The position vector of a particle is determined by the expression `vec r = 3t^2 hat i+ 4t^2 hat j + 7 hat k`. The displacement traversed in first `10` seconds is :

Options

A. 500m
B. 300m
C. 150m
D. 100m

Explanation

To solve the problem, we need to find the displacement traversed by the particle in the first 10 seconds given its position vector. The position vector is given by:

\[ \vec{r} = 3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k} \]

### Step 1: Find the initial position vector at \( t = 0 \)

To find the initial position vector \( \vec{r_1} \), we substitute \( t = 0 \) into the position vector equation:

\[ \vec{r_1} = 3(0)^2 \hat{i} + 4(0)^2 \hat{j} + 7 \hat{k} = 0 \hat{i} + 0 \hat{j} + 7 \hat{k} = 7 \hat{k} \]

### Step 2: Find the final position vector at \( t = 10 \)

Next, we find the final position vector \( \vec{r_2} \) by substituting \( t = 10 \):

\[ \vec{r_2} = 3(10)^2 \hat{i} + 4(10)^2 \hat{j} + 7 \hat{k} = 3(100) \hat{i} + 4(100) \hat{j} + 7 \hat{k} = 300 \hat{i} + 400 \hat{j} + 7 \hat{k} \]

### Step 3: Calculate the displacement vector

The displacement vector \( \Delta \vec{r} \) is given by the difference between the final and initial position vectors:

\[ \Delta \vec{r} = \vec{r_2} - \vec{r_1} = (300 \hat{i} + 400 \hat{j} + 7 \hat{k}) - (0 \hat{i} + 0 \hat{j} + 7 \hat{k}) \]

Simplifying this gives:

\[ \Delta \vec{r} = 300 \hat{i} + 400 \hat{j} + 7 \hat{k} - 7 \hat{k} = 300 \hat{i} + 400 \hat{j} \]

### Step 4: Find the magnitude of the displacement vector

To find the magnitude of the displacement vector \( \Delta \vec{r} \), we use the formula for the magnitude of a vector:

\[ |\Delta \vec{r}| = \sqrt{(300)^2 + (400)^2} \]

Calculating this:

\[ |\Delta \vec{r}| = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \]

### Final Answer

The displacement traversed in the first 10 seconds is:

\[ \boxed{500} \]