A particle has initial velocity `4i + 4j ms^(-1)` and an acceleration `-0.4i ms^(-2)`, at what time will its speed bbe `5ms^(-1)`?

Options

A. 2.5 s
B. 17.5 s
C. s
D. 8.5 s

Explanation

To solve the problem, we need to find the time at which the speed of the particle becomes \(5 \, \text{ms}^{-1}\). We start with the given initial velocity and acceleration.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Initial velocity, \( \mathbf{u} = 4\mathbf{i} + 4\mathbf{j} \, \text{ms}^{-1} \)
- Acceleration, \( \mathbf{a} = -0.4\mathbf{i} \, \text{ms}^{-2} \)

2. **Write the Expression for Final Velocity**:
The final velocity \( \mathbf{v} \) after time \( t \) can be expressed as:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a}t \]
Substituting the values:
\[ \mathbf{v} = (4\mathbf{i} + 4\mathbf{j}) + (-0.4\mathbf{i})t \]
This simplifies to:
\[ \mathbf{v} = (4 - 0.4t)\mathbf{i} + 4\mathbf{j} \]

3. **Calculate the Magnitude of the Velocity**:
The magnitude of the velocity \( |\mathbf{v}| \) is given by:
\[ |\mathbf{v}| = \sqrt{(4 - 0.4t)^2 + (4)^2} \]
We want this magnitude to equal \( 5 \, \text{ms}^{-1} \):
\[ \sqrt{(4 - 0.4t)^2 + 16} = 5 \]

4. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ (4 - 0.4t)^2 + 16 = 25 \]

5. **Simplify the Equation**:
Rearranging the equation:
\[ (4 - 0.4t)^2 = 25 - 16 \]
\[ (4 - 0.4t)^2 = 9 \]

6. **Take the Square Root**:
Taking the square root of both sides results in:
\[ 4 - 0.4t = 3 \quad \text{or} \quad 4 - 0.4t = -3 \]

7. **Solve for \( t \)**:
- For \( 4 - 0.4t = 3 \):
\[ 0.4t = 1 \implies t = \frac{1}{0.4} = 2.5 \, \text{s} \]
- For \( 4 - 0.4t = -3 \):
\[ 0.4t = 7 \implies t = \frac{7}{0.4} = 17.5 \, \text{s} \]

8. **Conclusion**:
The times at which the speed of the particle will be \( 5 \, \text{ms}^{-1} \) are \( t = 2.5 \, \text{s} \) and \( t = 17.5 \, \text{s} \).

### Final Answer:
The particle will have a speed of \( 5 \, \text{ms}^{-1} \) at \( t = 2.5 \, \text{s} \) and \( t = 17.5 \, \text{s} \).

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