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

Options

A. `sqrt3 m//s`
B. `2sqrt3 m//s`
C. `4 m//s`
D. `8 m//s`

Explanation

To solve the problem, we need to find the magnitude of the velocity of the particle after 1 second, given its initial velocity and the acceleration acting at an angle. Here’s the step-by-step solution:

### Step 1: Identify the Initial Conditions
- Initial velocity \( u = 2 \, \text{m/s} \)
- Acceleration \( a = 2 \, \text{m/s}^2 \)
- Angle of acceleration \( \theta = 60^\circ \)
- Time \( t = 1 \, \text{s} \)

### Step 2: Break Down the Initial Velocity
Since the initial velocity is given as a scalar, we assume it acts in the horizontal direction (x-direction):
- \( u_x = u = 2 \, \text{m/s} \)
- \( u_y = 0 \, \text{m/s} \) (since there is no vertical component initially)

### Step 3: Resolve the Acceleration into Components
The acceleration acts at an angle of \( 60^\circ \) to the initial velocity. We can resolve this acceleration into its x and y components:
- \( a_x = a \cos \theta = 2 \cos 60^\circ = 2 \times \frac{1}{2} = 1 \, \text{m/s}^2 \)
- \( a_y = a \sin \theta = 2 \sin 60^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m/s}^2 \)

### Step 4: Calculate the Final Velocity Components
Using the equations of motion, we calculate the final velocity components after 1 second:
- \( v_x = u_x + a_x t = 2 + 1 \times 1 = 3 \, \text{m/s} \)
- \( v_y = u_y + a_y t = 0 + \sqrt{3} \times 1 = \sqrt{3} \, \text{m/s} \)

### Step 5: Calculate the Magnitude of the Final Velocity
The magnitude of the resultant velocity \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \, \text{m/s} \]

### Final Answer
The magnitude of the velocity at the end of 1 second is \( 2\sqrt{3} \, \text{m/s} \).

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