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. If the total time of travel of the car is $t$, then the maximum velocity attained by the car is given by

• A. $\frac{\alpha \beta }{\alpha +\beta }t$
• B. $\frac{1}{2}\frac{\alpha \beta }{\alpha +\beta }{t}^2$
• C. $\frac{\alpha \beta }{\alpha -\beta }t$
• D. $\frac{1}{2}\frac{\alpha \beta }{\alpha -\beta }{t}^2$

Answer 1

Answer: (A)

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

let initial distance traveled by the car is $s_1,$ in time $t_1$, and velocity is $v_1$ then
$s_1=\frac{1}{2}\alpha t_1^2$
$v_1=\alpha t_1$
$v_1^2=2\alpha s_1$
let remaining distance traveled by the car is $s_2,$ in time $t_2$, and velocity is $v_2$ then

Now, $u_2=v_1$
So, $s_2=v_1{t}_2-\frac{1}{2}\beta t_2^2$
$v_2=0=\alpha t_1-\beta t_2\Rightarrow \frac{t_1}{t_2}=\frac{\beta }{\alpha }$
$t_1+t_2=t\Rightarrow t_2(\frac{\beta }{\alpha }+1)=t\Rightarrow t_2=\frac{\alpha t}{\beta +\alpha }$
So, $t_1=\left(\frac{\beta t}{\beta +\alpha }\right)$
Clearly maximum velocity attained is $v_1,$ so
$v_1=\alpha t_1=\left(\frac{\alpha \beta t}{\beta +\alpha }\right)$