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(C) Let the car start from point $A$ from rest and move up to point $B$ with acceleration $f$ over distance $S$.The velocity of the car at point $B$ i

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$S = \frac{1}{72}f{t^2}$

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(C) Let the car start from point $A$ from rest and move up to point $B$ with acceleration $f$ over distance $S$.The velocity of the car at point $B$ is $v = \sqrt{2fS}$ (using $v^2 = u^2 + 2as$).The car moves distance $BC = x$ with this constant velocity $v$ in time $t$,so $x = vt = \sqrt{2fS} \cdot t$ ... $(i)$.At point $C$,the car begins to decelerate at rate $a' = \frac{f}{2}$ until it comes to rest at point $D$. Let the distance $CD = y$.Using $v^2 = u^2 - 2a'y$,where final velocity is $0$: $0 = v^2 - 2(\frac{f}{2})y \implies y = \frac{v^2}{f} = \frac{2fS}{f} = 2S$ ... (ii).The total distance $AD = AB + BC + CD = S + x + 2S = 15S$.$3S + x = 15S \implies x = 12S$.Substituting $x = 12S$ into equation $(i)$: $12S = \sqrt{2fS} \cdot t$.Squaring both sides: $144S^2 = 2fS \cdot t^2$.Dividing by $2S$: $72S = f \cdot t^2 \implies S = \frac{1}{72}ft^2$.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

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

Similar Questions

Match the relations in Column-$I$ with the equations in Column-$II$ for uniformly accelerated motion.

Column-$I$ Column-$II$

$(1)$ Velocity-time relation $(a)$ $v = v_0 + at$

$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

$(c)$ $v^2 = v_0^2 + 2as$

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