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A car , starting from rest, accelerates ...

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 `15 S` , then

A

`S=(1)/(2)ft^2`

B

`S=(1)/(4)ft^2`

C

`S=(1)/(72)ft^2`

D

`S=(1)/(6)ft^2`

Text Solution

Verified by Experts

The correct Answer is:
C
Let car starts from point A from rest and moves up to point B with acceleration f
Velocity of car at point B, `v=sqrt(2fS)`
Car moves distance BC with this constant velocity in time t
`x=sqrt(2fS).t`
So the velocity of car at point C also will be `sqrt(2fs)` and finally car stops after covering distance y.
Distance `CDimplies=y=((sqrt(2fs))^2)/(2((f)/(2)))=(2fs)/(f)=2S`
So the total distance `AD=AB+BC+CD=15S` (given)
`impliesS+x+2S=15Simpliesx=12S`
Substituting the value of x in equation (i) we get
`x=sqrt(2fs).timplies12S=sqrt(2fs.t)implies144S^2=2fS.t^2`
`impliesS=(1)/(72)ft^2`
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