The momenta of a body in two perpendicular directions at any time 't' are given by `P_(x)=2t^(2) +6` and `P_(y)=(3t^(2))/(2)+3`. The force acting on the body at `t =2` sec is .

Options

A. `5 units`
B. `2 units`
C. `10 units`
D. `15 units`

Explanation

To find the force acting on the body at \( t = 2 \) seconds, we will use the relationship between momentum and force. The force is the rate of change of momentum with respect to time.

### Step-by-Step Solution:

1. **Identify the momentum equations**:
- The momentum in the x-direction is given by:
\[ P_x = 2t^2 + 6 \]
- The momentum in the y-direction is given by:
\[ P_y = \frac{3t^2}{2} + 3 \]

2. **Differentiate the momentum equations to find the force**:
- The force in the x-direction \( F_x \) is given by the derivative of \( P_x \) with respect to time \( t \):
\[ F_x = \frac{dP_x}{dt} = \frac{d}{dt}(2t^2 + 6) = 4t \]
- The force in the y-direction \( F_y \) is given by the derivative of \( P_y \) with respect to time \( t \):
\[ F_y = \frac{dP_y}{dt} = \frac{d}{dt}\left(\frac{3t^2}{2} + 3\right) = 3t \]

3. **Evaluate the forces at \( t = 2 \) seconds**:
- Calculate \( F_x \) at \( t = 2 \):
\[ F_x = 4(2) = 8 \, \text{N} \]
- Calculate \( F_y \) at \( t = 2 \):
\[ F_y = 3(2) = 6 \, \text{N} \]

4. **Combine the forces to find the total force**:
- The total force \( \vec{F} \) acting on the body can be represented as:
\[ \vec{F} = F_x \hat{i} + F_y \hat{j} = 8 \hat{i} + 6 \hat{j} \, \text{N} \]

5. **Calculate the magnitude of the total force**:
- The magnitude of the force \( F \) can be calculated using the Pythagorean theorem:
\[ F = \sqrt{F_x^2 + F_y^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{N} \]

### Final Answer:
The force acting on the body at \( t = 2 \) seconds is \( \vec{F} = 8 \hat{i} + 6 \hat{j} \, \text{N} \) or \( 10 \, \text{N} \) in magnitude.

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