A car starting from rest accelerates at the rate f through a...

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



And slope of $BC = \dfrac{f}{2}$

$v = f{t}_{1} = \dfrac{f}{2}{t}_{2}$

${t}_{2} = 2{t}_{1}$

In graph area of $∆ OAD$ gives

Distance, $S = \dfrac{1}{2}f{t}_{1}^{2} \ldots . \left(i\right)$

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} = \dfrac{1}{2}\dfrac{f}{2}{\left(2{t}_{1}\right)}^{2}$$<br /><br />Thus,${S}_{1} + {S}_{2} + {S}_{3} = 15S$<br /><br />$S + \left(f{t}_{1}\right)t + f{t}_{1}^{2} = 15S$<br /><br />$S + \left(f{t}_{1}\right)t + 2S = 15S \left(S = \dfrac{1}{2} f{t}_{1}^{2}\right)

\left(f{t}_{1}\right)t = 12S \ldots \left(ii\right)

From Eqs. (i) and (ii), we have

\dfrac{12S}{S} =\dfrac{\left(f{t}_{1}\right)t}{\dfrac{1}{2}\left(f{t}_{1}\right){t}_{1}}

$∴ {t}_{1} = \dfrac{t}{6}$$<br /><br />From Eq. (i), we get<br /><br />$∴ S = $\d$ dfrac{1}{2}f{ \left({t}_{1}\right) }^{2}$<br /><br />$∴ S = $\d$ dfrac{1}{2}f{ \left(\dfrac{t}{6}\right) }^{2} = \dfrac{1}{72} ft^{2}