Time Speed Distance

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).

Note: To convert km/h to m/s, multiply by \frac{5}{18}, and to convert m/s to km/h, multiply by \frac{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 solving 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:

1. 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 (S1xS2/(S1+S2)).

2. 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 Speed can be expressed as S1+S2 = (L1+L2)/t.

3. When two trains pass each other, the speed differential between them can be determined using the equation, 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.

S1-S2 = (L1+L2)/t

4. 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

5. 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.

Time Speed Distance - Questions and Answers

Question 1: A cyclist can complete a 1200 m race in three minutes. Will he be able to beat another cyclist who rides at 25 km/hr?

Solution:

We are given that the first cyclist can complete a 1200 m race in 3 minutes or 180 seconds.

=> Speed of the first cyclist = 1200 / 180 = 6.67 m/sec
We convert this speed to km/hr by multiplying it by 18/5.
=> Speed of the first cyclist = 24 km/hr
Also, we are given that the speed of the second cyclist is 25 km/hr.
Therefore, the first cyclist cannot beat the second cyclist.

Question 2: To cover a certain distance, a student has two options: to cycle or to walk. If he walks one way and cycles back, it takes him 5 hours. If he cycles both ways, it would take him 3 hours. How much time will he take if he walks both ways?

Solution:

To determine the time taken by the student if he walks both ways.

Time taken to walk one side + Time taken to cycle one side = 5 hours
Time taken to cycle both sides = 2 × Time taken to cycle one side = 3 hours
=> Time taken to cycle one side = 1.5 hours
Therefore, time taken to walk one side = 5 – 1.5 = 3.5 hours
Thus, time taken to walk both sides = 2 × 3.5 = 7 hours.

Question 3 : A postman traveled from his post office to a village in order to distribute mails. He started on his bicycle from the post office at the 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 a 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 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

Question 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.

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.

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

Related Articles:

• Practice Speed, Time and Distance Aptitude Quiz Questions
• Problem on Time Speed and Distance | Set-2