Question

a train is accelerating at a constant rate . it's speed is u when it s front end crosses a pole on the ground. it's speed becomes v when its rear end crosses the same pole. find it's speed when its mid point crosses the pole. pls help​

Answer 1

Given:

A train is accelerating at a constant rate . it's speed is u when it s front end crosses a pole on the ground. it's speed becomes v when its rear end crosses the same pole.

To find:

It's speed when its mid point crosses the pole ?

Calculation:

Let length of train be L .

Now , let acceleration be a .

$\therefore \: {v}^{2} = {u}^{2} + 2aL$

$\implies\: a = \dfrac{{v}^{2} - {u}^{2}}{ 2L} \: \: \: \: ......(1)$

Now , when midpoint of train passes the pole , the distance of train that passed the pole is L/2.

$\therefore \: {(v_{2})}^{2} = {u}^{2} + 2a (\dfrac{L}{2} )$

$\implies\: {(v_{2})}^{2} = {u}^{2} + 2 \bigg( \dfrac{ {v}^{2} - {u}^{2} }{2L} \bigg) \dfrac{L}{2}$

$\implies\: {(v_{2})}^{2} = {u}^{2} + \bigg( \dfrac{ {v}^{2} - {u}^{2} }{L} \bigg) \dfrac{L}{2}$

$\implies\: {(v_{2})}^{2} = {u}^{2} + \bigg( \dfrac{ {v}^{2} - {u}^{2} }{2} \bigg)$

$\implies\: {(v_{2})}^{2} = \dfrac{ {v}^{2} + {u}^{2} }{2}$

$\implies\: v_{2} = \sqrt{\dfrac{ {v}^{2} + {u}^{2} }{2} }$

So, velocity of train when midpoint passes the pole:

$\boxed{ \bold{\: v_{2} = \sqrt{\dfrac{ {v}^{2} + {u}^{2} }{2} }}}$