A particle is moving with a constant acceleration `vec(a) = hat(i) - 2 hat(j) m//s^(2)` and instantaneous velocity `vec(v) = hat(i) + 2 hat(j) + 2 hat(k) m//s` then rate of change of speed of particle 1 second after this instant will be :

Options

A. `sqrt(5) m//s^(2)`
B. `sqrt(3) m//s^(2)`
C. `sqrt(10) m//s^(2)`
D. None of these

Answer

To solve the problem, we need to find the rate of change of speed of the particle 1 second after the given instant. We are provided with the acceleration vector and the instantaneous velocity vector.

### Step-by-Step Solution:

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

2. **Understand the Concept of Rate of Change of Speed:**
- The rate of change of speed of a particle is given by the magnitude of the acceleration vector. This is because speed is a scalar quantity and its rate of change is determined by the vector components of acceleration.

3. **Calculate the Magnitude of the Acceleration Vector:**
- The acceleration vector components are:
- \(a_x = 1 \, \text{m/s}^2\) (from \(\hat{i}\))
- \(a_y = -2 \, \text{m/s}^2\) (from \(-2 \hat{j}\))
- The magnitude of the acceleration vector \(|\vec{a}|\) can be calculated using the formula:
\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2} \]
- Substituting the values:
\[ |\vec{a}| = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \, \text{m/s}^2 \]

4. **Determine the Rate of Change of Speed After 1 Second:**
- Since the acceleration is constant, the rate of change of speed remains the same over time. Thus, the rate of change of speed 1 second after the instant is still \(\sqrt{5} \, \text{m/s}^2\).

5. **Final Answer:**
- The rate of change of speed of the particle 1 second after the instant is \(\sqrt{5} \, \text{m/s}^2\).

### Conclusion:
The correct option is \(\sqrt{5} \, \text{m/s}^2\).