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 traversed is 5 S, then :

Options

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

Answer

c

Explanation

The velocity time graph of motion is shown below:

For same change in velocity .f.t t=constant.
`:. (1)/(2)ft_(1)^(2)+ft_(1)^(t)+(1)/(2)((f)/(2)).(2t_(1))^(2)=15 S`
`S+ft_(1) +2S=15 S`
`ft_(1)t=12 S`
Now `S=(1)/(2)ft_(1)^(2)=(1)/(2)f((t^(2))/(36))=(ft^(2))/(72)`