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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 $d=$
$d = \frac{1}{2}\, \alpha \, t^2$
$d = \frac{1}{4}\, \alpha \, t^2$
$d = \frac{1}{72}\, \alpha \, t^2$
$d = \frac{1}{6}\, \alpha \, t^2$
Solution
$A$ to $\mathrm{B}: \mathrm{v}^{2}=2 \alpha \mathrm{d}_{1}=2 \alpha \mathrm{d}$
$C$ to $D: 0=v^{2}-2 \frac{\alpha}{2} d_{3} \Rightarrow d_{3}=2 d$
${\mathrm{d}_{3}=2 \mathrm{d}}$
${\mathrm{d}_{1}+\mathrm{d}_{2}+\mathrm{d}_{3}=15 \mathrm{d} \Rightarrow \mathrm{d}_{2}=12 \mathrm{d}}$
$\mathrm{B}$ to $\mathrm{C}: \mathrm{d}_{2}=12 \mathrm{d}=\mathrm{vt} \Rightarrow \mathrm{v}=\frac{12 \mathrm{d}}{\mathrm{t}}$
$A$ to $\mathrm{B}: \mathrm{v}^{2}=2 \alpha \mathrm{d} \Rightarrow\left(\frac{12 \mathrm{d}}{\mathrm{t}}\right)^{2}=2 \alpha \mathrm{d}$
${\frac{144 \mathrm{d}^{2}}{\mathrm{t}^{2}}=2 \alpha \mathrm{d}}$
${\mathrm{d}=\frac{1}{72} \alpha \mathrm{t}^{2}}$
