A driver of a car sights a child on the road at a distance of 200 m. If the speed of the car is 108 kmph then calculate the retardation and time taken to stop the car just in front of the child.

Explanation

Given
`v=0, v_(0)="108 kmph, "=108x(5)/(18)=30ms^(-1)`
`x=200m`
Hence `v^(2)=v_(0)^(2)+2ax "gives "0=(30)^(2)+2a(200)`
`therefore" "a=(900)/(400)=-2.25ms^(-2)`
`therefore" "a=-2.25ms^(-2)`
Using the formula,`v=v_(0)+at`
`0=30-2.25t" "t=(30)/(2.25)=(30)/((2(1)/(4)))=(30xx4)/(9)=(10xx4)/(3)=(40)/(3)`
`therefore" "t = 13.33s`