Time, speed and Distance concepts

1. What are Speed, Time and Distance

Speed and Distance Time is one of the most popular and crucial topics in any competitive exam’s Mathematics or Quants portion. The concept of Speed, Time, and Distance is frequently used in motion in a straight line, circular motion, boats and streams, races, clocks, and so on

Speed is defined as the ratio of distance travelled to time spent travelling. Speed is a scalar quantity since it has just one direction and no magnitude.

2. Unit of Speed. Time and Distance

Time, Speed and Distance is expressed as a different unit such as

• Time: seconds(s), minutes (min), hours (hr)
• Distance: kilometres (km), (meters (m), miles, feet
• Speed: m/;s, km/hr

3. Relationship between Speed, Time and distance

• Speed = Distance/Time – This determines how slowly or quickly an object moves. It describes the distance travelled divided by the time required to complete the journey. 
• Distance is directly proportional to speed, while time is inversely proportional to speed.
• Distance = Speed X Time, and Time = Distance / Speed, as speed increases, time taken decreases, and vice versa. 

4. The formula of Speed, Time and Distance.

Some important formulas of Time, Speed and Distance is given below. 

5. Convert Speed, Time and distance into different units

You can convert the speed, time and distance unit into another measurable unit to calculate the exact speed, time and distance. 

For example,

• To convert km/hr to m/s, use the formula: a Km/hr = an x (5/18) m/s
• To convert m/sec to km/hr, use the formula: a m/s = an x (18/5) Km/hr
• If a person travels from point A to point B at S kilometres per hour (kmph) and then returns from point B to point A at S2 kmph, the total duration for the round journey is T hours. T (S1S2/(S1+S2)) is the distance between points A and B.
• If two moving trains, one of length L1 travelling at speed S2 and the other of length L2 travelling at speed S2, intersect in time t. Their Total Velocity can thus be written as S1+S2 = (L1+L2)/t.
• If a train of length L1 travels at S1 and can pass a platform, bridge, or tunnel of length L2 in time t, the speed is S1 = (L1+L2)/t.
• The speed differential between two trains can be calculated 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, L2 is the slower train’s length, and t is the time it takes for them to pass each other.
• If two people A and B both start from different points P and Q at the same time and cross each other in T1 and T2 hours, respectively, then \( \frac{{\text{A’s speed}}}{{\text{B’s speed}}} ={{\sqrt{T_2}}} \frac{.}{.} {{\sqrt{T_1}}} \)

6. Examples

Q1) A  can run 750 metres in two and a half minutes. Will he be able to catch another runner who averages 17.95 km/hr? 

Ans. A can run 750 metres in 2 minutes and 30 seconds, or 150 seconds. 

=> First runner’s speed = 750 / 150 = 5 m/sec 

We convert this speed to kilometres per hour by multiplying it by 18/5. 

=> The first runner’s speed = 18 km/hr 

, and the second runner’s speed is 17.95 km/hr. 

As a result, the first runner can defeat the second runner. 

Q2) Ankit has decided to travel 9 km in 84 minutes. He travelled 1/3 of distance in 4 km per hr and remaining distance at different speed. Find out the speed of Ankit after 1/3 distance is covered ?

Ans.  1/3 distance of 9 km is 3 km 

= Ankit travelled 3km with 4km/hr. Then time = D/S= 3km/4km/hr= 75 minutes

= time left = 84-75 minutes= 9 minutes

Now ankit have to cover the remaining 6 kms in 9 minutes or 9/60=0.15 hr then the speed will be 6km/0.15hr= 40 km/hr

Q3) Two stations X and Y are 465 km distant. A train starts from X towards Y at 10 AM with a speed of 65 km/hr. Another train leaves from Y towards X at 11 AM at a speed of 35 km/hr. Find out the time of train when they meet together.

Ans The train leaving from X leaves an hour earlier than the train that leaves from Y. 

=> Distance covered by train leaving from x = 65 km / hr x 1 hr = 65 km 

Distance left = 465 – 65 = 400 km 

Now, the train from Y 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 both the trains to meet = 400 km / 100 km / hr = 4 hours 

Hence, Train X and Y meet together at 4 hours after 11 AM, i.e., 3 PM. 

Q4) A and B must drive their own automobiles from Raipur to Hyderabad. A is travelling at 80 kilometres per hour, while B is travelling at 120 kilometres per hour. Determine B’s travel time to Hyderabad if A takes 9 hours.

Sol) In this case, the distance is the same which means the time consumed will be inversely proportional to the speed. The speed of A and B in the above question is 80:120 or 2:3.