A body has an initial velocity of `3 ms^-1` and has an acceleration of `1 ms^-2` normal to the direction of the initial velocity. Then its velocity `4 s` after the start is.

Options

A. `7 m s^-1` along the direction of initial velocity.
B. `7 m s^-1` along the normal to the direction of initial velocity.
C. `7 m s^-1` midway between the two directions.
D. `5 m s^-1` at an angle `tan^-1(4//3)` with the direction of initial velocity.

Explanation

To solve the problem, we need to determine the final velocity of a body after 4 seconds, given its initial velocity and acceleration. The initial velocity is \(3 \, \text{m/s}\) and the acceleration is \(1 \, \text{m/s}^2\) acting perpendicular to the initial velocity.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Initial velocity, \( \mathbf{u} = 3 \, \text{m/s} \) (along the x-axis).
- Acceleration, \( \mathbf{a} = 1 \, \text{m/s}^2 \) (along the y-axis).

2. **Represent the Initial Velocity and Acceleration as Vectors:**
- We can represent the initial velocity as \( \mathbf{u} = 3 \hat{i} + 0 \hat{j} \).
- The acceleration can be represented as \( \mathbf{a} = 0 \hat{i} + 1 \hat{j} \).

3. **Use the Formula for Final Velocity:**
- The formula to find the final velocity \( \mathbf{v} \) after time \( t \) is:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a} \cdot t \]
- Here, \( t = 4 \, \text{s} \).

4. **Calculate the Final Velocity Vector:**
- Substitute the values into the equation:
\[ \mathbf{v} = (3 \hat{i} + 0 \hat{j}) + (0 \hat{i} + 1 \hat{j}) \cdot 4 \]
- This simplifies to:
\[ \mathbf{v} = 3 \hat{i} + 4 \hat{j} \]

5. **Calculate the Magnitude of the Final Velocity:**
- The magnitude of the velocity vector \( \mathbf{v} \) is given by:
\[ |\mathbf{v}| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]

6. **Determine the Direction of the Final Velocity:**
- The angle \( \theta \) with respect to the x-axis can be found using:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3} \]
- Thus, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).

### Final Answer:
The final velocity of the body after 4 seconds is \( 5 \, \text{m/s} \) at an angle of \( \tan^{-1}\left(\frac{4}{3}\right) \) with respect to the initial velocity direction.