Question

A car is travelling with a velosity of 40 m/s.the driver applies brake with a uniform retardation of 5m/s.find the time required to bring the car to rest? Calculate the distance travelled by the car after applying the break?

Answer 1

To bring the car to rest,
v=0 m/s, u= 40 m/s, a= -5 m/s^2
v= u+ at
t= (v-u)/a
t= (0-40)/(-5)
t= 8 seconds
Distance travelled after applying brakes,
2as= v^2 - u^2
s= (v^2 - u^2)/(2a)
s= (0 - 1600)/ (-10)
s= 160m
Hope it helped! ;)

Answer 2

Given,

Velocity of a car=40m/s

Retardation of the car when break applies=5m/s²

To Find,

Time required to bring the car rest.

Distance travelled by the car.

Solution,

Final velocity,

v=0 m/s,

Initial velocity,u= 40 m/s,

Retardation,a= -5 m/s²

So by using law of motion,

v= u+ at

⇒t= (v-u)/a

⇒t= (0-40)/(-5)

⇒t= 8 seconds

Then, distance travelled after applying brakes,

By using third law of motion,

2as= v² - u²

⇒s= (v² - u²)/(2a)

⇒s= (0 - 1600)/ (-10)

⇒s= 160m

hence, time required to bring the car to rest is 8 seconds and distance travelled by the car after applying the break is 160m.