Question

A truck has to carry a load in the shortest time from one station to another station situated at a distance $L$ from the first. It can start up or slowdown at the same acceleration or deceleration $a$. What maximum velocity must the truck attain to satisfy this condition?

• A. $\sqrt{2La}$
• B. $\sqrt{5La}$
• C. $\sqrt{La}$
• D. $\sqrt{3La}$

Answer 1

The correct option is C $\sqrt{La}$

Let $v$ be the maximum velocity attained and $t$ be the total time of journey.
Also, let $t_1$ is the duration of acceleration and deceleration. Then,
$v=0+a{t}_1$
As per the question
$L=\frac{1}{2}a{t}_1^2+v(t-2t_1)+\frac{1}{2}a{t}_1^2$
$\Rightarrow L=a{\left(\frac{v}{a}\right)}^2+v(t-\frac{2v}{a})$
$\Rightarrow L=\frac{v^2}{a}+vt-\frac{2v^2}{a}=vt-\frac{v^2}{a}$
$\Rightarrow t=\frac{L}{v}+\frac{v}{a}$
or, $\Rightarrow \frac{dt}{dv}=-\frac{L}{v^2}+\frac{1}{a}$
As we know that when $t$ is minimum, $\frac{dt}{dv}=0$
$\Rightarrow v_{max}=\sqrt{La}$