with Solutions

1. Is there any other formula to relate the speed distance and time?

No, there is only one relation that one must remember to solve all speed distance time problems. The triangle method is the easiest and fastest way to remember these relations.

2. What are other speed measures, and how to convert them to km/hour?

The other speed measures include meters per second, denoted as m/sec or m/s and miles per hour, denoted as mph.

 To convert m/sec into km/hour, one must use the following formula—

$\frac{m}{s}\times \frac{5}{18}=\frac{km}{hr}$ ($\because $1 m/s = 3.6 km/h)

To convert mph into kmph one must use the following relation—

1 mile= approximately 1.609 km

1 km= 0.6214 miles

3. What is the device that records distance called?

The device that calculates the distance travelled is known as an odometer.

4. What other factors need to be considered when calculating real-life problems?

Simple formulas are not useful when solving speed distance time problems in real life since many other factors affect the outcome. One must consider air drag, frictional forces, repulsion, wear and tear and, most importantly, gravitational forces when solving these real-life problems. The formulas of relation, however, remain the same.

Distance Word Problems

In these lessons, we will learn how to solve rate time distance word problems where the objects are traveling in opposite directions. You may be required to find the time when the objects meet or the time when the objects are a certain distance apart.

Related Pages Rate, Time, Distance - Algebra Word Problems Distance Problems Distance Word Problems Average Speed Problems

Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of time .

The formula for distance problems is: distance = rate × time or d = r × t .

Things to watch out for:

Make sure that you change the units when necessary. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately.

It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.

The following diagrams give the steps to solve Rate Time Distance Word Problems. Scroll down the page for examples and solutions.

Distance Problems: Traveling In Opposite Directions

Example: A bus and a car leave the same place and traveled in opposite directions. If the bus is traveling at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?

Solution: Step 1: Set up a rtd table.

Step 2: Fill in the table with information given in the question.

If the bus is traveling at 50 mph and the car is traveling at 55 mph, in how many hours will they be 210 miles apart?

Let t = time when they are 210 miles apart.

Step 3: Fill in the values for d using the formula d = rt

Step 4: Since the total distance is 210, we get the equation:

Answer: They will be 210 miles apart in 2 hours.

Example of a distance word problem with vehicles moving in opposite directions

In this video, you will learn to solve introductory distance or motion word problems - for example, cars traveling in opposite directions, bikers traveling toward each other, or one plane overtaking another. You should first draw a diagram to represent the relationship between the distances involved in the problem, then set up a chart based on the formula rate times time = distance.

The chart is then used to set up the equation.

Example: Two cars leave from the same place at the same time and travel in opposite directions. One car travels at 55 mph and the other at 75 mph. After how many hours will they be 520 miles apart?

Rate-Time-Distance Problem

Solve this word problem using uniform motion rt = d formula:

Example: Two cyclists start at the same corner and ride in opposite directions. One cyclist rides twice as fast as the other. In 3 hours, they are 81 miles apart. Find the rate of each cyclist.

Distance - Opposite Directions

Example: Brian and Jennifer both leave the convention at the same time traveling in opposite directions. Brian drove at 35 mph and Jennifer drove at 50 mph. After how much time were they 340 miles apart?

Distance - Opposite Directions find t

Example: Two joggers start from opposite ends of an 8 mile course running towards each other. One jogger is running at a rate of 4 mph. The other is running at a rate of 6 mph. After how long will the joggers meet?

Distance - Opposite Directions find r

Example: Bob and Fred start from the same point and walk in opposite directions. Bob walks 2 mph faster than Fred. After 3 hours they are 30 miles apart. How fast does each walk?

GMAT Challenge Question: Distance/Rate/Time

Example: Trains A and B left stations R and S simultaneously on two separate parallel rail tracks that are 350 miles long. The trains pass each other at point X after traveling for a certain amount of time. How many miles of the rail tracks has train A traveled when the two trains passed each other?

• Up to point X, the average speed of train B was 25% less than the average speed of train A.
• Up to point X, the average speed of train B was 60 mph and it took two and a half hours for train B to arrive at point X.

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SPEED DISTANCE TIME PROBLEMS WITH SOLUTIONS

Problem 1 :

A passenger train takes 3 hours less than a slow train for journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the passenger train, find the speed of two trains.

Let x be the speed of the of the passenger train

Speed of the slow train is 10 km/hr less than that of the passenger train

So x-10 be the speed of the slow train

Distance has to be covered  =  600 km

Time  =  Distance/speed

Let T 1 be the time taken by passenger train

Let T 2 be the time taken by the slow train

The differences of time taken by both trains are 3 hours

T 1  =  600/x

T 2   =  600/(x-0)

T 2 –T 1   =  3 hours

(600/(x–10)) – (600/x)  =  3

600[(1/(x-10)-(1/x)]  =  3

x-(x-10)/(x 2 -10x)  =  1/200

(x-x+10)/(x 2 -10x)  =  1/200

2000  =  x 2 -10x

x 2 -10x-2000  =  0

x 2 –50x+40x-2000  =  0

x(x–50)+40(x–50)  =  0

(x+40) (x–50)  =  0

By solving, we get x  =  -40 and x  =  50

Therefore the speed of passenger train  =  50 km/hr

Speed of slow train  =  40 m/hr.

Problem 2 :

The distance between two stations A and B is 192 km. Traveling by a fast train takes 48 minutes less than another train. Calculate the speed of the fast train if the speeds of two trains differ by 20 km/hr

Distance between two stations A and B  =  192 km

Fast train takes 48 minutes less then the time taken by the slow train.

Let x be the speed taken by the fast train

Speeds of two trains differ by 20 km/hr

So speed of slow train is x – 20.

Let T 1 be the time taken by the fast train

T 1   =  192/x

T 2   =  192/(x–20)

48/60  =  4/5 hours

T 1 – T 2   =  4/5

[192/(x-20)-192/x]  =  4/5

192[(x–x+20)/x(x - 20)]  =  4/5

192(20)/x 2 –20 x  =  4/5

3840 (5)  =  4(x 2 –20 x)

19200  =  4x 2 – 80 x

4800  =  x 2 – 20 x

x 2 –20x–4800  =  0

x 2 –60x+40x-4800  =  0

(x–60) (x+40)  =  0

x  =  60 and x  =  -40

Speed of fast train is 60 km/hr.

Problem 3 :

A train covers a distance of 300 km at a certain average speed. If its speed was decreased by 10km/hr, the journey would take 1 hour longer. What is the average speed.

Let x be the average speed of the train

So x–10 be the decreased speed

Time  =  Distance/Speed

T 1  and T 2  be the time taken by the train to cover the distance with speed of x km/hr and (x-10) km/hr respectively.

T 1   =  300/x

T 2   =  300/(x–10)

T 1 – T 2   =  1

[300/x] - [300/(x-10)]  =  1

3000/(x 2 – 10x)  =  1

3000  =  x 2 – 10 x

x 2 – 10x  =  3000

x 2 –10x–3000  =  0

x 2 –60x+50x–3000  =  0

x (x – 60) + 50 (x – 60) = 0

(x + 50)(x – 60) = 0

By solving, we get

x  =  -50 and x  =  60

So speed of the 60 km/hr.

Problem 4 :

The time taken by a train to travel a distance of 250 km was reduced by 5/4 hours when average speed was increased by 10km/hr. Calculate the average speed.

Distance to be covered  =  250 km

Let x be the required average speed.

If the average speed was increased by 10 km/hr

x+10 be the increased speed

Let T 1 be the time taken to cover the distance in the average speed of x km/hr

Let T 2 be the time taken to cover the distance in the average speed of (x + 10) km/hr

T 1   =  250/x

T 2   =  250/(x+10)

T 1 – T 2   =  5/4

250/x – 250/(x + 10)  =  5/4

250 [(x+10–x)/x(x+10)]  =  5/4

2500/(x 2 + 10x)  =  5/4

2500 (4)  =  5(x 2 +10x)

10000  =  5x 2 +10 x

Now we are going to divide the whole equation by 5, so we get

x 2 +10x  =  2000

x 2 +10x–2000 = 0

  x 2 + 50x-40x-2000 = 0

x(x+50)–40(x+50) = 0

(x–40) (x+50) = 0

x  =  40 and x  =  -50

Therefore the required average speed = 40 km/hr

Increased speed  =  (40+10)

Problem 5 :

An express train makes run 240 km t a certain speed. Another train whose speed is 12 km/hr less takes an hour longer to make the same trip. Find the speed of the express train.

Let x be speed of express train

So x–12 be the speed of another train

Distance to be covered  =  240 km

Let T 1 be the time taken by the train to cover the distance 240 km at the speed of x km/hr

Let T 2 be the time taken by the train to cover the distance 240 km at the speed of (x + 12) km/hr

Time = Distance /speed

T 1   =  240/x

T 2   =  240/(x - 12)

T 2 - T 1 = 1 hour

[240/(x- 12)]  - [240/x]  =  1

240[(1/(x -12) - 1/x]  =  1

240[(x - x + 12)/x(x - 12)]  =  1

240[12/(x 2 - 12 x)]  =   1

2880  =  (x 2 -12 x)

x 2 -12x-2880  =  0

x 2 +60x-48x-2880  =  0

x(x+60)- 48 (x+60)  =  0

(x-48) (x+60) = 0

x = 48              x = -60

Speed of express train  =  48 km/hr

Speed of other train  =  (x - 12)

=  (48-12)

=  36 km/hr

Problem 6 :

A plane traveled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for

(i) the onward journey and

(ii) the return journey. If the journey took 30 minutes less than  onward journey, write down an equation in x and its value.

Let “x” be average speed of plane

On the return journey, the speed was increased by 40 km/hr

So “x + 40” be the speed of plane

Distance to be covered = 400 km

Let T 1 be the time taken for onward journey in the speed of x km/hr

Let T 2 be the time taken for downward journey to cover the same distance 400 km at the speed of (x + 40) km/hr

T 1   =  400/x

T 2   =  400/(x+40)

T 1 - T 2   =  30 minutes

[400/x]-[400/(x + 40)]  =  30/60

400[(1/x) - 1/(x+40)]  =  1/2

400[40/(x 2 + 40 x)]  =  1/2

16000 (2)  =  (x 2 +40 x)

x 2 +40x-32000  =  0

x 2 +160x-100x-32000  =  0

(x - 100) (x + 160) = 0

 x = 100              x = -160

Speed of the plane  =  48 km/hr

Increased speed  =  (x+40)

=  (48+40)

=  88 km/hr

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Speed, Time and Distance – Formulas & Aptitude Questions

For candidates appearing in competitive exams, mastering quantitative aptitude topics such as Speed, Time, and Distance is crucial. From calculating average speeds to solving complex distance-time problems, candidates must be prepared for a variety of questions that test their speed, time, and distance skills.

To help you stay ahead in the competition, this article provides an overview of the concepts and formulas related to these topics as well as some useful tricks, sample questions, and answers to help candidates prepare for this essential topic.

If you are preparing for competitive exams, it is essential to have a clear understanding of the quantitative aptitude syllabus and the topics covered in it. To help you navigate this crucial subject, we have compiled a comprehensive guide that covers the key topics and concepts related to quantitative aptitude.

Practice Quiz :

Practice Speed, Time and Distance Aptitude Quiz Questions

Speed, Time, and Distance Concepts

Speed, distance, and time are essential concepts of mathematics that are used in calculating rates and distances. This is one area every student preparing for competitive exams should be familiar with, as questions concerning motion in a straight line, circular motion, boats and streams, races, clocks, etc. often require knowledge of the relationship between speed, time, and distance. Understanding these inter-relationships will help aspirants interpret these questions accurately during the exams.

Units of Speed, Time, and Distance

The most commonly used units of speed, time, and distance are:

• Speed : kilometers per hour (km/h), meters per second (m/s), miles per hour (mph), feet per second (ft/s).
• Time : seconds (s), minutes (min), hours (h), days (d).
• Distance : kilometers (km), meters (m), miles (mi), feet (ft).

For example, to convert km/h to m/s, multiply by 5/18, and to convert m/s to km/h, multiply by 18/5. 

Being familiar with these units and their conversions can help in solving quantitative aptitude questions related to speed, time, and distance efficiently.

Relationship Between Speed, Time & Distance

Understanding the relationship between speed, time, and distance is essential to solve problems.

Speed, Time, and Distance 

• Speed = Distance/Time

The speed of an object describes how fast or slow it moves and is calculated as distance divided by time.

Speed is directly proportional to distance and inversely proportional to time.

• Distance = Speed X Time

The distance an object travels is directly proportional to its speed – the faster it moves, the greater the distance covered.

• Time = Distance / Speed

Time is inversely proportional to speed – the faster an object moves, the less time it takes to cover a certain distance. As speed increases, time taken decreases, and vice versa

Speed, Time, and Distance Formulas 

Some important speed, distance, and time formulas are given in the table below:-

Speed, Time, and Distance Conversions

The Speed, Time, and Distance Conversions into various units is important to understand for solving problems:-

To convert from km/hour to m / sec: a Km/hr = a x (5/18) m/s To convert from m / sec to km/hour: a m/s = a x (18/5) Km/hr If a person travels from point A to point B at a speed of S1 kilometers per hour (kmph) and returns back from point B to point A at a speed of S2 kmph, the total time taken for the round trip will be T hours. Distance between points A and B = T (S1S2/(S1+S2)). If two moving trains, one of length l1 traveling at speed S1 and the other of length l2 going at speed S2, intersect each other in a period of time t. Then their Total Velocity can be expressed as S1+S2 = (l1+l2)/t. When two trains pass each other, the speed differential between them can be determined using the equation S1-S2 = (l1+l2)/t, where S1 is the faster train’s speed, S2 is the slower train’s speed, l1 is the faster train’s length and l2 is the slower train’s length, and t is the time it takes for them to pass each other. If a train of length l1 is travelling at speed S1, it can cross a platform, bridge or tunnel of length l2 in time t, then the speed is expressed as  S1 = (l1+l2)/t If the train needs to pass a pole, pillar, or flag post while travelling at speed S, then S = l/t. If two people A and B both start from separate points P and Q at the same time and after crossing each other they take T1 and T2 hours respectively, then (A’s speed) / (B’s speed) = √T2 / √T1 

Applications of Speed, Time, and Distance

Average Speed = Total Distance Traveled/Total Time Taken

Case 1: when the same distance is covered at two separate speeds, x and y, then Average Speed is determined as 2xy/x+y. Case 2 : when two speeds are used over the same period of time, then Average Speed is calculated as (x + y)/2.

Relative speed: The rate at which two moving bodies are separating from or coming closer to each other. 

Case 1 : If two objects are moving in opposite directions, then their relative speed would be S1 + S2 Case 2 : If they were moving in the same direction, their relative speed would be S1 – S2

Inverse Proportionality of Speed & Time : When Distance is kept constant, Speed and Time are inversely proportional to each other. 

This relation can be mathematically expressed as S = D/T where S (Speed), D (Distance) and T (Time).  To solve problems based on this relationship, two methods are used: Inverse Proportionality Rule Constant Product Rule .

Sample Problems on Speed, Time, and Distance

Q 1. a runner can complete a 750 m race in two and a half minutes. will he be able to beat another runner who runs at 17.95 km/hr .

Solution: 

We are given that the first runner can complete a 750 m race in 2 minutes and 30 seconds or 150 seconds.  => Speed of the first runner = 750 / 150 = 5 m / sec  We convert this speed to km/hr by multiplying it by 18/5.  => Speed of the first runner = 18 km / hr  Also, we are given that the speed of the second runner is 17.95 km/hr.  Therefore, the first runner can beat the second runner. 

Q 2. A man decided to cover a distance of 6 km in 84 minutes. He decided to cover two-thirds of the distance at 4 km/hr and the remaining at some different speed. Find the speed after the two-third distance has been covered. 

We are given that two-thirds of the 6 km was covered at 4 km/hr.  => 4 km distance was covered at 4 km/hr.  => Time taken to cover 4 km = 4 km / 4 km / hr = 1 hr = 60 minutes  => Time left = 84 – 60 = 24 minutes  Now, the man has to cover the remaining 2 km in 24 minutes or 24 / 60 = 0.4 hours  => Speed required for remaining 2 km = 2 km / 0.4 hr = 5 km / hr 

Q 3. A postman traveled from his post office to a village in order to distribute mail. He started on his bicycle from the post office at a speed of 25 km/hr. But, when he was about to return, a thief stole his bicycle. As a result, he had to walk back to the post office on foot at the speed of 4 km/hr. If the traveling part of his day lasted for 2 hours and 54 minutes, find the distance between the post office and the village. 

Solution : 

Let the time taken by postman to travel from post office to village=t minutes.  According to the given situation, distance from post office to village, say d1=25/60*t km {25 km/hr = 25/60 km/minutes}  And  distance from village to post office, say d2=4/60*(174-t) km {2 hours 54 minutes = 174 minutes}  Since distance between village and post office will always remain same i.e. d1 = d2  => 25/60*t = 4/60*(174-t) => t = 24 minutes.  => Distance between post office and village = speed*time =>25/60*24 = 10km 

Q 4. Walking at the speed of 5 km/hr from his home, a geek misses his train by 7 minutes. Had he walked 1 km/hr faster, he would have reached the station 5 minutes before the actual departure time of the train. Find the distance between his home and the station. 

Let the distance between his home and the station be ‘d’ km.  => Time required to reach the station at 5 km / hr = d/5 hours  => Time required to reach the station at 6 km/hr = d/6 hours  Now, the difference between these times is 12 minutes = 0.2 hours. (7 minutes late – 5 minutes early = (7) – (-5) = 12 minutes)  Therefore, (d / 5) – (d / 6) = 0.2  => d / 30 = 0.2  => d = 6  Thus, the distance between his home and the station is 6 km. 

Q 5. Two stations B and M are 465 km distant. A train starts from B towards M at 10 AM with a speed of 65 km/hr. Another train leaves from M towards B at 11 AM at a speed of 35 km/hr. Find the time when both trains meet. 

The train leaving from B leaves an hour early than the train that leaves from M.  => Distance covered by train leaving from B = 65 km / hr x 1 hr = 65 km  Distance left = 465 – 65 = 400 km  Now, the train from M also gets moving and both are moving towards each other.  Applying the formula for relative speed,  Relative speed = 65 + 35 = 100 km / hr  => Time required by the trains to meet = 400 km / 100 km / hr = 4 hours  Thus, the trains meet at 4 hours after 11 AM, i.e., 3 PM. 

Q 6. A policeman sighted a robber from a distance of 300 m. The robber also noticed the policeman and started running at 8 km/hr. The policeman also started running after him at the speed of 10 km/hr. Find the distance that the robber would run before being caught. 

Since both are running in the same direction, relative speed = 10 – 8 = 2 km/hr  Now, to catch the robber if he were stagnant, the policeman would have to run 300 m. But since both are moving, the policeman needs to finish off this separation of 300 m.  => 300 m (or 0.3 km)is to be covered at the relative speed of 2 km/hr.  => Time taken = 0.3 / 2 = 0.15 hours  Therefore, distance run by robber before being caught = Distance run in 0.15 hours  => Distance run by the robber = 8 x 0.15 = 1.2 km     Another Solution :  Time of running for both the policeman and the robber is same.  We know that Distance = Speed x Time  => Time = Distance / Speed  Let the distance run by the robber be ‘x’ km at the speed of 8 km / hr.  => Distance run by policeman at the speed of 10 km / hr = x + 0.3  Therefore, x / 8 = (x + 0.3) / 10  => 10 x = 8 (x + 0.3)  => 10 x = 8 x + 2.4  => 2 x = 2.4  => x = 1.2  Therefore, Distance run by the robber before getting caught = 1.2 km 

Q 7. To cover a certain distance, a geek had two options, either to ride a horse or to walk. If he walked one side and rode back the other side, it would have taken 4 hours. If he had walked both ways, it would have taken 6 hours. How much time will he take if he rode the horse both ways? 

Time taken to walk one side + Time taken to ride one side = 4 hours  Time taken to walk both sides = 2 x Time taken to walk one side = 6 hours  => Time taken to walk one side = 3 hours  Therefore, time taken to ride one side = 4 – 3 = 1 hour  Thus, time taken to ride both sides = 2 x 1 = 2 hours 

FAQs on Speed, Time, and Distance

Q1. what is speed, time, and distance.

Answer : 

Speed, time and distance are the three major concepts in physics. Speed is the rate of motion of an object between two points over a particular period of time which is measured in metres per second (m/s). Time is calculated by reading a clock, and it is a scalar quantity that do not change with direction. Distance is the total amount of ground covered by an object.

Q2. What is the average speed?

Answer: 

The formula for speed, time and distance is a calculation of the total distance an object travels over a given amount of time. It is a scalar quantity, meaning it’s an absolute value with no direction. To calculate it, you need to divide the total distance traveled by the amount of time it took to cover that distance.

Q3. What is the formula of speed, distance, and time?

Speed = Distance/Time Time = Distance/Speed Distance = Speed x Time

Q4. What is the relationship between speed, distance, and time?

The relationship is given as follows: Distance = Speed x Time

Related Articles:

Problem on Time Speed and Distance | Set-2

Test your knowledge of Speed, Time and Distance in Quantitative Aptitude with the quiz linked below, containing numerous practice questions to help you master the topic:-

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Real-life Problems Based on Speed, Time and Distance: Formulas, Examples

• All About Real-life Problems Based on Speed, Time and Distance: Formulas, Examples

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• Written By SHWETHA B.R
• Last Modified 22-06-2023

When children or regular people apply mathematics to address Real-life Problems Based on Speed, Time and Distance , they learn that math is more than a task to accomplish for the purpose of the teacher. It also gives children vital skills for influencing their surroundings. The speed of a moving object is defined as the distance it travels in one unit of time. This article will teach us about the mathematical link between speed, distance, and time.

Although the concepts of speed, time, and distance remain the same, the types of questions presented in tests may vary. One of the most common quantitative aptitude topics asked in government tests are speed, time, and distance. This is one of those topics that students are already aware of before they begin studying for competitive exams.

It is essential for students to learn the concept of Speed, Time and Distance. With regular practice of problems, they can develop speed which will, in turn, help them score higher marks in the exam. Continue reading to know more.

The word speed refers to how quickly something or someone is moving. If we know the distance traveled and the time it took, we may estimate an object’s average speed. The rate at which an object travels is known as speed (covering a particular distance). It’s a scalar quantity because it only defines magnitude, not direction.

The formula used to find the speed is given by,

The meter per second \((\rm{m/s})\) is the SI unit for speed.

From the above example, as the speed increases, the time decreases.

Learn Formulas for Speed Time Distance

Time is a measured duration during which an action or event occurs. The time formula calculates how long an object takes to travel a certain distance at a given speed.

Seconds is the SI unit for time \((\rm{s}).\)

The length of the line segment that connects two points is called distance. The distance is the extent or amount of space between two things, points, lines, etc.

Meter is the SI unit for distance.

Relation of Speed, Time, and Distance

Now, we shall look at the mathematical relation between speed, distance, and time. The speed of a moving body is the distance it travels in a unit amount of time.

\({\text{Speed = }}\frac{{{\text{distance}}}}{{{\text{time}}}}\)

If the distance is in kilometers and the time is in hours, the speed is in kilometers per hour. The speed is \(\rm{m/sec}\) if the distance is measured in metres and the time is measured in seconds.

\({\rm{Distance}} = {\rm{speed}} \times {\rm{time}}\)

\({\text{Time = }}\frac{{{\text{distance}}}}{{{\text{speed}}}}\)

When the distance is constant, speed is inversely proportional to time. When \(D\) is constant, \(S\) is inversely proportional to \(T\). The time taken will be in the ratio \(n : m\) if the speeds are \(m : n.\)

When the distance traveled remains constant, speed is inversely related to the time required. As a result, as speed rises, time decreases, and vice versa.

Formulas of Speed, Distance and Time

All fundamental problems can be solved using these formulas. When applying the given formulas, you should ensure that the units are used correctly. When the distance traveled remains constant, speed is inversely related to the time required. As a result, as speed rises, time decreases, and vice versa.

Units of Speed, Distance and Time

Time : seconds \((\rm{s})\), minutes \((\rm{min})\), hours \((\rm{hr})\) Distance : metres \((\rm{m})\), kilometres \((\rm{km})\), miles, feet Speed : \({\rm{m/s}},\,{\rm{km/hr}}.\) If the distance is in \({\rm{km}}\) and the time is in \({\rm{hr}}\), then \({\text{Speed = }}\frac{{{\text{distance}}}}{{{\text{time}}}}\) and the unit of speed will be \(\rm{km/hr}.\)

Effect of Speed and Time on Distance

Understand that whatever speed we choose involves both distance and time. “Faster” can mean “far” (a greater distance) or “quicker” (a shorter distance) (less time). To double one’s travel distance at the same time, one must increase their speed. When one’s speed increases, it takes to cover the same distance is cut in half. Distance is unchanged by speed. It can affect the time it takes to cover or cross a certain distance.

Real-Life Problems Based on Speed, Time and Distance

1. A boy walks at a speed of \(5\,\rm{kmph}\). How much time does he take to walk \(20\,\rm{km}\)? We know, \({\text{Time = }}\frac{{{\text{distance}}}}{{{\text{speed}}}}\) So, the time required to walk \( = \frac {20}{5} = 4\,\rm{hours}\) So, the boy walks \(20\,\rm{km}\) in \(4\,\rm{hours}\).

2. A cyclist covers \(14\,\rm{miles}\) in \(2\,\rm{hours}\). Calculate his speed. We know, \({\text{Speed = }}\frac{{{\text{distance}}}}{{{\text{time}}}}\) Hence, the \({\text{Speed}} = \frac{14}{2} = 7\,{\text{miles}}\,{\text{per}}\,{\text{hour}}.\) So, the speed of cyclists is \(7\,{\text{miles}}\,{\text{per}}\,{\text{hour}}.\)

3. A cyclist travels at a speed of \(25\,\rm{km/hour}\). How far will he travel in \(50\,\rm{minutes}\)? We know, \({\rm{Distance}} = {\rm{speed}} \times {\rm{time}}\) So, the distance travelled in \(50\) minutes \(25 \times \frac{{50}}{{60}} = 20.83\,{\text{km}}\) So, the cyclist travels \(20.83\,{\text{km}}.\)

Solved Examples – Real Life Problems Based on Speed, Time and Distance

Q.1. A car travels \(320\,\rm{km}\) in \(4\,\rm{hours}.\) What is its speed in \(\rm{km/hr}\)? Ans: We know the formula for speed is given by, \({\text{Speed = }}\frac{{{\text{distance}}}}{{{\text{time}}}}\) \( \Rightarrow {\text{speed}} = \frac{{320}}{4} = 80\,{\text{km/hr}}.\) Therefore, the speed at which a car travels is \(80\,{\text{km/hr}}.\)

Q.2. Traveling at a speed of \(45\,\rm{kmph}\), how long will it take to travel \(135\,\rm{km}\)? Ans: Given, speed \(= 45\,{\text{kmph}}\), distance \(= 135\,{\text{km}}\) We know the relation between speed distance and time is, \({\text{speed}} = \frac{{{\text{distance}}}}{{{\text{time}}}}\) \( \Rightarrow 45\,{\text{kmph}} = \frac{{135\,{\text{km}}}}{{{\text{time}}}}\) \( \Rightarrow {\text{time}} = \frac{{135\,{\text{km}}}}{{45\,{\text{kmph}}}}\) \( = 3\,{\text{hours}}\) Therefore, the required time to complete \(135\,{\text{km}}\) is \( 3\,{\text{hours}}.\)

Q.3. A truck was running from a city at an initial speed of \(40\,\rm{kmph}\). The truck’s speed was increased by \(3\,\rm{kmph}\) at the end of every hour. Find the total distance covered by the truck in the first \(5\,\rm{hours}\)  of the journey. Ans: The total distance covered by the truck in the first \(5\,\rm{hours}\) \(= 40 + 43 + 46 + 49 + 52\) \(= 230\,\rm{kms}\) Therefore, \(230\,\rm{km}\) is the total distance covered by the truck in the first \(5\,\rm{hours}\) of the journey.

Q.4. Arun can run a distance of \(120\,\rm{m}\) in \(20\,\rm{seconds}\). Find the speed of Arun in \(\rm{m/s}.\) Ans: Given, time \(= 20\,\rm{seconds}\), distance \(= 120\,\rm{m}\), speed \(=\)? We know, \({\text{speed = }}\frac{{{\text{distance}}}}{{{\text{time}}}}\) \( \Rightarrow {\text{speed}} = \frac{{120}}{{20}} = 60\,{\text{m}}/{\text{s}}\) Hence, the speed of Arun is \(60\,{\text{m}}/{\text{s}}.\)

Q.5: Travelling at a speed of \(50\,\rm{kmph}\) , how long will it take to travel \(80\,\rm{km}\) ? Ans: Given, speed \(= 50\,\rm{kmph}\), distance \(= 80\,\rm{km}\) We know the relation between speed distance and time is \({\text{time = }}\frac{{{\text{distance}}}}{{{\text{speed}}}}\) \( \Rightarrow {\text{time}} = \frac {{80}}{{50}}\) \( \Rightarrow {\text{time}} = \frac {{8}}{{5}}\) \( \Rightarrow {\text{time}} = 1.6\;\rm{hours}\) \( \Rightarrow {\text{time}} = 1\,{\rm{hour}}\,36\,{\rm{minutes}}\) Therefore, \(1\,{\rm{hour}}\,36\,{\rm{minutes}}\) is going to take to travel \(80\,\rm{km}.\)

Q.6. If the distance travelled by train is \(405\,\rm{km}\) in \(4\,{\rm{hours}}\,30\,{\rm{minutes}}\), what is its speed? Ans: Given, time \(= 4\,{\rm{hours}}\,30\,{\rm{minutes}} = 4.5\,{\rm{hours}}\),distance \(= 500\,\rm{km}\),speed \(=\)? We know, \({\text{speed}} = \frac{{{\text{distance}}}}{{{\text{time}}}}\) \( \Rightarrow {\text{speed}} = \frac{{405}}{{4.5}}\) \( \Rightarrow {\text{speed}} = 90\,{\text{km}}/{\text{hr}}\) Hence, the obtained speed is \(90\,{\text{km}}/{\text{hr}}.\)

Q.7. Express the speed of \(90\,\rm{meters}\) per minute in kilometres per hour. Ans: Given, speed \( = 90\,{\text{meters}}/{\text{minutes}}\) We know \(1\,{\text{meter}} = \frac{1}{{1000}}\;{\text{km}}\) and \(1\,{\text{minute}} = \frac{1}{{60}}{\text{hour}}\) Hence, the speed \(= 90 \times \frac{{\frac{1}{{1000}}{\text{km}}}}{{\frac{1}{{60}}\;{\text{hr}}}}\) \( \Rightarrow {\text{speed}} = 90 \times \frac{{60}}{{1000}}\) \( \Rightarrow {\text{speed}} = 9 \times \frac{6}{{10}}\) \( \Rightarrow {\text{speed}} = \frac{{54}}{{10}}\) \( \Rightarrow {\text{speed}} = 5.4\,{\text{km}}/{\text{hr}}\) Therefore, the speed can be expressed as \(5.4\,{\text{km}}/{\text{hr}}.\)

Q.8. A car travels a distance of \(600\,{\text{km}}\) in \(10\,{\rm{hours}}\) . What is its speed? Ans: Given, time \(= 10\,\rm{hours}\), distance \(= 600\,\rm{km}\), speed \(=\)? We know that \({\text{speed}} = \frac{{{\text{distance}}}}{{{\text{time}}}}\) \(\Rightarrow {\text{speed}} = \frac{{600}}{{10}}\) \( \Rightarrow {\text{speed}} = 60\,{\text{km}}/{\text{hr}}\) Therefore, the obtained speed is \(60\,{\text{km}}/{\text{hr}}.\)

This article includes the definition of speed, distance and time, the relationship among those three formulas. It helps to solve various problems, including real-life problems, too quickly. This article helps better understand “Real Life Problems Based on Speed, Time and Distance”. This article’s outcome helps in applying the suitable formulas while solving the various problems based on them.

Solve Important Problems on Trains

Frequently Asked Questions (FAQs)

We have provided some frequently asked questions here:

Q.1. What is the formula for speed and distance? Ans: The formula to find speed is given by, \({\text{speed}} = \frac{{{\text{distance}}}}{{{\text{time}}}}\) The formula to find the distance is given by, \({\text{distance}} = {\text{speed}} \times {\text{time}}.\)

Q.2. What is the difference between speed and distance? Ans: The rate at which the distance is travelled in unit time is referred to as the speed. The speed is equal to \(S = \frac{D}{T}\) if ‘\(D\)’ is the distance travelled by an item in time ‘\(T\)’.

Q.3. What is the effect of speed and time on distance? Ans: Whatever speed we choose, understand that it involves both distance and time. Increasing one’s speed involves increasing one’s travel distance in the same amount of time. Increasing one’s speed also cuts the time it takes to cover the same distance.

Q.4. How do you solve problems involving speed, distance and time? Ans: Speed is calculated using the formula \({\text{speed}} = \frac{{{\text{distance}}}}{{{\text{time}}}}.\) We need to know the units for distance and time to figure out the units for speed. Because the distance is measured in metres \((\rm{m})\) and the time is measured in seconds \((\rm{s})\), the units will be metres per second \((\rm{m/s}).\)

Q.5. How do you find speed and distance in a math problem? Ans: Speed and distance can be calculated by using the formula, \({\text{speed}} = \frac{{{\text{distance}}}}{{{\text{time}}}}\) The formula to find the distance is given by, \({\text{distance}} = {\text{speed}} \times {\text{time}}.\)

Q.6. What is the mathematical relationship between speed, distance and time? Ans: The mathematical relation between speed, distance, and time is that in a moving body, the speed is the distance it travels in a unit amount of time. If the distance is in kilometres and the time is in hours, the speed is in kilometres per hour. The speed is \(\frac{{\text{m}}}{{{\text{sec}}}}\), if the distance is measured in metres and the time is measured in seconds.

We hope this detailed article on some real-life problems based on speed, distance and time helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!

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GMAT Time Speed Distance Question and Answers

The Quantitative Reasoning section of the GMAT tests your skills in two main types of multiple-choice questions: Data Sufficiency and Problem Solving. You'll need a solid grasp of basic algebraic concepts, arithmetic operations, and fundamental geometry but one area that often trips up test-takers is the time and distance questions. These simple-looking problems require you to think carefully and apply the right formulas and problem-solving strategies. To ace the Quantitative section mastering time and distance questions is crucial. This blog will dive into tips and examples to help you master time and distance problems on the GMAT. 

Scoring well in the GMAT does not end your journey. You apply to the university that matches your GMAT score , do the loan and visa applications, apply for scholarships, and many more. If you feel overwhelmed with all this information, worry not, Yocket is there to help you. With end-to-end assistance and personalized guidance, we will help you secure admissions to the top institutes. Connect with Yocket Premium today to learn more about it!

GMAT Time Speed Distance Concept

The Time, Speed, and Distance (TSD) concept is fundamental to solving various quantitative problems on the GMAT . Understanding the relationship between these three variables is crucial for tackling questions effectively. Here’s a breakdown of the key concepts, formulas, and strategies to approach TSD problems:

Key Concepts and Formulas

1. basic relationship:.

Distance=Speed×Time

Speed=Distance/Time

Time=Distance/Speed

Distance can be measured in meters (m), kilometers (km), miles, etc.

Time can be measured in seconds (s), minutes (min), hours (h), etc.

Speed is usually measured in units like meters per second (m/s), kilometers per hour (km/h), miles per hour (mph), etc.

3. Conversions:

To convert speed from km/h to m/s: multiply by 5/18

To convert speed from m/s to km/h: multiply by 18/5

4. Relative Speed:

When two objects move in the same direction, the relative speed is the difference between their speeds.

Relative Speed=Speed1−Speed2

When two objects move in opposite directions, the relative speed is the sum of their speeds.

Relative Speed=Speed1+Speed2

5. Average Speed:

For a trip with varying speeds, average speed is calculated by dividing the total distance by the total time taken. 

Average Speed=Total Distance/Total Time

If a round trip is made at different speeds, the average speed can be calculated using: 

Average Speed=(2×Speed1×Speed2)/(Speed1+Speed2)

6. Boat and Stream:

Speed downstream (with the current): 

Speedboat+Speedstream

Speed upstream (against the current): 

Speedboat−Speedstream

When a train passes a stationary object, the distance covered is the length of the train.

When a train passes a moving object (e.g., another train) relative speed and the combined lengths are considered.

Suggested: All GMAT Maths Formulas You Need in 2024

GMAT Time Speed Distance Questions and Answers

Below are practice questions for the time speed distance topic. It is divided based on difficulty levels.

GMAT Time Speed Distance Questions: Difficulty Level I

Two trains move in opposite directions at constant speeds of 70 km/h and 40 km/h respectively. They take 6 seconds to pass each other. If the trains were moving in the same direction, a passenger in the faster train would overtake the slower train in 20 seconds. What are the lengths of the trains (in meters)?

(a) 130, 80

(b) 110, 60

(c) 116.67, 66.67

(d) 126.67, 76.67

Rajesh drove from City A to City B without stopping. The average speed for the entire journey was 50 km/h. What was the average speed from City A to City B?

The distance from City B to City C is 0.25 times from City A to City B.

The average speed from City B to City C was twice that of the average speed from City A to City B.

What is the time difference between Tokyo and New York?

The departure time in Tokyo is exactly 10:00 a.m. local time and the arrival time in New York is at 11:00 a.m. local time.

The flight time is 14 hours.

A ship sets sail on a long voyage. When it is 22 miles from the shore, a seaplane, whose speed is eight times that of the ship, is sent to deliver mail. How far from the shore does the seaplane catch up with the ship?

(a) 26 miles

(b) 28 miles

(c) 24 miles

(d) 30 miles

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GMAT Time Speed Distance Questions: Difficulty Level II

Reena plans to drive from City A to Station C at 60 km/h to catch a train that arrives from City B. She must reach C at least 20 minutes before the train arrives. The train leaves B, 600 km south of A, at 9:00 am and travels 55 km/h. C is located northwest of B at an angle of 45 degrees to AB and west of A at an angle of 30 degrees to AB. When is the latest time by which Reena must leave A?

(a) 7:00 am

(b) 7:20 am

(c) 7:30 am

(d) 7:45 am

(e) 8:00 am

John, Mike, and Steve start running from the same point in the same direction along a straight line at 06:00 a.m., 07:00 a.m., and 09:00 a.m. with speeds of 3 km/h, 4 km/h, and 5 km/h respectively. When Mike catches John, he sends John back immediately to deliver a message to Steve. When will John meet Steve?

(a) 09:30 a.m.

(b) 10:00 a.m.

(c) 10:40 a.m.

(d) 11:00 a.m.

P and Q are two points 150 km apart. Runner A starts from P towards Q at 12 km/h. At the same time, runner B starts from Q but in the same direction as A at 24 km/h. After an hour, B turns back and changes his speed to 12 km/h. After another hour, B returns and changes his speed to 24 km/h again. He keeps on changing his speed and direction in this manner till he meets A. After how much time will A and B meet for the first time?

(a) 25 hours

(b) 15 hours

(c) 8 hours

(d) 12 hours

A jungle king planned a 3-kilometre race between a fox and a turtle. Soon after the start, the fox took a significant lead. He decided to nap for ‘x’ minutes, knowing he would still win by 15 minutes. He stopped under a tree to nap. Meanwhile, the turtle kept moving. When the fox woke up, he realized that he had slept for (15+x) minutes, and immediately started running at 3/2  times his original speed. The race ended in a dead heat. If the original speed ratio of the fox to the turtle was 5:1, and the fox overslept by (5/4)x minutes, how long did the turtle take to complete the race?

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GMAT Time Speed Distance Questions: Difficulty Level III

A cyclist and a runner start from point A and head towards point B, which is 90 km away. The cyclist travels at 30 km/h, while the runner travels at 10 km/h speed. Every hour, the cyclist stops for 10 minutes. When the cyclist reaches point B, he immediately turns around and heads towards point A. After how many hours will the cyclist meet the runner for the first time?

Answer: 2.57 hours

A boat travels from point X to point Y and back to point X. The boat speed is 8km/h in still water and the speed of the current is 2 km/h. The total time for the round trip is 5 hours. What is the distance between X and Y?

Answer: 18.75 km

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Tips for GMAT Time Speed Distance Questions

To ace the GMAT Time Speed Distance (TSD) problems follow the below 5 tips.

Suggested: Colleges Accepting GMAT Score 700 or above

From the Desk of Yocket

To ace the quantitative reasoning section of GMAT it is crucial to master time, speed, and distance problems. You can approach even the trickiest TSD questions confidently by understanding the fundamental concepts, formulas, and strategies outlined in this blog. Practice with various difficulty levels to enhance your ability to approach these questions confidently on the test day. Remember, to conquer the time and distance problems in the GMAT perseverance and a focused mindset are key.

Are you wondering about the requirements for GMAT scores for universities abroad? Reach out to Yocket professionals and resolve all your queries, alongside getting a thorough profile evaluation and end-to-end assistance on everything studying abroad. Book A 15-minute Consultation Call Today!

Frequently Asked Questions About GMAT Time Speed Distance Questions

What is the formula of time and distance?

The formula is as follows: Speed = distance/time. This implies that Time = distance/speed and Distance = Speed x Time.

How to convert km/h to m/s?

To convert km/h to m/s, we multiply by 5 / 18. So, 1 km/hour = 5 / 18 m / sec.

What is the concept of relative speed?

Relative speed is used when two objects move relative to each other: When moving in opposite directions, relative speed = Speed1 + Speed2. When moving in the same direction, relative speed = |Speed1 - Speed2|.

What is the key to solving complex time, speed, and distance problems?

Break the problem into smaller, manageable parts, solve for one variable at a time, and ensure all units are consistent. Using diagrams can also help visualize the problem.

How do you approach problems involving changing speeds and directions?

Track the changes in speed and direction step-by-step, calculating the distance covered in each segment and the cumulative time taken until the meeting point or endpoint is reached.

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