Velocity of particle at time t=0 is `2ms^(-1)`.A constant acceleration of `2ms^(-2)` act on the particle for 1 second at an angle of `60^(@)` with its initial velocity .Find the magnitude velocity and displacement of the particle at the end of t=1s.

Options

A. `sqrt(3) m//s`
B. `2sqrt(3) m//s`
C. `4 m//s`
D. `8 m//s`

Answer

To solve the problem, we need to find the magnitude of the velocity and the displacement of the particle at the end of \( t = 1 \) second, given that the initial velocity is \( 2 \, \text{m/s} \) and a constant acceleration of \( 2 \, \text{m/s}^2 \) acts at an angle of \( 60^\circ \) with the initial velocity.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Initial velocity \( \mathbf{u} = 2 \, \text{m/s} \)
- Acceleration \( \mathbf{a} = 2 \, \text{m/s}^2 \)
- Angle \( \theta = 60^\circ \)
- Time \( t = 1 \, \text{s} \)

2. **Resolve the acceleration into components:**
- Acceleration in the x-direction:
\[ a_x = a \cos(60^\circ) = 2 \cdot \frac{1}{2} = 1 \, \text{m/s}^2 \]
- Acceleration in the y-direction:
\[ a_y = a \sin(60^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m/s}^2 \]

3. **Calculate the final velocity in the x-direction:**
- The initial velocity in the x-direction \( u_x = 2 \, \text{m/s} \)
- Using the equation:
\[ v_x = u_x + a_x t = 2 + 1 \cdot 1 = 3 \, \text{m/s} \]

4. **Calculate the final velocity in the y-direction:**
- The initial velocity in the y-direction \( u_y = 0 \, \text{m/s} \)
- Using the equation:
\[ v_y = u_y + a_y t = 0 + \sqrt{3} \cdot 1 = \sqrt{3} \, \text{m/s} \]

5. **Calculate the magnitude of the resultant velocity:**
- The magnitude of the velocity \( v \) is given by:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \, \text{m/s} \]

6. **Calculate the displacement in the x-direction:**
- Using the formula for displacement:
\[ s_x = u_x t + \frac{1}{2} a_x t^2 = 2 \cdot 1 + \frac{1}{2} \cdot 1 \cdot 1^2 = 2 + \frac{1}{2} = 2.5 \, \text{m} \]

7. **Calculate the displacement in the y-direction:**
- Using the formula for displacement:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2} \cdot \sqrt{3} \cdot 1^2 = \frac{\sqrt{3}}{2} \, \text{m} \]

8. **Calculate the magnitude of the resultant displacement:**
- The magnitude of the displacement \( s \) is given by:
\[ s = \sqrt{s_x^2 + s_y^2} = \sqrt{(2.5)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{6.25 + \frac{3}{4}} = \sqrt{6.25 + 0.75} = \sqrt{7} \, \text{m} \]

### Final Answers:
- Magnitude of velocity at \( t = 1 \, \text{s} \): \( 2\sqrt{3} \, \text{m/s} \)
- Displacement at \( t = 1 \, \text{s} \): \( \sqrt{7} \, \text{m} \)