Question

A car is moving in a straight line with speed 18 km $h^{-1}$. It is stopped in 5 s by applying the brakes. Find:

(i) the speed of car in m $s^{-1}$,
(ii) the retardation of car
(iii) the speed of car after 2 s of applying the brakes.

Answer 1

(i) Initial speed of the car, u=18km/h

To convert km/h into m/s we need to multiply it by 5/18.

Therefore speed in m/s = $18\times \frac{5}{18}$ =5m/s

(ii) Final speed, $v=0$ , u = 5 m/s , time taken to stop t = 5s
From the first equation of motion, $v=u+at\Rightarrow 0=5+a\times 5\Rightarrow a=-1$
Hence, the retardation is $1m/s^2$.

(iii) Velocity after 2 seconds of applying brakes, u = 5 m/s , a = -1 m/$s^2$ , t = 2s

using first equation of motion we get:

$v^{\prime }=u+a{t}^{\prime }=5-1\times 2=5-2=3m/s$