Question

A car moving at the rate of 72 km/h and applies brake which provides a retardation of 5 m/s2. ( I ) How much time does 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 ? ​

Answer 1

Answer:

Time=4 seconds

distance before coming to rest=40 m

Stopping distance if speed is doubled = 80m

Explanation:

given data

initial speed(u) 72kmph convert to m/s it will be 20m/s. consider u=20m/s

and it stops after some time so final speed v=0

Acceleration= -5m/s^2

i) t=v-u/a

t= 0-20/-5

t= 4 s

ii) s=(u+v/2) Xt

s=(20+0/2)X4

s=10x4=40m

iii) if u= 2u then

u = 2x20

u= 40 m/s then find s=?

s=(u+v/2) Xt

s=(40+0/2)X4

s=20x4=80m