Question

A car starts from rest and travels with uniform acceleration "$\alpha$" for some time and then with uniform retardation "$\beta$" and comes to rest. The time of motion is "$t$". Find the maximum velocity attained by it.

• A. $\alpha \beta t/(\alpha /2+\beta )$
• B. $\alpha \beta t/(\alpha +\beta )$
• C. $\alpha \beta t/(\alpha +\beta /2)$
• D. $\alpha /2\beta t/(\alpha +\beta )$

Answer 1

The correct option is B $\alpha \beta t/(\alpha +\beta )$

let it accelerate for time $t_1$ and retard for time $t_2$

$t_1+t_2=t\dots \left(1\right)$

using $v=u+at$

initially $u=0$, $v_{final}=\alpha t_1$

finally $v=0,u=\alpha t_1$

$0=\alpha t_1-\beta t_2$

$\frac{t_2}{\alpha }=\frac{t_1}{\beta }\dots \left(2\right)$

using $\left(1\right)\&\left(2\right)$

$t_1=\frac{\beta t}{\alpha +\beta }$

maximum velocity is :$\alpha t_1$

$v_{max}=\frac{\alpha \beta t}{\alpha +\beta }$