Question

An engine of a train, moving with uniform acceleration, passes the signal post with velocity $u$ and the last compartment with velocity $v$. The velocity with which middle point of the train passes the signal post is:

• A. $\sqrt{\frac{v^2-u^2}{2}}$
• B. $\frac{v-u}{2}$
• C. $\sqrt{\frac{v^2+u^2}{2}}$
• D. $\frac{v+u}{2}$

Answer 1

The correct option is C

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

Step 1: Given data and assumptions.

Uniform acceleration$=a$

Initial velocity passes the signal pole by the engine$=u$

The final velocity passes the signal pole by the last compartment$=v$

Let the length of the train be $l$

And, velocity passes the signal pole by the middle point of the train$=v_m$

Step 2: Finding the velocity past the signal pole by the middle point of the train.

Formula used:

$v^2-u^2=2as$

Where $s$ is distance, $v$ is final velocity, $u$ is the initial velocity.

Now,

When the last compartment passes through the pole distance covered by the engine is $l$ then,

$v^2-u^2=2al$

$\Rightarrow$$a=\frac{v^2-u^2}{2l}$ …..$\left(i\right)$

When the middle point of the train passes the pole, the distance covered by the engine is $\raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ then,

$v_m^2-u^2=2a\frac{l}{2}$

Where $v_m$ is velocity at the middle point of the train

$\Rightarrow$$a=\frac{v_m^2-u^2}{l}$ ….$\left(ii\right)$

From the equation $\left(i\right)$ and $\left(ii\right)$ we get.

$\frac{v^2-u^2}{2l}=\frac{v_m^2-u^2}{l}$

$\Rightarrow$$v_m^2=\frac{v^2-u^2}{2}+u^2$

$\Rightarrow$$v_m=\sqrt{\frac{v^2+u^2}{2}}$

Hence, option C is correct.