A force `vecF=6t^(2)hati+4thatj` is acting on a particle of mass 3 kg then what will be velocity of particle at t=3 second and if at t=0, particle is at rest:-

Options

A. If its speed changes, its velocity must change and it must have some acceleration
B. If its velocity changes, its speeds must change and it must have some acceleration
C. If its velocity changes, its speed may or may not change, and it must have some acceleration
D. If its speed changes but direction of motion does not change, its velocity may remain constant

Explanation

To solve the problem, we need to find the velocity of a particle under the influence of a given force over a specific time interval. Let's break it down step by step.

### Step 1: Identify the Given Information
We have:
- Force \( \vec{F} = 6t^2 \hat{i} + 4t \hat{j} \)
- Mass \( m = 3 \, \text{kg} \)
- Initial velocity \( \vec{v}(0) = 0 \) (the particle is at rest)
- Time \( t = 3 \, \text{s} \)

### Step 2: Use Newton's Second Law
According to Newton's second law, the force acting on an object is equal to the mass of the object multiplied by its acceleration:
\[ \vec{F} = m \vec{a} \]
From this, we can express acceleration \( \vec{a} \) as:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{6t^2 \hat{i} + 4t \hat{j}}{3} \]
This simplifies to:
\[ \vec{a} = 2t^2 \hat{i} + \frac{4}{3}t \hat{j} \]

### Step 3: Relate Acceleration to Velocity
Acceleration is the rate of change of velocity:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Thus, we can write:
\[ d\vec{v} = \vec{a} dt = (2t^2 \hat{i} + \frac{4}{3}t \hat{j}) dt \]

### Step 4: Integrate to Find Velocity
To find the velocity, we integrate \( d\vec{v} \):
\[ \vec{v} = \int (2t^2 \hat{i} + \frac{4}{3}t \hat{j}) dt \]
This results in:
\[ \vec{v} = \left( \frac{2}{3}t^3 \hat{i} + \frac{2}{3}t^2 \hat{j} \right) + \vec{C} \]
Since the particle is at rest at \( t = 0 \), we can find the constant of integration \( \vec{C} \):
\[ \vec{v}(0) = 0 \implies \vec{C} = 0 \]
Thus, the velocity function simplifies to:
\[ \vec{v} = \frac{2}{3}t^3 \hat{i} + \frac{2}{3}t^2 \hat{j} \]

### Step 5: Evaluate Velocity at \( t = 3 \, \text{s} \)
Now, we substitute \( t = 3 \) into the velocity equation:
\[ \vec{v}(3) = \frac{2}{3}(3^3) \hat{i} + \frac{2}{3}(3^2) \hat{j} \]
Calculating each component:
\[ \vec{v}(3) = \frac{2}{3}(27) \hat{i} + \frac{2}{3}(9) \hat{j} \]
\[ \vec{v}(3) = 18 \hat{i} + 6 \hat{j} \]

### Final Answer
The velocity of the particle at \( t = 3 \, \text{s} \) is:
\[ \vec{v}(3) = 18 \hat{i} + 6 \hat{j} \]

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