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

The Correct Option is D

Let maximum velocity $= v$ Now, $v = 0 + \alpha t _{1}$
Similarly, $0= v - \beta t _{ 2 }$
From the above equations we get,
$t _{1}=\frac{ v }{ \alpha } \quad \& \quad t _{2}=\frac{ v }{ \beta }$
$t _{1}+ t _{ 2 }= t =\frac{ v }{ \alpha }+\frac{ v }{ \beta }$
$\Rightarrow v =\frac{\alpha \beta}{ \alpha + \beta } t$