Explanation

(d) The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of $OA = f$
[image:3_image202.emf]
And slope of $BC = \frac{f}{2}$
$v = f\ t_{1} = \frac{f}{2}t_{2}$
$t_{2} = 2t_{1}$
In graph area of $\mathrm{\Delta}OAD$ gives
Distance, $S = \frac{1}{2}\ f\ t_{1}^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Area of rectangle $ABED$ gives distance travelled in time $t.$
$S_{2} = \left( f\ t_{1} \right)t$
Distance travelled in time $t_{2} =$
$S_{3} = \frac{1}{2}\frac{f}{2}\left( 2t_{1} \right)^{2}$
Thus, $S_{1} + S_{2} + S_{3} = 15\ S$
$S + \left( ft_{1} \right)t + ft_{1}^{2} = 15\ S\ $
$S + \left( ft_{1} \right)t + 2S = 15\ S\ \ \ \ \ \ \ \ \ \ \left( S = \frac{1}{2}\ ft_{1}^{2} \right)$
$\left( ft_{1} \right)t = 12\ S\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we have
$\frac{12\ S}{S} = \frac{\left( ft_{1} \right)t}{\frac{1}{2}\left( ft_{1} \right)t_{1}}$
$\therefore t_{1} = \frac{t}{6}$
From Eq. (i), we get
$\therefore S = \frac{1}{2}\ f\left( t_{1} \right)^{2}$
$\therefore S = \frac{1}{2}\ f\left( \frac{t}{6} \right)^{2} = \frac{1}{72}\ ft^{2}$