Question

A car moving at speed of 72km/h and applies brakes which provide a retardation of 5m/s2. I how much time does the car take to stop? Ii how much distance does the car cover before coming to rest? Iii what would be the stopping distance needed if speed of the car is doubled?

Answer 1

Initial velocity, u= 72km/h = 72×5/18= 20m/s

Final velocity,v= 0 m/s

Acceleration,a = -5 m/s^2

Using first equation of motion,

v=u+at

0=20+(-5)t

-20=-5t

t= 4 s, where t is the time taken to stop the car.

Using third equation of motion,

v^2-u^2 = 2as

(0)^2 - (-20)^2=2(-5)s

0-400=-10s

s= -400/-10= 40m, where s is the distance.

We know,

s is directly proportional to v

If the speed of car will be doubled,

Then s will also become double..

Therefore the new distance, s' is

s' = 2×40= 80 m