At time t second, a particle of mass 3 kg has position vector r metre, where `r = 3that(i) - 4 cos t hat(j)`. Find the impulse of the force during the time interval `0 le t le(pi)/(2)`

Options

A. (a)`12hatjNs`
B. (b)`9hatjNs`
C. (c)`4hatjNs`
D. (d)`14hatjNs`

Answer

To solve the problem step by step, we will follow the physics principles related to motion, specifically focusing on position, velocity, acceleration, and impulse.

### Step 1: Identify the Position Vector
The position vector of the particle is given as:
\[ \mathbf{r} = 3t \hat{i} - 4 \cos(t) \hat{j} \]

### Step 2: Calculate the Velocity Vector
The velocity vector \(\mathbf{v}\) is the time derivative of the position vector \(\mathbf{r}\):
\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(3t \hat{i} - 4 \cos(t) \hat{j}) \]
Calculating the derivatives:
\[ \mathbf{v} = 3 \hat{i} + 4 \sin(t) \hat{j} \]

### Step 3: Calculate the Acceleration Vector
The acceleration vector \(\mathbf{a}\) is the time derivative of the velocity vector \(\mathbf{v}\):
\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3 \hat{i} + 4 \sin(t) \hat{j}) \]
Calculating the derivatives:
\[ \mathbf{a} = 0 \hat{i} + 4 \cos(t) \hat{j} \]

### Step 4: Calculate the Force Vector
Using Newton's second law, the force \(\mathbf{F}\) is given by:
\[ \mathbf{F} = m \mathbf{a} \]
Given that the mass \(m = 3 \, \text{kg}\):
\[ \mathbf{F} = 3 \cdot (0 \hat{i} + 4 \cos(t) \hat{j}) = 12 \cos(t) \hat{j} \]

### Step 5: Calculate the Impulse
Impulse \(\mathbf{I}\) is defined as the integral of force over time:
\[ \mathbf{I} = \int_{0}^{\frac{\pi}{2}} \mathbf{F} \, dt \]
Substituting the force:
\[ \mathbf{I} = \int_{0}^{\frac{\pi}{2}} 12 \cos(t) \hat{j} \, dt \]
Factoring out the constant:
\[ \mathbf{I} = 12 \hat{j} \int_{0}^{\frac{\pi}{2}} \cos(t) \, dt \]

### Step 6: Evaluate the Integral
The integral of \(\cos(t)\) from \(0\) to \(\frac{\pi}{2}\) is:
\[ \int_{0}^{\frac{\pi}{2}} \cos(t) \, dt = [\sin(t)]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]
Thus, we have:
\[ \mathbf{I} = 12 \hat{j} \cdot 1 = 12 \hat{j} \]

### Step 7: Final Result
The impulse of the force during the time interval \(0 \leq t \leq \frac{\pi}{2}\) is:
\[ \mathbf{I} = 12 \hat{j} \, \text{Ns} \]

### Summary
The impulse is \(12 \, \text{Ns}\) in the direction of \(\hat{j}\).

---