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