A tain moving at `30 m//s` reduces its speed to `10 m//s` due to braking in a distance of 240m. Now the braking force is increased by `12(1)/(2)%`. Find the stoping distance in the new situation if the train is mobing with the same initial speed `30 m//s`

Options

A. `120 m`
B. `240 m`
C. `360 m`
D. `420 m`

Answer

B

Explanation

`u = 30 " " v = 10 '' " s= 240`
`a = (v^(2) - u^(2))/(2S) = -(5)/(3)`
Let the retardation be a' and distance
`a' = a + a (12(1)/(2))/(100) = (9a)/(8) = (9)/(8) xx (5)/(3) = (15)/(8)`
`v^(2) - u^(2) = 2as`
`0 = 30^(2) - 2 ((15)/(8)) S_(2) rArr S_(2) = 240`