Question

). A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate (i) the
average retardation of the car (ii) time
taken by the car to come to rest.​

Answer 1

Answer:

1 the average retardation of the car

Answer 2

Explanation:

Given that,

Initial speed of the car, u = 120 km/h = 33.33 m/s

Finally, it comes to rest, v = 0

Distance, s = 100 m

(a) Let a is the average retardation. Using third equation of motion as :

$v^2-u^2=2as$

$a=\dfrac{-u^2}{2s}$

$a=\dfrac{-(33.33)^2}{2\times 100}$

$a=-5.55\ m/s^2$

(b) Let t is the time taken by the car to come to rest. It is given by :

$t=\dfrac{v-u}{a}$

$t=\dfrac{-33.33}{-5.55}$

t = 6 seconds

Hence, this is the required solution.