The momentum of a body in two perpendicular direction 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

Answer

To find the force acting on the body at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Understand the given momentum equations
The momentum of the body in the x-direction is given by:
\[ P_x = 2t^2 + 6 \]
The momentum of the body in the y-direction is given by:
\[ P_y = \frac{3t^2}{2} + 3 \]

### Step 2: Differentiate the momentum equations to find force
The force acting on the body can be found using the relationship between force and momentum:
\[ F = \frac{dP}{dt} \]

#### Step 2.1: Calculate the force in the x-direction
Differentiate \( P_x \) with respect to \( t \):
\[ F_x = \frac{dP_x}{dt} = \frac{d}{dt}(2t^2 + 6) = 4t \]

#### Step 2.2: Calculate the force in the y-direction
Differentiate \( P_y \) with respect to \( t \):
\[ F_y = \frac{dP_y}{dt} = \frac{d}{dt}\left(\frac{3t^2}{2} + 3\right) = 3t \]

### Step 3: Evaluate the forces at \( t = 2 \) seconds
#### Step 3.1: Calculate \( F_x \) at \( t = 2 \)
\[ F_x = 4(2) = 8 \, \text{N} \]

#### Step 3.2: Calculate \( F_y \) at \( t = 2 \)
\[ F_y = 3(2) = 6 \, \text{N} \]

### Step 4: Calculate the resultant force
The resultant force \( F \) can be calculated using the Pythagorean theorem:
\[ F = \sqrt{F_x^2 + F_y^2} \]
Substituting the values:
\[ F = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{N} \]

### Final Answer
The force acting on the body at \( t = 2 \) seconds is \( 10 \, \text{N} \).

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