Two particles start moving from the same point along the same straight line. The first moves with constant velocity `v` and the second with constant acceleration `a`. During the time that elapses before the sound catches the first, the greatest distance between the particle is.

Options

A. `(v^(2))/a`
B. `(v^(2))/(2a)`
C. `(2v^(2))/a`
D. `(v^(2))/(4a)`

Answer

To solve the problem step by step, we will analyze the motion of both particles and find the maximum distance between them before one catches up to the other.

### Step 1: Define the motion of both particles
- **Particle 1** moves with a constant velocity \( v \).
- **Particle 2** starts from rest and moves with a constant acceleration \( a \).

### Step 2: Write the equations of motion
- The position of Particle 1 after time \( t \) is given by:
\[ s_1 = vt \]
- The position of Particle 2 after time \( t \) is given by:
\[ s_2 = \frac{1}{2} a t^2 \]

### Step 3: Find the relative displacement
- The distance between the two particles at any time \( t \) is:
\[ s = s_1 - s_2 = vt - \frac{1}{2} a t^2 \]

### Step 4: Maximize the distance
- To find the maximum distance, differentiate \( s \) with respect to \( t \) and set the derivative equal to zero:
\[ \frac{ds}{dt} = v - at = 0 \]
- Solving for \( t \):
\[ at = v \implies t = \frac{v}{a} \]

### Step 5: Substitute \( t \) back into the distance equation
- Now substitute \( t = \frac{v}{a} \) back into the distance equation:
\[ s_{\text{max}} = v\left(\frac{v}{a}\right) - \frac{1}{2} a \left(\frac{v}{a}\right)^2 \]
- Simplifying this:
\[ s_{\text{max}} = \frac{v^2}{a} - \frac{1}{2} a \cdot \frac{v^2}{a^2} = \frac{v^2}{a} - \frac{v^2}{2a} = \frac{v^2}{2a} \]

### Step 6: Conclusion
- The maximum distance between the two particles before the second catches up to the first is:
\[ s_{\text{max}} = \frac{v^2}{2a} \]