A car moving with a velocity of `20 ms^(-1)` is stopped in a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is

Options

A. 640 m
B. 320 m
C. 1280 m
D. 160 m

Answer

D

Explanation

Initial velocity, `u - 20 m s^(-1)`,
Final velocity, v = 0 m s^(-1),
Distance travelled, s = 40 m
Using third equation of motion,
`v^(2)- u^(2) = 2as : 0 - 20^(2) = 2 x a x 40`
We get, `a = -5 m s^(-2)`
For second case, `u = 2 xx 20 m s^(-1) = 40 m s^(-1)`
`V = 0 ms^(-1)` and `a = -5 m s^(-2)`
Using the third equation of motion, `s= (0- (40)^(2))/(2(-5))= 160m`
`therefore` Distance travelled by the same car travelling at double velocity is 160 m,