Question

A train is travelling with uniform acceleration 'a' along a straight line. The front end of the train crosses a tree, next to
the track, with a speed u and the back end of the train crosses same tree with a speed v. The speed of the train when
the midpoint of the train crosses the tree is

Answer 1

Answer:

I have taken v2 as v while v1 as u.

Velocity of mid point = v1

We all know that,

v2-u2=2as or v2=u2+2as -----> (1)

a=v2-u2/2s

Now for the mid-point of the train

s1=s/2

a1=v12-u2/2s1

a1=v12-u2/2s/2

a1=v12-u2/s or v12=u2 + as ----> (2)

from (1) and (2) you will get

as= v2-u2/2

v12=u2 + as

v12= v2+u2 /2

V=under root v2+u2/2.

Answer 2

The speed of the train when  the midpoint of the train crosses the tree is

$\boxed{\sqrt{\frac{u^2+v^2}{2}}}$

Explanation:

Given

The acceleration of train is $a$ and the velocity of the front end of the train is  $u$ and the back end is  $v$

To find out

The speed of the train when  the midpoint of the train crosses the tree

Solution

Let the length of the train is $d$

In crossing $d$ the speed changes from $u$ to $v$ with an acceleration $a$

Therefore, from the third equation of motion

$v^2=u^2+2ad$

$\implies d=\frac{v^2-u^2}{2a}$

The length of the front end to the midpoint will be $d/2$

Let the velocity of the train be v' when the mid-point crosses the tree

Therefore,

Again from third equation of motion

$v'^2=u^2+2a\frac{d}{2}$

$\implies v'^2=u^2+2a\times\frac{v^2-u^2}{4a}$

$\implies v'^2=\frac{4au^2+2av^2-2au^2}{4a}$

$\implies v'^2=\frac{2a(u^2+v^2)}{4a}$

$\implies v'^2=\frac{u^2+v^2}{2}$

$\implies v'=\sqrt{\frac{u^2+v^2}{2}}$

Hope this answer is helpful.

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