Power supplied to a mass `2 kg` varies with time as `P = (3t^(2))/(2)` watt. Here `t` is in second . If velocity of particle at `t = 0 is v = 0`, the velocity of particle at time `t = 2s` will be:

Options

A. `1 ms^(-1)`
B. `4 ms^(-1)`
C. `2 ms^(-1)`
D. `2sqrt2 ms^(-1)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Mass (m) = 2 kg
- Power (P) as a function of time (t): \( P = \frac{3t^2}{2} \) watts
- Initial velocity at \( t = 0 \): \( v = 0 \)

### Step 2: Calculate the work done (W)
The work done can be calculated by integrating the power over the time interval from 0 to 2 seconds.

\[ W = \int_{0}^{2} P \, dt = \int_{0}^{2} \frac{3t^2}{2} \, dt \]

### Step 3: Solve the integral
We can take the constant \( \frac{3}{2} \) out of the integral:

\[ W = \frac{3}{2} \int_{0}^{2} t^2 \, dt \]

Now, we compute the integral \( \int t^2 \, dt \):

\[ \int t^2 \, dt = \frac{t^3}{3} \]

Evaluating this from 0 to 2:

\[ \int_{0}^{2} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]

### Step 4: Substitute back to find work done
Now substituting back into the work done equation:

\[ W = \frac{3}{2} \cdot \frac{8}{3} = \frac{24}{6} = 4 \, \text{joules} \]

### Step 5: Relate work done to kinetic energy
The work done on the particle is equal to the change in kinetic energy. Since the initial velocity is 0, the work done is equal to the kinetic energy at \( t = 2 \):

\[ W = \Delta KE = \frac{1}{2} m v^2 \]

### Step 6: Solve for velocity (v)
Substituting the known values into the kinetic energy equation:

\[ 4 = \frac{1}{2} \cdot 2 \cdot v^2 \]

This simplifies to:

\[ 4 = 1 \cdot v^2 \implies v^2 = 4 \]

Taking the square root:

\[ v = 2 \, \text{m/s} \]

### Final Answer
The velocity of the particle at time \( t = 2 \) seconds is \( 2 \, \text{m/s} \).

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