Question

A car accelerates form rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$, the maximum velocity acquired by the car is given by

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

Answer 1

Answer: (A)

Let $t_{1}$ be the time during which the car accelerates at a rate $\alpha$.
The velocity at the end of time $t_{1}$ will be $v$ at $t_{1}=u+\alpha t_{1}=0+\alpha t_{1}=\alpha t_{1}$ $(\because u=0)$
The time during which the car decelerates is $t_{2}=t-t_{1}$.
For this time $t_{2^{\prime}}$ the initial velocity is $\alpha t_{1}$ and the final velocity is zero and the acceleration is $-\beta$ Therefore $0=\alpha t_{1}-\beta\left(t-t_{1}\right)$
which gives $t_{1}=\frac{\beta t}{\alpha+\beta} .$
Therefore, maximum velocity $=v$ at
$t_{1}=\alpha t_{1}=\frac{\alpha \beta t}{\alpha+\beta}$