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 will follow these steps:

### Step 1: Identify the accelerations of both particles
- 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, \( a_{cQ} = 1 \, \text{m/s}^2 \)
- (No tangential acceleration)

### Step 2: Represent the accelerations as vectors
- Since P and Q are diametrically opposite, we can represent their accelerations in a coordinate system.
- Let's assume:
- The tangential acceleration of P acts in the positive y-direction.
- The centripetal acceleration of P acts in the negative x-direction.
- The centripetal acceleration of Q acts in the positive x-direction.

Thus, we can write:
- Acceleration of P, \( \vec{a_P} = (-5 \hat{i} + 8 \hat{j}) \, \text{m/s}^2 \)
- Acceleration of Q, \( \vec{a_Q} = (1 \hat{i}) \, \text{m/s}^2 \)

### Step 3: Calculate the relative acceleration of P with respect to Q
The relative acceleration of P with respect to Q is given by:
\[ \vec{a_{PQ}} = \vec{a_P} - \vec{a_Q} \]
Substituting the values:
\[ \vec{a_{PQ}} = (-5 \hat{i} + 8 \hat{j}) - (1 \hat{i}) = (-5 - 1) \hat{i} + 8 \hat{j} = (-6 \hat{i} + 8 \hat{j}) \, \text{m/s}^2 \]

### Step 4: Calculate the magnitude of the relative acceleration
To find the magnitude of \( \vec{a_{PQ}} \), we use the Pythagorean theorem:
\[ |\vec{a_{PQ}}| = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s}^2 \]

### Final Answer
The acceleration of P with respect to Q is \( 10 \, \text{m/s}^2 \).

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