A car moving along a straight highway with speed of ` 126 km 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 ?

Options

A. `2.06m//s^2` `1.44s`
B. `3.06m//s^2` `11.44s`
C. `3.06m//s^2` `21.44s`
D. `13.06m//s^2` `11.44s`

Answer

B

Explanation

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
Retardation produced in the car = a
From third equation of motion, a can be calculated as:
`v^2-u^2=2as`
`(0)^2-(35)^2=2xxaxx200`
`a=-(35xx35)/(2xx200)=-3.06m//s^2`
From first equation of motion, time (t) taken by the car to stop can be obtained as:
v=u+at
`t=(v-u)/a=(-35)/(-3.06)`=11.44s