The two ends of a train moving with constant acceleration pass a certain point with velocities u and 3u. The velocity with which the middle point of the train passes the same point is

Options

A. 2u
B. `(3)/(2)u`
C. `sqrt5u`
D. `sqrt(10)u`

Answer

To find the velocity with which the middle point of the train passes a certain point, we can follow these steps:

### Step 1: Understand the problem
We have a train moving with constant acceleration. The front end of the train passes a point with velocity \( u \), and the back end passes the same point with velocity \( 3u \). We need to find the velocity of the middle point of the train as it passes the same point.

### Step 2: Define the variables
Let:
- \( u \) = velocity of the front end of the train when it passes the point
- \( 3u \) = velocity of the back end of the train when it passes the point
- \( L \) = length of the train
- \( a \) = acceleration of the train
- \( v_m \) = velocity of the middle point of the train when it passes the point

### Step 3: Use the equations of motion
We can use the third equation of motion, which states:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = displacement

### Step 4: Apply the equation for the front end and back end
1. For the front end of the train:
- Initial velocity \( u \)
- Final velocity \( 3u \)
- Displacement \( L \)

Applying the equation:
\[ (3u)^2 = u^2 + 2aL \]
Simplifying:
\[ 9u^2 = u^2 + 2aL \implies 8u^2 = 2aL \implies a = \frac{4u^2}{L} \]

### Step 5: Find the velocity of the middle point
The middle point of the train will have an initial velocity of \( u \) and will travel a distance of \( \frac{L}{2} \) to reach the same point. We can use the same equation of motion:
\[ v_m^2 = u^2 + 2a\left(\frac{L}{2}\right) \]
Substituting \( a = \frac{4u^2}{L} \):
\[ v_m^2 = u^2 + 2\left(\frac{4u^2}{L}\right)\left(\frac{L}{2}\right) \]
This simplifies to:
\[ v_m^2 = u^2 + 4u^2 = 5u^2 \]
Taking the square root:
\[ v_m = \sqrt{5}u \]

### Final Answer
The velocity with which the middle point of the train passes the same point is \( \sqrt{5}u \).

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