Two particles P and Q simultaneously start moving from point A with velocities `15 m//s` and `20 m//s` respectively. The two particles move with acceleration equal in magnitude but opposite in direction. When P overtakes Q at point B then its velocity is `30 m//s`, the velocity of Q at point B will be

Answer

To find the velocity of particle Q at point B when particle P overtakes it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Particle P starts with an initial velocity \( u_P = 15 \, \text{m/s} \).
- Particle Q starts with an initial velocity \( u_Q = 20 \, \text{m/s} \).
- Both particles have equal magnitudes of acceleration but in opposite directions. Let the acceleration of P be \( a \) and the acceleration of Q be \( -a \).

2. **Use the final velocity of particle P:**
- When particle P overtakes particle Q at point B, its final velocity is given as \( v_P = 30 \, \text{m/s} \).

3. **Apply the equation of motion for particle P:**
- We can use the equation of motion:
\[ v = u + at \]
- For particle P:
\[ 30 = 15 + at_P \]
- Rearranging gives:
\[ at_P = 30 - 15 = 15 \quad \text{(1)} \]

4. **Apply the equation of motion for particle Q:**
- For particle Q, we use the same equation of motion:
\[ v = u + at \]
- Since the acceleration is in the opposite direction, we have:
\[ v_Q = 20 - at_Q \]
- We need to find the time \( t_Q \) when P overtakes Q. Since both particles start at the same time and P overtakes Q, we can assume \( t_P = t_Q = t \).

5. **Substituting \( t \) into the equations:**
- From equation (1), we have \( at = 15 \). Therefore, we can express \( a \) as:
\[ a = \frac{15}{t} \quad \text{(2)} \]
- Substitute \( a \) from (2) into the equation for Q:
\[ v_Q = 20 - \left(\frac{15}{t}\right)t \]
- Simplifying gives:
\[ v_Q = 20 - 15 = 5 \, \text{m/s} \]

### Final Answer:
The velocity of particle Q at point B is \( 5 \, \text{m/s} \).