Question

a car moving at the rate of 72 kilometre per hour and applies break which provide retardation of 5 metre per second square
{1}how much time does it takes to stop?
{2}How much distance does the car cover before coming to rest?
{3}What would be the stopping distance needed if speed of the car is doubled?

Answer 1

v= 20 m/s
a=- 5 m/ s^2

1.
v- u = a t
t = -20 / -5
t = 4s

2.
v^2 - u^2 = 2 a s
s = -400 / ( 2 x - 5 )
s = 40 m

3.

s'= -1600 / ( 2 x -5 )
s '= 160 m
s'= 4s