A car start moving with uniform acceleration from its position of rest and it moves 100 m is 10 s. On applying brakes, it stops after covering 50 m. Then the magnitude of acceleration in the second part of its motion is ______ m `s^(-2)`.

Answer

4

Explanation

Case I : `ArarrB`
Initial velocity = u = 0
t = 10 s S = 100 m
By `s=ut+1//2at^(2)`
`100=1/(2)xxaxx100`
`a=2 m s^(-2)`
`V_(B)=u+at`
`=0+2xx10=20 m s^(-1)`
Case II : `BrarrC`
`c=20 m s^(-1)v=0 m s^(-1)` s = 50 m
By `veca=(v^(2)-u^(2))/(2s)=(0^(2)-20^(2))/(2xx50)`
`(-400)/(100)=-4`
`a=-4 m s^(-2),|veca|=4 m s^(-2)`
Hence, the correct answer si 4.