Question

A car moving along a straight highway with speed of 126 $\text{km}\;\text h^{-1}$ is brought to a stop within a distance of 200 $\text m$. 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, $\text u$ = 126 $\text{km}/ \text h$ = 35 $\text{m}/ \text s$
Final velocity of the car, $\text v$ = 0
Distance covered by the car before coming to rest, s = 200 $\text m$
Retardation produced in the car = $\text a$
From third equation of motion, a can be calculated as :
$\text v^2-\text u^2$ = 2$\text {as}$
$(0)^2-(35)^2$ = 2 × $\text a$ × 200
$\text a$ = $-\frac{35 \times 35}{2 \times 200}$= - 3.06 $\text m$ / $\text s^2$
From first equation of motion, time ($\text t$) is taken by the car to stop can be obtained as :
$\text v$ = $\text u$ + $\text {at}$
$\text t$ = $\frac{\text v-\text u}{\text a}$= $\frac{-35}{-3.06}$ = 11.44 s