Power supplied to a particle of mass 3 kg varies with time as `p = 4t ^(3)` watt, where t in time is seconds. If velocity of particle at `t =0s ` is ` u=2 m//s.` The velocity of particle at time `t =3s` will be :

Options

A. `sqrt58 m//s`
B. `sqrt29m//s`
C. `sqrt87 m//s`
D. `sqrt27m//s`

Explanation

To solve the problem, we need to find the velocity of a particle at time \( t = 3 \) seconds given that the power supplied to the particle varies with time as \( P = 4t^3 \) watts, and the initial velocity at \( t = 0 \) seconds is \( u = 2 \) m/s.

### Step-by-step Solution:

1. **Understand the relationship between power and work done:**
The power supplied to the particle is given by \( P = \frac{dW}{dt} \), where \( W \) is the work done. Therefore, the work done can be found by integrating power over time:
\[ W = \int P \, dt \]

2. **Set up the integral for work done:**
Given \( P = 4t^3 \), we can write:
\[ W = \int_0^3 4t^3 \, dt \]

3. **Calculate the integral:**
To evaluate the integral:
\[ W = 4 \int_0^3 t^3 \, dt \]
The integral of \( t^3 \) is:
\[ \int t^3 \, dt = \frac{t^4}{4} \]
Therefore,
\[ W = 4 \left[ \frac{t^4}{4} \right]_0^3 = \left[ t^4 \right]_0^3 = 3^4 - 0^4 = 81 \, \text{joules} \]

4. **Apply the work-energy theorem:**
The work-energy theorem states that the work done on an object is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_f - KE_i \]
Where \( KE_f \) is the final kinetic energy and \( KE_i \) is the initial kinetic energy. The initial kinetic energy is given by:
\[ KE_i = \frac{1}{2} m u^2 \]
Substituting \( m = 3 \, \text{kg} \) and \( u = 2 \, \text{m/s} \):
\[ KE_i = \frac{1}{2} \times 3 \times (2)^2 = \frac{1}{2} \times 3 \times 4 = 6 \, \text{joules} \]

5. **Set up the equation for final kinetic energy:**
Let \( v \) be the final velocity at \( t = 3 \) seconds. The final kinetic energy is:
\[ KE_f = \frac{1}{2} m v^2 \]
Therefore, we can write:
\[ W = KE_f - KE_i \implies 81 = \frac{1}{2} \times 3 v^2 - 6 \]

6. **Solve for \( v^2 \):**
Rearranging the equation gives:
\[ 81 + 6 = \frac{3}{2} v^2 \]
\[ 87 = \frac{3}{2} v^2 \]
Multiplying both sides by \( \frac{2}{3} \):
\[ v^2 = \frac{2 \times 87}{3} = \frac{174}{3} = 58 \]

7. **Find \( v \):**
Taking the square root gives:
\[ v = \sqrt{58} \, \text{m/s} \]

### Final Answer:
The velocity of the particle at \( t = 3 \) seconds is \( v = \sqrt{58} \, \text{m/s} \).

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