Assume that a car is able to stop with a retardation of 8 `ms^(–2)` and that a driver can react to an emergency in 0.5 sec. Calculate the overall stopping distance of the car for a speed of 60 `km^(–1)` of the car.

Explanation

Here, `u=60km h^(-1)=60xx5/(18)=(50)/3ms^(-1)`
Since the application of brakes takes 0.5 s, before this the car was moving with uniform speed of
`(50)/3ms^(-1)`
`:.` Distance covered in 0.5 sec, with a uniform speed is
`S_1=uxxt=(50)/(3)xx0.5=(25)/3m=8.33m`
Now car begins to move with a retardation of `8ms^(-2)`
`:.` Distance covered before coming to rest,
`2aS_2=v^2-u^2`
or `S_2=(v^2-u^2)/(2a)=(0-((50)/(3))^2)/(-2xx8)=(50xx50)/(9xx2xx8)`
`=(2500)/(144)=17.36m`
`:.` Total (overall) distance`=S_1+S_2=8.33+17.36`
`impliesS=25.69m`