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 travell

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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

TestHub · Accepted Answer

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

And slope of $B C = \frac{f}{2}$

$v = f t_{1} = \frac{f}{2} t_{2}$

$t_{2} = 2 t_{1}$

In graph area of $∆ O A D$ gives

Distance, $S = \frac{1}{2} f t_{1}^{2} \dots . (i)$

Area of rectangle $A B E D$ gives distance travelled in time $t .$

$S_{2} = (f t_{1}) t$

Distance travelled in time $t_{2} =$

$S_{3} = \frac{1}{2} \frac{f}{2} {(2 t_{1})}^{2}$

Thus, $S_{1} + S_{2} + S_{3} = 15 S$

$S + (f t_{1}) t + f t_{1}^{2} = 15 S$

$S + (f t_{1}) t + 2 S = 15 S (S = \frac{1}{2} f t_{1}^{2})$

$(f t_{1}) t = 12 S \dots (i i)$

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

$\frac{12 S}{S} = \frac{(f t_{1}) t}{\frac{1}{2} (f t_{1}) t_{1}}$

$∴ t_{1} = \frac{t}{6}$

From Eq. (i), we get

$∴ S = \frac{1}{2} f {(t_{1})}^{2}$

$∴ S = \frac{1}{2} f {(\frac{t}{6})}^{2} = \frac{1}{72} f t^{2}$

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