Question

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 as the 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. $S = \frac{1}{72} ft^2$
• D. $S = \frac{1}{4} ft^2$

Answer 1

Answer: (C)

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

and slope of $B C=\frac{f}{2}$
$v=f t_{1}=\frac{f}{2} t_{2}$
$\therefore t_{2}=2 t_{1}$
In graph area of $\triangle O A D$ gives
distances, $ S=\frac{1}{2} f t_{1}^{2}$ ... (i)
Area of rectangle $A B E D$ gives distance travelled in time $t$.
$S_{2}=\left(f t_{1}\right) t$
Distance travelled in time $t_{2}$
$=S_{3}=\frac{1}{2} f_{2}\left(2 t_{1}\right)^{2}$
Thus, $ S_{1}+S_{2}+S_{3}=15 S$
$S+\left(f t_{1}\right) t+f t_{1}^{2}=15 S$
$S+\left(f t_{1}\right) t+2 S =15 S \left(S=\frac{1}{2} f t_{1}^{2}\right) $
$\left(f t_{1}\right) t =12 S $ ... (ii)
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}} $
$t_{1}=\frac{t}{6}$
From Eq. (i), we get
$\therefore S=\frac{1}{2} f\left(t_{1}\right)^{2} $
$S=\frac{1}{2} f\left(\frac{t}{6}\right)^{2}=\frac{1}{72} f t^{2}$