Question

A car is traveling with a speed of 36km/h. the driver applies the brakes and retards the car uniformly. The car is stopped in 5s. Find (i) the retardation of the car
(ii) distance traveled before it is stopped after applying the brakes.

Answer 1

Answer:

First of all convert 36km/h into m/s

36×5/18=10m/s

Given time=5 second

Final velocity =0m/s

Acceleration =final velocity - initial velocity /time

O-10/5=-2m/s^2

Since acceleration is in negative hence it is called retardation

Distance S=ut +1/2at^2

10×5+1/2×-2×(5)^2

50-25=25 metre

Answer 2

Given:

u=36km/h=10m/s

t=5s

v=0m/s

1). a=(v-u)/t

a=(0-10)/5

a=-10/5=-2m/s²

so,retardation is 2m/s².

2). v²=u²+2aS

0²=10²+2*(-2)*S

0=100-4*S

S=-100/-4=25m Ans.