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(C) Let the motion be divided into three parts: $A$ to $B$,$B$ to $C$,and $C$ to $D$.For part $A$ to $B$: Initial velocity $u = 0$,acceleration $a = \

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$d = \frac{1}{72}\, \alpha \, t^2$

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(C) Let the motion be divided into three parts: $A$ to $B$,$B$ to $C$,and $C$ to $D$.For part $A$ to $B$: Initial velocity $u = 0$,acceleration $a = \alpha$,distance $d_1 = d$. Using $v^2 = u^2 + 2ad$,we get $v^2 = 2\alpha d$.For part $C$ to $D$: Final velocity $v' = 0$,acceleration $a = -\alpha/2$,distance $d_3$. Using $v'^2 = v^2 + 2ad_3$,we get $0 = v^2 - 2(\alpha/2)d_3$,which implies $v^2 = \alpha d_3$. Since $v^2 = 2\alpha d$,we have $2\alpha d = \alpha d_3$,so $d_3 = 2d$.Given total distance $d_1 + d_2 + d_3 = 15d$. Substituting $d_1 = d$ and $d_3 = 2d$,we get $d + d_2 + 2d = 15d$,which gives $d_2 = 12d$.For part $B$ to $C$: The car moves with constant speed $v$ for time $t$. So,$d_2 = v \cdot t$,which means $12d = vt$,or $v = 12d/t$.Substituting $v$ into the equation $v^2 = 2\alpha d$: $(12d/t)^2 = 2\alpha d$.$144d^2 / t^2 = 2\alpha d$.$d = (2\alpha t^2) / 144 = \frac{1}{72} \alpha t^2$.

$A$ car,starting from rest,accelerates at the rate $\alpha$ through a distance $d$,then continues at a constant speed for time $t$ and then decelerates at the rate of $\alpha/2$ to come to rest. If the total distance traveled is $15\, d$,then $d=$

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