Question

If the engine of a long train moving with constant acleration crosses a tree with velocity 'u' and the last compartment of the train crosses the same tree with velocity 'v', then the velocity with which the middle compartment crosses the same tree is

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

Hint:

Use of $v^2 = u^2 + 2as$ in train problems:

• Consider the train as a rigid body with constant acceleration.
• Apply motion equation to segments (e.g., front to middle, front to back).

Answer 1

The Correct Option is C

From kinematics: \[ v^2 = u^2 + 2aL \Rightarrow aL = \frac{v^2 - u^2}{2} \] For middle point (distance $L/2$): \[ v_m^2 = u^2 + 2a \cdot \frac{L}{2} = u^2 + aL \] Substitute $aL$: \[ v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{v^2 + u^2}{2} \Rightarrow v_m = \sqrt{\frac{v^2 + u^2}{2}} \]

Answer 2

Step 1: Understand the problem
A train is moving with constant acceleration. When the engine passes a tree, its velocity is \(u\), and when the last compartment passes the same tree, its velocity is \(v\). We need to find the velocity of the middle compartment as it crosses the tree.

Step 2: Use the concept of constant acceleration
Since the train moves with constant acceleration \(a\), velocity varies uniformly along the length of the train.
- Let the train length be \(L\).
- The time difference between the engine crossing and last compartment crossing the tree is the time taken by the train to move length \(L\).

Step 3: Velocity relation at any point
Velocity changes uniformly from \(u\) to \(v\) along the length.
Velocity squared relation under constant acceleration:
\[ v^2 = u^2 + 2aL \]
For the middle compartment (at distance \(L/2\) from the engine), velocity squared is:
\[ v_{\text{middle}}^2 = u^2 + 2a \cdot \frac{L}{2} = u^2 + aL \]

Step 4: Express \(aL\) in terms of \(u\) and \(v\)
From step 3:
\[ v^2 = u^2 + 2aL \implies aL = \frac{v^2 - u^2}{2} \]

Step 5: Calculate velocity of middle compartment
\[ v_{\text{middle}}^2 = u^2 + aL = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2} \]
Therefore,
\[ v_{\text{middle}} = \sqrt{\frac{u^2 + v^2}{2}} \]

Step 6: Conclusion
The velocity with which the middle compartment crosses the tree is \(\sqrt{\frac{u^2 + v^2}{2}}\).