Kinematics Question 381

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 what maximum velocity must the truck attain to satisfy this condition?

Options:

A) $ \sqrt{La} $

B) $ \sqrt{2La} $

C) $ \sqrt{3La} $

D) $ \sqrt{5La} $

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Answer:

Correct Answer: A

Solution:

Let v be the maximum velocity attained and t the total time of journey. t’ it’s the duration of acceleration and retardation. Then $ v=0+at. $
$ \therefore \text{L=}\frac{1}{2}at{{’}^{2}}+v( t-2t’ )\text{+}\frac{1}{2}at{{’}^{2}} $

$ \text{=a}{{( \frac{v}{a} )}^{2}}\text{+v}( t-\frac{2v}{a} ) $

$ =\frac{{{v}^{2}}}{a}\text{+vt}-\frac{2{{v}^{2}}}{a}\text{=vt}-\frac{{{v}^{2}}}{a} $

$ \text{=t=}\frac{L}{v}\text{+}\frac{v}{a}\Rightarrow \frac{dv}{dt}=0\text{ }\therefore {{v} _{\max }}=\sqrt{La} $



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