Two particles P and Q are moving on circle. At a certain instant of time both the particles and diametrically opposite and P has tangential acceleration `8 m//s^(2)` and centripetal acceleration `5 m//s^(2)` whereas Q has only e centripetal acceleration of`1 m//s^(2)`. At that instant acceleration `("in" m//s^2)` of P with respect to Q is

Options

A. `14`
B. `sqrt(80)`
C. `10`
D. `12`

Answer

To find the acceleration of particle P with respect to particle Q, we need to consider the components of their accelerations.

### Step-by-Step Solution:

1. **Identify the Given Accelerations:**
- Particle P has:
- Tangential acceleration, \( a_{tP} = 8 \, \text{m/s}^2 \)
- Centripetal acceleration, \( a_{cP} = 5 \, \text{m/s}^2 \)
- Particle Q has:
- Centripetal acceleration, \( a_{cQ} = 1 \, \text{m/s}^2 \)
- Tangential acceleration, \( a_{tQ} = 0 \, \text{m/s}^2 \) (since it has only centripetal acceleration)

2. **Determine the Direction of Accelerations:**
- For particle P:
- Centripetal acceleration points towards the center of the circle (let's assume it points in the negative x-direction).
- Tangential acceleration points in the direction of motion (let's assume it points in the positive y-direction).
- For particle Q:
- Centripetal acceleration also points towards the center of the circle (which is in the positive x-direction, since Q is diametrically opposite to P).

3. **Express the Accelerations in Vector Form:**
- The acceleration vector of P can be expressed as:
\[ \vec{a_P} = -5 \hat{i} + 8 \hat{j} \]
- The acceleration vector of Q can be expressed as:
\[ \vec{a_Q} = -1 \hat{i} + 0 \hat{j} \]

4. **Calculate the Acceleration of P with Respect to Q:**
- The acceleration of P with respect to Q is given by:
\[ \vec{a_{PQ}} = \vec{a_P} - \vec{a_Q} \]
- Substituting the values:
\[ \vec{a_{PQ}} = (-5 \hat{i} + 8 \hat{j}) - (-1 \hat{i} + 0 \hat{j}) \]
\[ = (-5 + 1) \hat{i} + (8 - 0) \hat{j} \]
\[ = -4 \hat{i} + 8 \hat{j} \]

5. **Find the Magnitude of the Relative Acceleration:**
- The magnitude of the acceleration vector \( \vec{a_{PQ}} \) is given by:
\[ |\vec{a_{PQ}}| = \sqrt{(-4)^2 + (8)^2} \]
\[ = \sqrt{16 + 64} \]
\[ = \sqrt{80} \]
\[ = 4\sqrt{5} \, \text{m/s}^2 \]

### Final Answer:
The acceleration of P with respect to Q is \( 4\sqrt{5} \, \text{m/s}^2 \).