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