A car, starting from rest, accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate f/ 2 to come to rest. If the total distance travelled is 15 s, then

Options

A. `S = 1/2 ft^2`
B. `S = 1/4 ft^2`
C. `S = 1/72 ft^2`
D. `S = 1/6 ft^2`

Answer

C

Explanation

When accelerate
`S =1/2 v_(max) t_1 = 1/2 (v_(max)^2)/(f) rArr v_(max) = sqrt(2fS)`

When decelerate
`Y = 1/2 v_(max) t_2 = 1/2 (v_(max)^2 xx 2)/(f) = 2S`
Total distances = Area under `v -t graph = 15S
`rArr S xx x + 2S = 15S`
`rArr S + v_(max) t + 2S = 15S`
`rArr S + sqrt(2fS) t + 2S = 15S`
`rArr 2fS t^2 = 144 S^2 rArr S = 1/72 ft^2 `