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. 1) S=ft
B. 2) `S=frac(1)(6)ft^(2)`
C. 3) `S=frac(1)(6)ft^(2)`
D. 4) `S=frac(1)(4)ft^(2)

Answer

C

Explanation

The velocity -time graph for the given situation can be drwan as below . Magnitudes of slope of OA= f

and Slope of `BC = (f)/(2)`
`n=ft_(1)=(f)/(t)t_(2)`
`t_(2)=2t_(1)`
In graph area of DOAD gives
distance , `S =(1)/(2)ft_(1)^(22)`
in time t. `=S_(3)=(ft_(1)t`
Distance travalled in the time t2
`=S_(3)= (1)/(2)f(2t_(1))^(2)`
THus , `S_(1)+S_(2)+S_(3)=15S`
`S+(ft_(1))t+f_(1)^(2)=15S`
`S+(ft_(1))t+2S=15S` `(S=(1)/(2)ft_(1)^(2))`
`(ft_(1))t=12S`
From Eqs(i) and (ii) we have
`(125)/(S) =(ft_(1)t)/((1)/(2)(ft_(1))t_(1)`
`therefore t_(1)=(t)/(6)`
From Eq (i) we get
`therefore S =(1)/(2)f(t_(1))^(2)`
`therefore S =(1)/(2f((t)/(6))^(2)=(1)/(72)ft^(2)`