The position of a particle is defined by `vec r = 2sin((pi)/(4))t hati + 2cos((pi)/(4))t hatj+3t hatk` , where 't' is in seconds. Then distance travelled by particle in 2 second motions is. `[use pi^(2) = 10]`

Options

A. `sqrt(32)m`
B. `sqrt(46)m`
C. `sqrt(56)m`
D. None of these

Explanation

To find the distance travelled by the particle in 2 seconds, we first need to determine the velocity of the particle from the given position vector. The position vector is given by:

\[ \vec{r} = 2 \sin\left(\frac{\pi}{4}\right)t \hat{i} + 2 \cos\left(\frac{\pi}{4}\right)t \hat{j} + 3t \hat{k} \]

### Step 1: Differentiate the position vector to find the velocity vector

The velocity vector \(\vec{v}\) is the derivative of the position vector \(\vec{r}\) with respect to time \(t\):

\[ \vec{v} = \frac{d\vec{r}}{dt} \]

Calculating the derivatives:

1. For the \(x\) component:
\[ v_x = \frac{d}{dt}\left(2 \sin\left(\frac{\pi}{4}\right)t\right) = 2 \cos\left(\frac{\pi}{4}\right) \cdot \frac{\pi}{4} = \frac{\pi}{2} \sqrt{2} \]

2. For the \(y\) component:
\[ v_y = \frac{d}{dt}\left(2 \cos\left(\frac{\pi}{4}\right)t\right) = -2 \sin\left(\frac{\pi}{4}\right) \cdot \frac{\pi}{4} = -\frac{\pi}{2} \sqrt{2} \]

3. For the \(z\) component:
\[ v_z = \frac{d}{dt}(3t) = 3 \]

Thus, the velocity vector is:

\[ \vec{v} = \left(\frac{\pi}{2} \sqrt{2}\right) \hat{i} + \left(-\frac{\pi}{2} \sqrt{2}\right) \hat{j} + 3 \hat{k} \]

### Step 2: Calculate the magnitude of the velocity vector

The magnitude of the velocity vector \(|\vec{v}|\) is given by:

\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

Substituting the values:

\[ |\vec{v}| = \sqrt{\left(\frac{\pi}{2} \sqrt{2}\right)^2 + \left(-\frac{\pi}{2} \sqrt{2}\right)^2 + 3^2} \]

Calculating each term:

1. \(\left(\frac{\pi}{2} \sqrt{2}\right)^2 = \frac{\pi^2}{4} \cdot 2 = \frac{\pi^2}{2}\)
2. \(\left(-\frac{\pi}{2} \sqrt{2}\right)^2 = \frac{\pi^2}{4} \cdot 2 = \frac{\pi^2}{2}\)
3. \(3^2 = 9\)

Now, summing these:

\[ |\vec{v}| = \sqrt{\frac{\pi^2}{2} + \frac{\pi^2}{2} + 9} = \sqrt{\pi^2 + 9} \]

Given that \(\pi^2 \approx 10\):

\[ |\vec{v}| = \sqrt{10 + 9} = \sqrt{19} \]

### Step 3: Calculate the distance travelled in 2 seconds

The distance travelled \(d\) in time \(t = 2\) seconds is given by:

\[ d = |\vec{v}| \cdot t \]

Substituting the values:

\[ d = \sqrt{19} \cdot 2 = 2\sqrt{19} \]

### Final Result

Thus, the distance travelled by the particle in 2 seconds is:

\[ d = 2\sqrt{19} \]