Q. A car accelerates from rest at a constant rate $a$ for some time, after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t,$ then the maximum velocity acquired by the car is
A) $\left( \begin{array}{r}
\frac{\alpha t + \beta^{2}}{a\beta} \\
\
\end{array} \right)t$
B) $\left( \begin{array}{r}
\frac{\alpha^{2} - \beta^{2}}{a\beta} \\
\
\end{array} \right)t$
C) $\frac{(\alpha + \beta)t}{\alpha\beta}$
D) $\frac{\alpha\beta t}{\alpha + \beta}$
Explanation
(d) Let the car accelerate at rate $\alpha\ $for time $t_{1}$ then maximum velocity attained,
$v = 0 + at_{1} = at_{1}$
Now, the car decelerates at a rate $\beta\ $for time $(t - t_{1})$ and finally comes to rest. Then,
$0 = v - \beta\left( t - t_{1} \right) \Rightarrow 0 = \alpha t_{1} - \beta t + \beta t_{1}$
$\Rightarrow t_{1} = \frac{\beta}{\alpha + \beta}t$
$\therefore v = \frac{\alpha\beta}{\alpha + \beta}t$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.