Direction : The table given below shows the data about track record of car…
2021
Direction : The table given below shows the data about track record of car racing game, where 4 cars P, Q, R and S are competing to reach finishing line which is at a distance of 3000 m from their initial starting point.
1. Drift refers to over starting intentionally done by driver.
2. While calculating average speed, do not consider time taken in drift/crash.
3. Points = 50 × drifts – 20 × crashed
How much speed does car R need to gain in order to cover twice the distance in 5/3 of the original time taken?
- A.
16 m/s
- B.
28 m/s
- C.
12 m/s
- D.
24 m/s
- E.
10 m/s
Show answer & explanation
Correct answer: D
Concept
Speed is the ratio of distance to time: speed = distance ÷ time. So the speed required to travel a given distance in a given time is found by dividing the (new) distance by the (new) time. When distance and time both scale, the required speed scales by the factor (distance-factor ÷ time-factor).
Application
Read car R's data from the table and work step by step:
Original distance to the finishing line = 3000 m; original time taken by R = 150 s.
Original speed of R = 3000 ÷ 150 = 20 m/s.
New distance = twice the original = 2 × 3000 = 6000 m.
New time = 5/3 of the original = (5/3) × 150 = 250 s.
Required new speed = new distance ÷ new time = 6000 ÷ 250 = 24 m/s.
Among the offered values, the speed R must attain is 24 m/s. The stem asks for the speed needed for the new run; the bare numerical increase over the original 20 m/s would be 4 m/s, but that value is not offered, so the answer is the new speed itself, 24 m/s.
Cross-check
Verify with the scaling shortcut: distance-factor = 2, time-factor = 5/3, so the speed multiplies by 2 ÷ (5/3) = 6/5 = 1.2. Then new speed = 20 × 1.2 = 24 m/s, which matches.