An object of mass 2 kg at rest at origin starts moving under the action of a force
`vec(F)=(3t^(2)hat(i)+4that(j))N` The velocity of the object at t = 2 s will be -

Options

A. `(3hat(i)+2hat(j))m//s`
B. `(2hat(i)+4hat(j))m//s`
C. `(4hat(i)+4hat(j))m//s`
D. `(3hat(i)-4hat(j))m//s`

Answer

To find the velocity of the object at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Identify the given information
- Mass of the object, \( m = 2 \, \text{kg} \)
- Force acting on the object, \( \vec{F} = (3t^2 \hat{i} + 4t \hat{j}) \, \text{N} \)

### Step 2: Calculate the acceleration
Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration \( \vec{a} \):
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{(3t^2 \hat{i} + 4t \hat{j})}{2} \]
This gives us:
\[ \vec{a} = \left(\frac{3t^2}{2} \hat{i} + 2t \hat{j}\right) \, \text{m/s}^2 \]

### Step 3: Determine the velocity components
To find the velocity, we need to integrate the acceleration with respect to time.

#### For the x-component of velocity (\( v_x \)):
\[ \frac{dv_x}{dt} = \frac{3t^2}{2} \]
Integrating with respect to \( t \):
\[ dv_x = \frac{3t^2}{2} dt \]
Integrating from \( t = 0 \) to \( t = 2 \):
\[ v_x = \int_0^2 \frac{3t^2}{2} dt = \frac{3}{2} \cdot \left[\frac{t^3}{3}\right]_0^2 = \frac{3}{2} \cdot \left(\frac{2^3}{3} - 0\right) = \frac{3}{2} \cdot \frac{8}{3} = 4 \, \text{m/s} \]

#### For the y-component of velocity (\( v_y \)):
\[ \frac{dv_y}{dt} = 2t \]
Integrating with respect to \( t \):
\[ dv_y = 2t dt \]
Integrating from \( t = 0 \) to \( t = 2 \):
\[ v_y = \int_0^2 2t dt = \left[t^2\right]_0^2 = 2^2 - 0 = 4 \, \text{m/s} \]

### Step 4: Combine the velocity components
Now we can write the total velocity vector at \( t = 2 \) seconds:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = 4 \hat{i} + 4 \hat{j} \, \text{m/s} \]

### Final Answer
The velocity of the object at \( t = 2 \) seconds is:
\[ \vec{v} = 4 \hat{i} + 4 \hat{j} \, \text{m/s} \]

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