problem solving in speed time and distance

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

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.

Share This Book

TIME SPEED AND DISTANCE PROBLEMS

Formulas .

To know the shortcuts required to solve problems on time, speed and distance,

please click here

Problem 1 :

If a person drives his car in the speed 50 miles per hour, how far can he cover in 2.5 hours?

Given :  Speed is 50 miles per hour.

So, the distance covered in 1 hour is

Then, the distance covered in 2.5 hours is

= 2.5 ⋅ 50 miles

= 125 miles

So, the person can cover 125 miles of distance in 2.5 hours.

Problem 2 :

If a person travels at a speed of 40 miles per hour. At the same rate, how long will he take to cover 160 miles distance?

Given :  Speed is 40 miles per hour.

The formula to find the time when distance and speed are given is

Time taken to cover the distance of 160 miles is

So, the person will take 4 hours to cover 160 miles distance at the rate of 40 miles per hour.

Problem 3 :

A person travels at a speed of 60 miles per hour. How far will he travel in 4.5 hours?

Given :  Speed is 60 miles per hour.

The distance covered in 1 hour is

Then, the distance covered in 4.5 hours is

= 4.5 ⋅ 60 miles

= 270 miles

So, the person will travel 270 miles distance in 4.5 hours.

Problem 4 :

A person travels at a speed of 60 kms per hour. Then how many meters can he travel in 5 minutes?

Given :  Speed is 60 kms per hour.

The distance covered in 1 hour or 60 minutes is

= 60 ⋅ 1000 meters

= 60000 meters

Then the distance covered in 1 minute is

The distance covered in 5 minutes is

= 5 ⋅ 1000

= 5000 meters

So, the person can cover 5000 meters distance in 5 minutes.

Problem 5 :

A person covers 108 kms in 3 hours. What is his speed in meter per second?

Given :  Distance is 108 kms and time is 3 hours.

The given distance in meters :

= 108 ⋅ 1000

= 108,000 meters

The given time in seconds :

= 3 ⋅ 60 minutes

= 180 minutes

= 180 ⋅ 60 seconds

= 10,800 seconds

The formula to find the speed is

Speed in meter per second is

= ¹⁰⁸⁰⁰⁰⁄₁₀₈₀₀

So, his speed in meter per second is 10.

Problem 6 :

A person covers 90 kms in 2 hours 30 minutes. Find the speed in meter per second.

Given :  Distance is 90 kms and time is 2 hrs 30 min.

= 90 ⋅ 1000

= 90,000 meters

= 2 hrs 30 min

= (120 + 30) min

= 150 minutes

= 150 ⋅ 60 seconds

= 9,000 seconds

= ⁹⁰⁰⁰⁰⁄₉₀₀₀

Problem 7 :

A person travels at the rate of 60 miles per hour and covers 300 miles in 5 hours. If he reduces his speed by 10 miles per hour, how long will he take to cover the same distance?

Original speed is 60 miles per hour.

If the speed is reduced by 10 miles per hour,  then the new speed is

= 50 miles per hour

Distance to be covered is 300 miles.

The formula to find time is

Time taken to cover 300 miles distance at the speed of 50 miles per hour is

So, if the person reduces his speed by 10 miles per hour, he will take 6 hours to cover 300 miles distance.

Problem 8 :

A person travels 50 kms per hour. If he increases his speed by 10 kms per hour, how many minutes will he take to cover 8000 meters?

Original speed is 50 kms per hour.

If the speed is increased by 10 kms per hour,  then the new speed is

= 60 kms per hour

Because, we have to find the time in minutes for the distance given in meters, let us change the speed from kms per hour into meters per minute.

1 hour ----> 60 kms

1 ⋅ 60 minutes ----> 60 ⋅ 1000 meters

60 minutes ----> 60000 meters

1 minute ----> ⁶⁰⁰⁰⁰⁄₆₀ meters

1 minute ----> 1000 meters

So, the speed is 1000 meters/minute.

Time = Distance/Speed

Time taken to cover 8000 meters distance at the speed of 1000 meters per minute is

= ⁸⁰⁰⁰⁄₁₀₀₀

= 8 minutes

So, if the person increases his speed by 10 kms per hour, he will take 8 minutes to cover 8000 meters distance.

Problem 9 :

A person can travel at the speed of 40 miles per hour. If the speed is increased by 50%, how long will it take to cover 330 miles?

Original speed is 40 miles per hour.

If the speed is increased by 50%, then the new speed is

= 150% of 40

= 1.5 ⋅ 40

= 60 miles per hour

Distance to be covered is 330 miles.

Time taken to cover 330 miles distance at the speed of 60 miles per hour is

= 5.5 hours

= 5 hrs 30 minutes

So, if the person is increased by 50%, it will take 5 hrs 30 minutes to cover 330 miles distance.

Problem 10 :

A person speed at a rate of 40 kms per hour. If he increases his speed by 20%, what is his new speed in meter per minute?

Original speed is 40 kms per hour

If the speed is increased by 20%, then the new speed is

= 120 % of 40

= 1.2 ⋅ 40

= 48 kms per hour

Now, let us change the speed from kms per hour into meters per minute.

1 hour ----> 48 kms

1 ⋅ 60 minutes ----> 48 ⋅ 1000 meters

60 minutes ----> 48,000 meters

1 minute ----> ⁴⁸⁰⁰⁰⁄₆₀ meters

1 minute ----> 800 meters

So, the speed is 800 meters/minute.

If the person increases his speed by 20%, his new speed will be 800 meter per minute.

Kindly mail your feedback to   [email protected]

We always appreciate your feedback.

© All rights reserved. onlinemath4all.com

• Sat Math Practice
• SAT Math Worksheets
• PEMDAS Rule
• BODMAS rule
• GEMDAS Order of Operations
• Math Calculators
• Transformations of Functions
• Order of rotational symmetry
• Lines of symmetry
• Compound Angles
• Quantitative Aptitude Tricks
• Trigonometric ratio table
• Word Problems
• Times Table Shortcuts
• 10th CBSE solution
• PSAT Math Preparation
• Privacy Policy
• Laws of Exponents

Recent Articles

Honors Algebra 2 Problems on Solving Exponential Equations

Mar 30, 24 11:45 PM

Properties of Parallelograms Worksheet

Mar 30, 24 09:11 PM

Division by Zero

Mar 30, 24 08:11 AM

Solving Problems Involving Distance, Rate, and Time

• Pre Algebra & Algebra
• Math Tutorials
• Exponential Decay
• Worksheets By Grade

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)

• What Is Velocity in Physics?
• Distance, Rate, and Time Worksheets
• Realistic Math Problems Help 6th-graders Solve Real-Life Questions
• 4th-Grade Math Word Problems
• Circumference of a Circle
• What Speed Actually Means in Physics
• Understanding the Distance Formula
• Help Kids Calculate the Area and Circumference of Circles
• How to Calculate Commissions Using Percents
• Rate of Change Worksheet with Solutions
• How to Derive the Formula for Combinations
• How to Determine the Geometry of a Circle
• Dimensional Analysis: Know Your Units
• Perimeter and Surface Area Formulas
• Problem Solving in Mathematics
• Mathematical Properties of Waves

Module 9: Multi-Step Linear Equations

Using the distance, rate, and time formula, learning outcomes.

• Use the problem-solving method to solve problems using the distance, rate, and time formula

One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you traveled if you drove at a steady rate of [latex]60[/latex] miles per hour for [latex]2[/latex] hours? (This might happen if you use your car’s cruise control while driving on the Interstate.) If you said [latex]120[/latex] miles, you already know how to use this formula!

The math to calculate the distance might look like this:

[latex]\begin{array}{}\\ \text{distance}=\left(\Large\frac{60\text{ miles}}{1\text{ hour}}\normalsize\right)\left(2\text{ hours}\right)\hfill \\ \text{distance}=120\text{ miles}\hfill \end{array}[/latex]

In general, the formula relating distance, rate, and time is

[latex]\text{distance}\text{=}\text{rate}\cdot \text{time}[/latex]

Distance, Rate, and Time

For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula

[latex]d=rt[/latex]

where [latex]d=[/latex] distance, [latex]r=[/latex] rate, and [latex]t=[/latex] time.

Notice that the units we used above for the rate were miles per hour, which we can write as a ratio [latex]\Large\frac{miles}{hour}[/latex]. Then when we multiplied by the time, in hours, the common units “hour” divided out. The answer was in miles.

Jamal rides his bike at a uniform rate of [latex]12[/latex] miles per hour for [latex]3\Large\frac{1}{2}[/latex] hours. How much distance has he traveled?

In the following video we provide another example of how to solve for distance given rate and time.

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of [latex]520[/latex] miles. If he can drive at a steady rate of [latex]65[/latex] miles per hour, how many hours will the trip take?

Show Solution

In the following video we show another example of how to find rate given distance and time.

• Question ID 145550, 145553,145619,145620. Authored by : Lumen Learning. License : CC BY: Attribution
• Ex: Find the Rate Given Distance and Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/3rYh32ErDaE . License : CC BY: Attribution
• Example: Solve a Problem using Distance = Rate x Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/lMO1L_CvH4Y . License : CC BY: Attribution
• Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

Forgot password? New user? Sign up

Existing user? Log in

Speed, Distance, and Time

Already have an account? Log in here.

Recommended Course

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)?

Master concepts like these

Learn more in our Classical Mechanics course, built by experts for you.

Problem Loading...

Note Loading...

Set Loading...

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

• Direct and inverse proportions

Add comment

Name (required)

E-mail (required, but will not display)

Notify me of follow-up comments

• You are here:  
• Home >>
• 26. Distance, speed, time
• 2. Basic operations
• 3. Expressions
• 4. Substitutions in expressions
• 5. Place value for beginners
• 6. Multiplication
• 7. Division
• 8. Order of operations
• Basic Laws of Math
• 9. Basic Laws of Math
• 10. Divisors and multiples
• 11. GCD and LCM
• 12. Fractions
• 13. Actions with fractions
• 14. Mixed numbers
• 15. Comparing Fractions
• 16. Units of Measurement
• 17. Ways to use fractions
• 18. Decimals
• 19. Operations with decimals
• 20. Using decimals
• 21. Rounding numbers
• 22. Repeating Decimals
• 23. Unit conversion
• Ratios & Proportion
• 25. Proportion
• 27. Direct and inverse proportions
• 28. What are percentages?
• Additional topic & Number line
• 29. Negative numbers
• 30. Absolute value or modulus
• 31. What is a set?
• 32. Adding and subtracting integers
• 33. Multiplication and division of integers
• Rational numbers
• 34. Rational numbers
• 35. Comparing rational numbers
• 36. Adding and subtracting rational numbers
• 37. Multiplying and dividing rational numbers
• 38. More information about fractions
• 39. Algebraic expressions
• 40. Taking out a common factor
• 41. Expanding brackets
• 42. Simple math problems
• 43. Tasks on fractions
• 44. Percent worksheets for practice
• 45. Speed, distance, time tasks
• 46. Productivity problem
• 47. Statistics in Math
• 48. Basic theory of equation
• 49. Problem solving with equations
• 50. Proportions: exercises
• 51. Systems of linear equations
• 52. Introduction to Inequalities
• 53. Systems of linear inequalities with one variable
• 54. Set Operations | Union | Intersection
• 55. Power with a natural exponent
• 56. Power with integer exponent
• 57. Perimeter, area and volume
• 58. Monomials
• 59. Polynomials
• 60. Reduced multiplication formulas
• 61. Polynomial decomposition
• 62. Dividing polynomials
• 63. Identical transformations of polynomials
• 64. Square root
• 65. Algorithm for extracting the square root
• 66. Quadratic equation
• 67. A quadratic equation with an even second coefficient
• 68. Viette's Theorem
• 69. Factoring a trinomial using decomposition
• 70. More about modules
• 71. Absolute value equations
• 72. Solving equations with module by method of intervals
• 73. Inequalities with module
• 74. Solving inequalities with module by method intervals
• 75. Square root from both parts of an equation
• Functions and graphs
• 76. What is a function?
• 77. Linear function and its graph
• 78. Power function
• 79. Direct Proportion and its Graph
• 80. Function y=x² and its graph
• 81. Function y = √x its properties and graph
• 82. Function y=k*x² its properties and graph
• 83. Function y=k/x its properties and graph
• 84. Graphical solution of equations and inequalities
• coming soon... (new lessons every month)

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

Relationship between Speed, Distance and Time

Conversion of Units of Speed

Problems on Calculating Distance

Problems on Calculating Time

Two Objects Move in Same Direction

Two Objects Move in Opposite Direction

Train Passes a Moving Object in the Same Direction

Train Passes a Moving Object in the Opposite Direction

Train Passes through a Pole

Train Passes through a Bridge

Two Trains Passes in the Same Direction

Two Trains Passes in the Opposite Direction

8th Grade Math Practice From Problems on Calculating Speed to HOME PAGE

New! Comments

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

• Preschool Activities
• Kindergarten Math
• 1st Grade Math
• 2nd Grade Math
• 3rd Grade Math
• 4th Grade Math
• 5th Grade Math
• 6th Grade Math
• 7th Grade Math
• 8th Grade Math
• 9th Grade Math
• 10th Grade Math
• 11 & 12 Grade Math
• Concepts of Sets
• Probability
• Boolean Algebra
• Math Coloring Pages
• Multiplication Table
• Cool Maths Games
• Math Flash Cards
• Online Math Quiz
• Math Puzzles
• Binary System
• Math Dictionary
• Conversion Chart
• Homework Sheets
• Math Problem Ans
• Free Math Answers
• Printable Math Sheet
• Funny Math Answers
• Employment Test
• Math Patterns
• Link Partners
• Privacy Policy

Recent Articles

Time Duration |How to Calculate the Time Duration (in Hours & Minutes)

Mar 31, 24 05:49 PM

Conversion of Rupees and Paise | How to convert rupees into paise?

Mar 31, 24 05:34 PM

4th Grade Multiplication Worksheet | Math Multiplication Worksheets

Mar 31, 24 05:22 PM

Word Problems on Multiplication |Multiplication Word Problem Worksheet

Mar 31, 24 11:57 AM

Multiplication by Ten, Hundred and Thousand |Multiply by 10, 100 &1000

Mar 31, 24 10:09 AM

Worksheet on Conversion of Units of Speed

Worksheet on Calculating Time

Worksheet on Calculating Speed

Worksheet on Calculating Distance

Worksheet on Train Passes through a Pole

Worksheet on Train Passes through a Bridge  

Worksheet on Relative Speed

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Chapter 6.8: 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 .

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.

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:

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

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:

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

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:

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

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?

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.

Answers to odd questions

Pre-Calculus Copyright © 2022 by St. Clair College is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

• Toggle navigation

• CREST Mathematics Olympiad (CMO)
• CREST Science Olympiad (CSO)
• CREST English Olympiad (CEO)
• CREST Reasoning Olympiad (CRO)
• CREST Cyber Olympiad (CCO)
• CREST Mental Maths Olympiad (CMMO)
• International Green Warrior Olympiad (IGWO)
• CREST International Drawing Olympiad (CIDO)
• CREST International Spell Bee Summer (CSB)
• CREST International Spell Bee Winter (CSBW)
• International Teacher Olympiads
• Teacher Mathematics Olympiad
• Teacher Science Olympiad
• Teacher English Olympiad
• Exam Syllabus
• Sample Papers
• Previous Year Papers
• Marking Scheme
• Cut-Offs & Ranking Criteria
• Awards & Recognition
• Subject Rankers
• Subject Cut-Off
• Zone Definition
• Green Warrior Initiatives
• Rankholder's Gallery
• Testimonials
• Olympiad Exam Blog
• WhatsApp Channel
• Olympiad Books
• Live Classes
• Mental Mathematics Olympiad (CMMO)
• Green Warrior Olympiad (IGWO)
• Country Wise Olympiads
• Student Registration
• Teacher Registration
• School Registration
• Become a Coordinator
• Special Needs Kids
• CMO arrow_drop_down
• CSO arrow_drop_down
• CEO arrow_drop_down
• CRO arrow_drop_down
• CCO arrow_drop_down
• CMMO arrow_drop_down
• IGWO arrow_drop_down
• CIDO arrow_drop_down
• CSB arrow_drop_down
• CSBW arrow_drop_down
• Individual Registration
• Register Your School
• Exam Schedule
• Ranking Criteria
• Become a Co-ordinator
• assignment_ind
• account_balance

Speed, Distance and Time - Concept, Formulas & Questions

• About Topic
• Practice Questions
• More Topics
• Register for CREST Olympiads
• Online Classes

Speed, Distance and Time - Sub Topics

• Introduction & Concept

Formula of Speed

Average Speed

Formula of Average Speed

Formula of Time

Formula of Distance

Relationship between Speed, Time and Distance

The reading material provided on this page for Speed, Distance and Time is specifically designed for students in grades 5 to 12. So, let's begin!

Introduction & Concept of Speed, Distance & Time

The concept of speed, time and distance holds significant importance in the mathematics section of various competitive exams. It is a fundamental topic that finds applications in different scenarios such as motion in a straight line, circular motion, boats and streams, races, clocks, and more. Understanding the interplay between speed, distance, and time is crucial for solving problems related to these areas. Aspirants should aim to grasp the relationship between these factors to excel in their exam preparation.

It can be defined as the rate at which an object covers distance. It represents how fast or slow an object is moving. Mathematically, speed is calculated by dividing the distance travelled by the time taken. The standard unit of speed is meters per second (m/s), although other units such as kilometres per hour (km/h) or miles per hour (mph) are commonly used in everyday situations.

Example:  A train covers a distance of 240 km in 3 hours. What is the speed of the train?

Solution: Distance = 240 km Time = 3 hours

Speed = Distance / Time = 240/3 = 80 km/h

Average speed in mathematics refers to the overall rate at which an object or individual covers a certain distance over a specific period of time. It is determined by dividing the total distance travelled by the total time taken.

Example: A cyclist travels 60 kilometres at a speed of 30km/h hours and then continues for another 40 kilometres at a speed of 20 km/h. What is the cyclist's average speed for the entire journey?

Solution: To find the average speed, we need to determine the total distance travelled and the total time taken for the entire journey.

First, let's calculate the time taken for the first part of the journey: Time is taken for the first part = Distance / Speed = 60 km / 30 km/h = 2 hours

Now, let's calculate the time taken for the second part of the journey: Time is taken for the second part = Distance / Speed = 40 km / 20 km/h = 2 hours

To find the total time taken for the entire journey, we sum up the individual times: Total time taken = 2 hours + 2 hours = 4 hours

Next, we calculate the total distance travelled: Total distance = 60 km + 40 km = 100 km

Finally, we can determine the average speed: Average speed = Total distance / Total time taken = 100 km / 4 hours = 25 km/h

Therefore, the cyclist's average speed for the entire journey is 25 km/h.

It refers to the measurement of the duration of an event or the interval between two events. It is typically measured in units of seconds, minutes, hours, days, weeks, months, and years.

Example: A man walks at a speed of 6 km/h and covers a distance of 24 km. How long does it take to cover the distance?

Solution: Speed = 6 km/h Distance = 24 km Time = Distance/Speed = 24/6 = 4 hours

It refers to the measurement of the separation between two points in space. It is typically measured in units of meters, kilometres, miles, or feet. Distance can also refer to the amount of space covered by a moving object, such as the distance travelled by car or the distance covered by a runner in a race.

Example: A car is travelling at a speed of 60 km/h. How far will it travel in 3 hours?

Solution: Distance = 60 km/h x 3 hours = 180 km

The relationship between speed, time, and distance can be described using the formula:

Share Your Feedback

CREST Olympiads has launched this initiative to provide free reading and practice material. In order to make this content more useful, we solicit your feedback.

Do share improvements at [email protected]. Please mention the URL of the page and topic name with improvements needed. You may include screenshots, URLs of other sites, etc. which can help our Subject Experts to understand your suggestions easily.

Mental Maths Related Topics

• What are Natural Numbers?
• Formula for Perimeter of a Square
• Perimeter of a Rectangle - Sub Topics
• Triangle and its Types
• Composite Numbers from 1 to 100
• List of Prime Numbers Between 1 to 100
• Estimation of Numbers - Sub Topics
• What is a Prime Factorisation?
• Lines and Angles - Sub Topics
• Perimeter of Closed Figures - Sub Topics
• Classification of Triangles, Circles and Quadrilaterals
• Questions for Problems on Ages with Solutions
• Learn Ascending and Descending Order of Decimals
• Solved Questions on BODMAS Rule

Speed, Distance, Time Textbook Exercise

Click here for questions, gcse revision cards.

5-a-day Workbooks

Primary Study Cards

Privacy Policy

Terms and Conditions

Corbettmaths © 2012 – 2024

Distance, Time and Speed Word Problems | GMAT GRE Maths

Before you get into distance, time and speed word problems, take a few minutes to read this first and understand: How to build your credit score in USA as an international student .

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

Mini-MBA | Start here | Success stories | Reality check | Knowledgebase | Scholarships | Services Serious about higher ed? Follow us:                

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?

Leave a Comment Cancel reply

Average Speed Problems

Related Pages Rate, Time, Distance Solving Speed, Time, Distance Problems Using Algebra More Algebra Lessons

In these lessons, we will learn how to solve word problems involving average speed.

There are three main types of average problems commonly encountered in school algebra: Average (Arithmetic Mean) Weighted Average and Average Speed.

How to calculate Average Speed?

The following diagram shows the formula for average speed. Scroll down the page for more examples and solutions on calculating the average speed.

Examples Of Average Speed Problems

Example: John drove for 3 hours at a rate of 50 miles per hour and for 2 hours at 60 miles per hour. What was his average speed for the whole journey?

Solution: Step 1: The formula for distance is

Distance = Rate × Time Total distance = 50 × 3 + 60 × 2 = 270

Step 2: Total time = 3 + 2 = 5

Step 3: Using the formula:

Answer: The average speed is 54 miles per hour.

Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.

How To Solve The Average Speed Problem?

How to calculate the average speed?

Example: The speed paradox: If I drive from Oxford to Cambridge at 40 miles per hour and then from Cambridge to Oxford at 60 miles per hour, what is my average speed for the whole journey?

How To Find The Average Speed For A Round Trip?

Example: On Alberto’s drive to his aunt’s house, the traffic was light, and he drove the 45-mile trip in one hour. However, the return trip took his two hours. What was his average trip for the round trip?

How To Find The Average Speed Of An Airplane With Good And Bad Weather?

Example: Mae took a non-stop flight to visit her grandmother. The 750-mile trip took three hours and 45 minutes. Because of the bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip?

How To Relate Speed To Distance And Time?

If you are traveling in a car that travels 80km along a road in one hour, we say that you are traveling at an average of 80kn/h.

Average speed is the total distance divided by the total time for the trip. Therefore, speed is distance divided by time.

Instantaneous speed is the speed at which an object is traveling at any particular instant.

If the instantaneous speed of a car remains the same over a period of time, then we say that the car is traveling with constant speed.

The average speed of an object is the same as its instantaneous speed if that object is traveling at a constant speed.

How To Calculate Average Speed In Word Problems?

Example: Keri rollerblades to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the journey takes her 0.65h. What is Keri’s average speed during the trip?

How To Use Average Speed To Calculate The Distance Traveled?

Example: Elle drives 169 miles from Sheffield to London. Her average speed is 65 mph. She leaves Sheffield at 6:30 a.m. Does she arrive in London by 9:00 a.m.?

How To Use Average Speed To Calculate The Time Taken?

Example: Marie Ann is trying to predict the time required to ride her bike to the nearby beach. She knows that the distance is 45 km and, from other trips, that she can usually average about 20 km/h. Predict how long the trip will take.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

• Trending Now
• Foundational Courses
• Data Science
• Practice Problem
• Machine Learning
• System Design
• DevOps Tutorial
• Problem on Trains, Boat and streams
• Work and Wages - Aptitude Questions and Answers
• Algebra | Set -1
• Problem on Pipes and Cisterns
• Problems on Work and Wages
• Mensuration 2D Formula & Aptitude Questions
• Problem on HCF and LCM
• Problem on Numbers
• Percentages - Aptitude Questions and Answers
• Height and Distances - Aptitude Questions and Answers
• Problems on Percentage
• Probability - Aptitude Questions and Answers
• Ratio proportion and partnership | Set-2
• Mensuration 2D | Set 2
• Mixture and Alligation | Set 2
• Permutation and Combination - Aptitude Questions and Answers
• Complete Interview Preparation For Product and Service Based Companies
• Quantitative Aptitude: Maths Tricks | Set 1 (Squares of numbers)
• Mensuration 3D Formulas and Aptitude Questions

Problem on Time Speed and Distance

Question 1: A racing car covers a certain distance at a speed of 320 km/hr in 5 hours. To cover the same distance in 5/3 hours it must travel at a speed of:  Solution: Given Distance is constant.  So, Speed is inversely proportional to time.   

1 unit -> 320 km/hr  3 unit -> 320 x 3 = 960 km/hr is required speed 

Question 2: A train running at a speed of 36 km/hr and 100 meter long. Find the time in which it passes a man standing near the railway line is :  Solution: Speed = 36 km/hr  Change in m/s  So, speed = 36 * 5/18 = 10 m/s  Time required = Distance/speed  = 100/10  = 10 second  

Question 3: If an employee walks 10 km at a speed of 3 km/hr, he will be late by 20 minutes. If he walks at 4 km/hr, how early from the fixed time he will reach ?  Solution: Time taken at 3 km/hr = Distance/speed  = 10/3  Actual time is obtained by subtracting the late time  So, Actual time = 10/3 – 1/3 = 9/3 = 3 hour  Time taken at 4 km/hr = 10/4 hr  Time difference = Actual time – time taken at 4 km/hr  = 3 – 10/4  = 1/2 hour  Hence, he will be early by 30 minutes . 

Question 4: The diameter of each wheel of a truck is 140 cm, If each wheel rotates 300 times per minute then the speed of the truck (in km/hr) (take pi=22/7)  Solution: Circumference of the wheel= 2 * 22/7 * r  = 2 * 22/7 * 140/2  = 440 cm  Speed of the car = (440 * 300 * 60)/(1000 * 100 )  = 79.2 km/hr  

Question 5: A man drives at the rate of 18 km/hr, but stops at red light for 6 minutes at the end of every 7 km. The time that he will take to cover a distance of 90 km is  Solution: Total Red light at the end of 90 km = 90/7 = 12 Red light + 6 km  Time taken in 12 stops= 12 x 6 = 72 minutes  Time taken by the man to cover the 90 km with 18 km/hr without stops = 90/18 = 5 hours  Total time to cover total distance = 5 hour + 1 hour 12 minute  = 6 hour 12 minute  

Question 6: Two jeep start from a police station with a speed of 20 km/hr at intervals of 10 minutes.A man coming from opposite direction towards the police station meets the jeep at an interval of 8 minutes.Find the speed of the man.  Solution:   

Here, 4 units -> 20 km/hr  1 unit -> 5 km/hr  Speed of the man = 1 unit = 1 x 5 = 5 km/hr  

Question 7: Two city A and B are 27 km away. Two buses start from A and B in the same direction with speed 24 km/hr and 18 km/hr respectively. Both meet at point C beyond B. Find the distance BC.  Solution: Relative speed = 24 – 18  = 6 km/hr  Time required by faster bus to overtake the slower bus = Distance/time  =27/6 hr  Distance between B and C= 18*(27/6)= 81 km  

Question 8: A man travels 800 km by train at 80 km/hr, 420 km by car at 60 km/hr and 200 km by cycle at 20 km/hr. What is the average speed of the journey?  Solution: Avg. Speed = Total distance/time taken  (800 + 420 + 200) / [(800/80) + (420/60) + (200/20)]  =>1420 / (10 + 7 + 10)  =>1420/27  => 1420/27 km/hr  

Question 9: Ram and Shyam start at the same with speed 10 km/hr and 11 km/hr respectively.  If Ram takes 48 minutes longer than Shyam in covering the journey, then find the total distance of the journey.  Solution: 

Ram takes 1 hour means 60 minutes more than Shyam.  But actual more time = 48 minute.  60 unit -> 48 min  1 unit -> 4/5  Distance travelled by them= Speed x time  = 11 x 10 = 110 unit  Actual distance travelled = 110 x 4/5  = 88 km  

Question 10: A person covered a certain distance at some speed. Had he moved 4 km/hr faster, he would have taken 30 minutes less. If he had moved 3 km/hr slower, he would have taken 30 minutes more. Find the distance (in km)  Solution: Distance = [S 1 S 2 / (S 1 – S 2 )] x T   S 1 = initial speed  S 2 = new speed  Distance travelled by both are same so put equal  [S (S + 4) / 4 ] * (30/60) =[ S (S – 3)/ 3 ]* (30/60)  S = 24  Put in 1st  Distance=(24 * 28) / 4 * (30/60) = 84 km  

Question 11: Ram and Shyam start from the same place P at same time towards Q, which are 60km apart. Ram’s speed is 4 km/hr more than that of Shyam.Ram turns back after reaching Q and meet Shyam at 12 km distance from Q.Find the speed of Shyam.  Solution: Let the speed of the Shyam = x km/hr  Then Ram speed will be = (x + 4) km/hr  Total distance covered by Ram = 60 + 12 = 72 km  Total distance covered by Shyam = 60 – 12 = 48 km  Acc. to question, their run time are same.  72/ (x + 4) = 48/ x  72x = 48x + 192  24x= 192  x= 8  Shyam speed is 8 km/hr  

Question 12: A and B run a kilometre and A wins by 20 second. A and C run a kilometre and A wins by 250 m. When B and C run the same distance, B wins by 25 second. The time taken by A to run a kilometre is  Solution: Let the time taken by A to cover 1 km = x sec  Time taken by B and C to cover the same distance are x + 20 and x + 45 respectively  Given A travels 1000 then C covers only 750.   

A/C = 3/4 = x/(x+45)  3x + 135 = 4x  x =135  Time taken by A is 2 min 15 second  

Please Login to comment...

• How to Delete Whatsapp Business Account?
• Discord vs Zoom: Select The Efficienct One for Virtual Meetings?
• Otter AI vs Dragon Speech Recognition: Which is the best AI Transcription Tool?
• Google Messages To Let You Send Multiple Photos
• 30 OOPs Interview Questions and Answers (2024)

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

Baltimore bridge collapse: What happened and what is the death toll?

What is the death toll so far, when did the baltimore bridge collapse, why did the bridge collapse, who will pay for the damage and how much will the bridge cost.

HOW LONG WILL IT TAKE TO REBUILD THE BRIDGE?

What ship hit the baltimore bridge, what do we know about the bridge that collapsed.

HOW WILL THE BRIDGE COLLAPSE IMPACT THE BALTIMORE PORT?

Get weekly news and analysis on the U.S. elections and how it matters to the world with the newsletter On the Campaign Trail. Sign up here.

Writing by Lisa Shumaker; Editing by Daniel Wallis and Bill Berkrot

Our Standards: The Thomson Reuters Trust Principles. , opens new tab

Thomson Reuters

Lisa's journalism career spans two decades, and she currently serves as the Americas Day Editor for the Global News Desk. She played a pivotal role in tracking the COVID pandemic and leading initiatives in speed, headline writing and multimedia. She has worked closely with the finance and company news teams on major stories, such as the departures of Twitter CEO Jack Dorsey and Amazon’s Jeff Bezos and significant developments at Apple, Alphabet, Facebook and Tesla. Her dedication and hard work have been recognized with the 2010 Desk Editor of the Year award and a Journalist of the Year nomination in 2020. Lisa is passionate about visual and long-form storytelling. She holds a degree in both psychology and journalism from Penn State University.

Israeli PM Netanyahu’s hernia operation was successful, hospital says

Israeli Prime Minister Benjamin Netanyahu was conscious and conversing with family on Monday after undergoing a successful hernia operation, the hospital treating him said.