Question

A car, starting from rest, is accelerated at a constant rate $\alpha$ until it attains a speed $v$. It is then retarded at a constant rate $\beta$ until it comes to rest. The average speed of the car during its entire journey is

• A. zero
• B. $\frac{\alpha v}{2\beta }$
• C. $\frac{\beta v}{2\alpha }$
• D. $\frac{v}{2}$

Answer 1

The correct option is D

$\frac{v}{2}$

The distance $s_1$ covered by the car during the time it is accelerated is given by $2\alpha s_1=v^2$, which gives $s_1=\frac{v^2}{2\alpha }$. The distance $s_2$ covered during the time the car is decelerated is similarly given by $s_2=\frac{v^2}{2\beta }$.
Therefore, the total distance covered is $s=s_1+s_2=\frac{v^2}{2}(\frac{1}{\alpha }+\frac{1}{\beta })........\left(i\right)$
If $t_1$ is the time of acceleration and $t_2$ that of deceleration, then $v=\alpha t_1=\beta t_2$
$\Rightarrow t_1=\frac{v}{\alpha }$ and $t_2=\frac{v}{\beta }$
Therefore, the total time taken is $t=t_1+t_2=v(\frac{1}{\alpha }+\frac{1}{\beta })........\left(ii\right)$
From (i) and (ii), the average speed of the car is given by
$\frac{Totaldistance}{Totaltime}=\frac{s}{t}=\frac{v}{2}$
Hence, the correct choice is (d).