A car starting from rest accelerates at the rate $f$ through a distance $S$, then continues at constant…

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 $15 \mathrm{~S}$, then
  1. $S=f t$
  2. $\mathrm{S}=1 / 6 \mathrm{ft}^2$
  3. $\mathrm{S}=1 / 2 \mathrm{ft}^2$
  4. None of these

Solution


$ \begin{aligned} & \mathrm{S}=\frac{\mathrm{ft}_1^2}{2} \\ & \mathrm{v}_0=\sqrt{2 \mathrm{Sf}} \end{aligned} $ During retardation $ S_2=2 S $ During constant velocity $ \begin{aligned} & 15 \mathrm{~S}-3 \mathrm{~S}=12 \mathrm{~S}=\mathrm{v}_0 \mathrm{t} \\ & \Rightarrow \mathrm{S}=\frac{\mathrm{ft}^2}{72} \end{aligned} $

Asked in: JEE Main 2005

Practice more Motion In One Dimension questions on Aicharya