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} $
Show Answer
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} $
