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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:-

<< Practice Speed, Time and Distance Aptitude Questions >>

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  • Average Speed

Average Speed is important to understand the rate at which a journey takes place. Throughout a journey, the speed may vary from time to time. In that case, finding the average speed becomes important to have an estimate of the rate at which the journey is completed. Here we will see some special cases and shortcuts that we can use to find the average speed in less than a minute. Let us begin by calculating the formula for the average speed and then solve some examples with the help of this formula.

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The speed of an object is found out by dividing the distance that the object covers by the time in which the object takes to cover this distance. If ‘D’ is the distance traveled in some time ‘T’ then the speed of the object for this journey or ‘s’ is equal to s = D/T.

Average Speed

This is the simplest case. But what if there is an object that covers part of its journey with one speed and the other part with another speed? Let us say a train goes from station A to B in 2 minutes and from B to C in 3 minutes. If the distance between any two stations is 6km, what is the speed of the train?

Answer: Here you can’t use the formula directly because as you may have observed the train has a different speed from A to B than from B to C. In this case when a body has different speed during different parts of the journey, we define the average speed as:

Average Speed = (Total distance covered throughout the journey)/(Total time taken for the journey)

Let the body cover a distance ‘a’ in time t 1 , then a distance b in time t 2 , c in time t 3 and so on, then the average speed of the body is found out by the following ratio :

Average speed = (a + b + c + …)/ (t 1 + t 2 + t 3  + …). Let us see some examples now.

Browse more Topics Under Averages

  • Data Sufficiency
  • Average of Number Series
  • Average Practice Questions

Some Solved Examples

Example 1: A truck goes from Haryana to Bangalore with an average speed of 60 km/hr. The journey takes 30 hours. It returns from Bangalore to Haryana on the same road with an average speed of 40 km/hr. What was the average speed of the truck during the roundtrip?

A) 50 km/hr              B) 55 km/hr            C) 48 km/hr              D) 52 km/hr

Answer: You may be tempted to find the average speed by summing up the speed and dividing by 2. That is wrong because the average speed is the total distance divided by the total time. Let us find the distance first from Haryana to Bangalore. This can be done as follows:

Distance = (Speed)×(Time). Thus we can say that Distance = 60×30 = 1800 kilometers. Now we have to find the time taken to travel from Bangalore to Haryana. We can write:

Time = Distance / speed = 1800/40 = 45 hours. Thus we can find the average speed as:

Average Speed = (Total distance)/(Total Time) = (1800 + 1800)/(30 + 45) = 3600/75 = 48 km/hr. Hence the correct option here is C) 48 km/hr.

So to find the average speed never use the formula of the averages but try to find the total distance covered and the total time taken. Let us see a few more examples where the journey is completed in three stages.

Practice  Average Practice Questions here

Three Stage Journey

Example 2: A person travels from Mumbai to Goa in exactly 8 hours with a speed of 120 km/hr. From Goa, the person travels to Delhi in 34 hours. If the distance from Goa to Delhi was 2000 km, how long will it take to get from Delhi to Bombay if the person maintains the speed equal to the average speed? The distance from Delhi to Bombay is 1400 km.

Answer: This is a three-stage journey. We will form a distance-speed-time table for every stage of the journey. It should make it easier for us.

We don’t have to find the individual speeds for the two stages of the journey. We just want the average speed. To find that we have to find the total distance and the total time. We do have the distance and the time for the second stage so let us get it for the first stage too.

For the first stage, the time can be found by Distance = Speed × Time = 120×8 = 960 km. So the total distance is = (960 + 2000) = 2960 km.

Now the total time taken for the journey = 8 + 34 hours = 42 hours. Always make sure that you check the units. The units throughout the question should be uniform.

Thus the average speed = 2960/42 = 70.48 km/hr. Thus this is the speed that the person has to maintain from Delhi to Bombay. The time taken for this journey is = Distance / Speed. In other words , we can write:

Time taken = 2960/70.48 = 41.99 hours ≈ 42 hours.

Practice Questions:

Q 1: A speeding car traveling at 40 km/hr decides to slow down to 30 km/hr after covering half the distance. What is the average speed?

A) 34.29 km/h               B) 35.36 km/hr           C) 36.32 km/hr                D) 33.21 km/hr

Ans: A) 34.29 km/h

Q 2: A car travels with the following speed and distance:

A) 115 km/hr        B) 112 km/hr              C) 1.17 km/hr                  D) 2.71 km/hr

Ans: C) 1.17 km/hr

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5 responses to “Average Speed”

But i think in the second example, the distance between delhi to mumbai is given 1400 km then why did you used the distance 2960 to calculate the time.

exactly..the question itself is wrong

Exactly! I wondered the same thing

why did they give the distance for the third journey 1400 km if we didn’t use it in the equation nor answer

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  17. Speed, Time and Distance

    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 ...

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  20. Average Speed: Formula, Videos, Examples and Practice Questions

    Average Speed is important to understand the rate at which a journey takes place. Throughout a journey, the speed may vary from time to time. In that case, finding the average speed becomes important to have an estimate of the rate at which the journey is completed. Here we will see some special cases and shortcuts that we can use to find the average speed in less than a minute.

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