Question

A 4m car travelling 72km/hr comes behind a 20 m truck. The car pulls out from behind the truck with the front of the car 10 m behind the rear of the truck and accelerates at 3.5 m/s^2 until the rear of the car is 10 m in front of the truck. Time? Dist?

How far did the car travel and how long did it take?

Answer 1

`5.01` seconds and `44` meters.

Explanation:

As the car was `10` meter behind, it covers `20` meter length of truck and comes `10` meters in front (we should also add `4` meters of car length as now its rear is `10` meters in front.

Hence the distance covered is `10+20+10+4=44` meters

Acceleration is `3.5m/s^2` and initial velocity is `72` kmph or `72×1000/3600=20` meters per second.

However, here what is important is relative velocity of car (relative to truck), which we consider to be travelling at same velocity as that of car and hence is `0`.

As equation is `S=ut+1/2at^2`, where `S` is distance covered, `t` is time taken, `u` is initial velocity and `a` is acceleration, we have

`44=1/2×3.5×t^2`

or `t^2=(44×2)/3.5`

or `t~=5.01`