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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 S 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 S2 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:

1. Inverse Proportionality Rule
2. Constant Product Rule.

## Sample Problems on Speed, Time, and Distance

### Q1: 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.

### Q2: 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.

Solution:

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

### Q3: 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

### Q4: 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.

Solution:

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.

### Q5 : 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.

Solution:

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.

### Q6: 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.

Solution:

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

### Q7: 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?

Solution :

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?

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?

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