Position vector of a particle moving in space is given by :
`vec(r)=3sin t hat i+3 cos t hatj+4 t hatk`
Distance travelled by the particle in `2s` is :

Options

A. 5 m
B. 10 m
C. 20 m
D. 50 m

Explanation

To find the distance traveled by the particle in 2 seconds, we will follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given by:
\[ \vec{r}(t) = 3\sin(t) \hat{i} + 3\cos(t) \hat{j} + 4t \hat{k} \]

### Step 2: Differentiate the position vector to find the velocity vector
To find the velocity vector \(\vec{v}(t)\), we differentiate the position vector \(\vec{r}(t)\) with respect to time \(t\):
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3\sin(t) \hat{i} + 3\cos(t) \hat{j} + 4t \hat{k}) \]
Differentiating each component:
- The derivative of \(3\sin(t)\) is \(3\cos(t)\).
- The derivative of \(3\cos(t)\) is \(-3\sin(t)\).
- The derivative of \(4t\) is \(4\).

Thus, the velocity vector is:
\[ \vec{v}(t) = 3\cos(t) \hat{i} - 3\sin(t) \hat{j} + 4 \hat{k} \]

### Step 3: Calculate the magnitude of the velocity vector
The magnitude of the velocity vector \(|\vec{v}(t)|\) is given by:
\[ |\vec{v}(t)| = \sqrt{(3\cos(t))^2 + (-3\sin(t))^2 + (4)^2} \]
Calculating each term:
- \((3\cos(t))^2 = 9\cos^2(t)\)
- \((-3\sin(t))^2 = 9\sin^2(t)\)
- \((4)^2 = 16\)

Combining these:
\[ |\vec{v}(t)| = \sqrt{9\cos^2(t) + 9\sin^2(t) + 16} \]
Using the identity \(\cos^2(t) + \sin^2(t) = 1\):
\[ |\vec{v}(t)| = \sqrt{9(1) + 16} = \sqrt{25} = 5 \]

### Step 4: Calculate the distance traveled in 2 seconds
The distance \(d\) traveled by the particle in time \(t\) is given by:
\[ d = |\vec{v}(t)| \cdot t \]
Substituting \(t = 2\) seconds:
\[ d = 5 \cdot 2 = 10 \text{ meters} \]

### Final Answer
The distance traveled by the particle in 2 seconds is:
\[ \boxed{10 \text{ meters}} \]