Question

A car moving along a straight highway with a speed of $126km{h}^{–1}$ is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Answer 1

Initial velocity of the car, u = 126 km/h = 35 m/s
Final velocity of the car, v = 0
Distance covered by the car before coming to rest, s = 200 m
Acceleration produced in the car = a
From third equation of motion, a can be calculated
$v^2-u^2=2as(0{)}^2-(35{)}^2=2\times a\times 200a=-\frac{35\times 35}{2\times 200}=-3.06m/s^2$
So, the retardation is 3.06 $m/s^2$.

From first equation of motion, time (t) taken by the car to stop can be obtained as:
$v=v+att=\frac{v-u}{a}=\frac{-35}{-3.06}=11.44s$