Question

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ and comes to rest. If total time elapsed is $t$, then maximum velocity acquired by car will be

• A. $\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) t$
• B. $\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right) t$
• C. $\left(\frac{\alpha+\beta}{\alpha \beta}\right) t$
• D. $\left(\frac{\alpha \beta}{\alpha+\beta}\right) t$

Answer 1

Answer: (D)

Initial velocity $(u)=0$; Acceleration in the first phase $=\alpha$;
Deceleration in the second phase $=\beta$ and total time $=t$.
When car is accelerating then
final velocity $(v)=u+\alpha t=0+\alpha t_{1}$
or $t_{1}=\frac{v}{\alpha}$ and when car is decelerating,
then final velocity $0=v-\beta t$ or $t_{2}=\frac{v}{\beta}$.
Therefore total time $(t)=t_{1}+t_{2}=\frac{v}{\alpha}+\frac{v}{\beta}$
$t=v\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=v\left(\frac{\beta+\alpha}{\alpha \beta}\right)$
or $v=\frac{\alpha \beta t}{\alpha+\beta}$.