Question

Q. A motor car slows down from 72 km/h to 36 km/h at a distance of 25 m. If the brakes are applied with the same force calculate :-

1. Total time in which the car comes to rest .
2. Distance travelled by it.

Answer 1

Acceleration = -6 m/s
Time to come to rest =3.33s
Distance travelled = 6.66m

Answer 2

Answer : (1)  time, t = 3.33 s  (2) Distance, s = 33.3 m

Explanation :

It is given that,

Initial velocity motor car, u = 72 km/h = 20 m/s

Final velocity of the motor car, v = 36 km/h = 10 m/s

Distance covered by motor car is, s = 25 m

(1) Using the third equation of motion :

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

$(10\ m/s)^2-(20\ m/s)^2=2\times 25\ m\times a$

$a=-6\ m/s^2$

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

When car comes to rest final velocity of the car becomes zero i.e. v = 0

$t=\dfrac{0-20\ m/s}{-6\ m/s^2}$

$t = 3.33\ s$

(2) Again using third equation of motion :

$0-(20\ m/s)^2=2\times -6\ m/s^2\times s$

$s=33.3\ m$

Hence, this is the required solution.