Question

Please Answer this.

A car a moving at rate of 72km/h and applies brakes which provide a retardation of 5ms-2.

(i) How much time does the car takes 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?

Thanks In Advance.

Answer 1

i) GIVEN

u=72km/h

     =72x5/18=20m/s

v=0

a=-5m/s²

Using 1st Eq of motion

v=u+at

0=20+(-5xt)

-5t=-20

t=5 sec

ii)using 3rd Eq of motion

2as = v²-u²

2x-5xs=0²-20²

-10s=-400

s=40m

iii)u=20x2=40m/s

2as=v²-u²

-10s=-1600

s=160m

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