A particle is moving with a constant acceleration `veca = hati-2hatj+2hatk m//s^(2)` and instantaneous velocity `vecv = hati+2hatj+2hatk m//s`, then rate of change of speed of particle 2 second after this instant will be :

Options

A. `(19)/(3)m//s^(2)`
B. `3m//s^(2)`
C. `(19)/(7)m//s^(2)`
D. None of these

Explanation

To solve the problem, we need to determine the rate of change of speed of the particle 2 seconds after the given instant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Acceleration vector: \(\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \, \text{m/s}^2\)
- Initial velocity vector: \(\vec{v} = \hat{i} + 2\hat{j} + 2\hat{k} \, \text{m/s}\)

2. **Understand Rate of Change of Speed:**
- The rate of change of speed is essentially the acceleration. Since the acceleration is constant, it will not change over time.

3. **Calculate the Speed:**
- The speed of the particle is the magnitude of the velocity vector:
\[ |\vec{v}| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \, \text{m/s} \]

4. **Acceleration is Constant:**
- Given that the acceleration is constant, it does not change with time. Therefore, the acceleration at any time, including 2 seconds later, remains:
\[ \vec{a} = \hat{i} - 2\hat{j} + 2\hat{k} \, \text{m/s}^2 \]

5. **Conclusion:**
- Since the acceleration is constant, the rate of change of speed after 2 seconds will still be equal to the magnitude of the acceleration vector:
\[ |\vec{a}| = \sqrt{(1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \, \text{m/s}^2 \]

### Final Answer:
The rate of change of speed of the particle 2 seconds after this instant will be \(3 \, \text{m/s}^2\).

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