Question

A car travels with a uniform velocity of 25 $m{s}^{-1}$ for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.Find:(i) the distance which the car travels before the brakes are applied,(ii) the retardation,and (iii) the distance travelled by the car after applying the brakes.

Answer 1

u=25m/sec
Distance it travels in 5sec
(i) Distance travelled before the brakes are applied i.e. in 5sec

D = u*t = 25*5 = 125m
(ii) After 5 sec brakes are applied
It then comes to rest in 10sec
u = 25 m/s
v = 0 m/s (as it comes to rest)
t = 10 s
a = change in velocity/time = $\frac{(v-u)}{t}=\frac{(0-25)}{10}$ = -2.5 $m/s^2$
(iii) Distance travelled by the car after applying the brakes, s=?

using eq. of motion: $s=ut+\frac{1}{2}a{t}^2$

$s=25\times 10+\frac{1}{2}(-2.5){\left(10\right)}^2=250-125=125m$
The distance it travels after the brakes are applied is 125 m