Power supplied to a particle of mass 4 kg varies with time as `P=(3t^(2))/(2)` watt. Here t in second. If velocity of particle at t = 0 is v = 0, the velocity of particle at time t = 2s will be

Options

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

Answer

To find the velocity of the particle at time \( t = 2 \) seconds, we can follow these steps:

### Step 1: Understand the relationship between power and work
Power \( P \) is defined as the rate of work done, which can be expressed mathematically as:
\[ P = \frac{dW}{dt} \]
where \( dW \) is the differential work done over a small time interval \( dt \).

### Step 2: Express work done in terms of power
From the power equation given in the problem:
\[ P = \frac{3t^2}{2} \text{ watts} \]
we can express the differential work done as:
\[ dW = P \cdot dt = \frac{3t^2}{2} dt \]

### Step 3: Integrate to find total work done
To find the total work done from \( t = 0 \) to \( t = 2 \) seconds, we integrate:
\[ W = \int_0^2 dW = \int_0^2 \frac{3t^2}{2} dt \]

### Step 4: Perform the integration
Calculating the integral:
\[ W = \frac{3}{2} \int_0^2 t^2 dt \]
The integral of \( t^2 \) is:
\[ \int t^2 dt = \frac{t^3}{3} \]
So we evaluate:
\[ W = \frac{3}{2} \left[ \frac{t^3}{3} \right]_0^2 = \frac{3}{2} \left( \frac{2^3}{3} - 0 \right) = \frac{3}{2} \cdot \frac{8}{3} = 4 \text{ joules} \]

### Step 5: Apply the work-energy theorem
According to the work-energy theorem, the work done on the particle is equal to the change in kinetic energy:
\[ W = K_f - K_i \]
where \( K_f \) is the final kinetic energy and \( K_i \) is the initial kinetic energy. Given that the initial velocity \( v(0) = 0 \), we have:
\[ K_i = 0 \quad \text{and} \quad W = K_f \]
Thus:
\[ K_f = 4 \text{ joules} \]

### Step 6: Relate kinetic energy to velocity
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle. Rearranging for \( v \):
\[ 4 = \frac{1}{2} \cdot 4 \cdot v^2 \]
This simplifies to:
\[ 4 = 2v^2 \]
Dividing both sides by 2:
\[ 2 = v^2 \]
Taking the square root:
\[ v = \sqrt{2} \text{ m/s} \]

### Conclusion
The velocity of the particle at time \( t = 2 \) seconds is:
\[ v = \sqrt{2} \text{ m/s} \]

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