Question

A car starting from rest accelerates at the rate $f$ through a distance $S,$ then continues at a constant speed for time $t$ and then decelerates at rate $f/2$ to come to rest. If the total distance travelled is $15 \, S,$ then

• A. $S=ft$
• B. $S=\frac{1}{6}ft^{2}$
• C. $S=\frac{1}{2} \, ft^{2}$
• D. $S=\frac{1}{72} \, ft^{2}$

Answer 1

Answer: (D)

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

And slope of $BC=\frac{f}{2}$
$v=f \, t_{1}=\frac{f}{2}t_{2}$
$t_{2}=2t_{1}$
In graph area of $\Delta OAD$ gives
Distance, $S=\frac{1}{2} \, f \, t_{1}^{2} \, \, \ldots .\left(i\right)$
Area of rectangle $ABED$ gives distance travelled in time $t.$
$S_{2}=\left(f \, t_{1}\right)t$
Distance travelled in time $t_{2}=$
$S_{3}=\frac{1}{2}\frac{f}{2}\left(2 t_{1}\right)^{2}$
Thus, $S_{1}+S_{2}+S_{3}=15 \, S$
$S+\left(f t_{1}\right)t+ft_{1}^{2}=15 \, S \, $
$S+\left(f t_{1}\right)t+2S=15 \, S \, \left(S = \frac{1}{2} \, f t_{1}^{2}\right)$
$\left(f t_{1}\right)t=12 \, S \, \, \ldots \left(i i\right)$
From Eqs. (i) and (ii), we have
$\frac{12 \, S}{S}=\frac{\left(f t_{1}\right) t}{\frac{1}{2} \left(f t_{1}\right) t_{1}}$
$\therefore t_{1}=\frac{t}{6}$
From Eq. (i), we get
$\therefore S=\frac{1}{2} \, f\left(t_{1}\right)^{2}$
$\therefore S=\frac{1}{2} \, f\left(\frac{t}{6}\right)^{2}=\frac{1}{72} \, ft^{2}$