Two objects are moving along the same straight line. They cross a point A With an acceleration a, 2a and velocity 2u, u at time `t = 0.` The distance moved by the object when one overtakes the

Options

A. `(6u^2)/a`
B. `(2u^2)/a`
C. `(4u^2)/a`
D. `(8u^2)/a`

Explanation

To solve the problem of two objects moving along the same straight line and determining the distance at which one overtakes the other, we can follow these steps:

### Step 1: Identify the initial conditions
- Object 1 has:
- Initial velocity \( u_1 = 2u \)
- Acceleration \( a_1 = a \)

- Object 2 has:
- Initial velocity \( u_2 = u \)
- Acceleration \( a_2 = 2a \)

### Step 2: Write the equations of motion for both objects
Using the equation of motion \( S = ut + \frac{1}{2}at^2 \):

- For Object 1:
\[ S_1 = u_1 t + \frac{1}{2} a_1 t^2 = 2u t + \frac{1}{2} a t^2 \]

- For Object 2:
\[ S_2 = u_2 t + \frac{1}{2} a_2 t^2 = u t + \frac{1}{2} (2a) t^2 = ut + at^2 \]

### Step 3: Set the distances equal to find the time of overtaking
Since the two objects will overtake each other at the same distance, we set \( S_1 = S_2 \):
\[ 2ut + \frac{1}{2} a t^2 = ut + at^2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 2ut - ut + \frac{1}{2} a t^2 - at^2 = 0 \]
\[ ut - \frac{1}{2} at^2 = 0 \]

### Step 5: Factor out common terms
Factoring out \( t \):
\[ t \left( u - \frac{1}{2} at \right) = 0 \]
This gives us two solutions: \( t = 0 \) (initial condition) or \( u - \frac{1}{2} at = 0 \).

### Step 6: Solve for time \( t \)
Setting the second factor to zero:
\[ u = \frac{1}{2} at \implies t = \frac{2u}{a} \]

### Step 7: Substitute \( t \) back to find the distance
Now we can substitute \( t = \frac{2u}{a} \) back into either distance equation. Let's use \( S_1 \):
\[ S_1 = 2u \left(\frac{2u}{a}\right) + \frac{1}{2} a \left(\frac{2u}{a}\right)^2 \]
\[ S_1 = \frac{4u^2}{a} + \frac{1}{2} a \cdot \frac{4u^2}{a^2} \]
\[ S_1 = \frac{4u^2}{a} + \frac{2u^2}{a} = \frac{6u^2}{a} \]

### Final Answer
The distance moved by the object when one overtakes the other is:
\[ \boxed{\frac{6u^2}{a}} \]