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

Velocity of car `(u)=20 ms^(-1)`
From Newton third equation,
`v^(2)-u^(2)=2as`
`(0)^(2)-(20)^(2)=2(a)xx40`
`400=2xx40xxa`
`a=(400)/(2xx40)`
`a=5 ms^(-2)`
In the second condition, the velocity becomes twive i.e., `u^(1)=2u`
Again from Newton.s third equation, we get
`(0)^(2)-(2u)^(2)=2xx(5)xx s`
`s=((2u)^(2))/(2xx5)`
`s=(4u^(2))/(2xx5)`
`s=(4xx20xx20)/(2xx5)`
`s=(4xx20xx20)/(2xx5)`
s = 160 m.