Question

A car a moving at rate of 72km/h and applies brakes which provide a retardation of
5ms -2 .
(i) How much time does the 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

v = 72 km/h

u = 0

a = 5m/s²

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

$t = \dfrac{0 - 72}{5}$

$t = \dfrac{ - 72}{5}$

$t = - 14.4 \: seconds$

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

$2 \times 5 \times s = {72}^{2} - {0}^{2}$

$10s = 5184$

$s = 510.4 \: km$