The position vector of a particle is determined by the expression ` vecr=2t^(2) hati+ 3thatj + 9hatk`. The magnitude of its displacement ( in m) in first two seconds is

Answer

To solve the problem of finding the magnitude of the displacement of the particle in the first two seconds, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Position Vector**: The position vector of the particle is given by:
\[ \vec{r} = 2t^2 \hat{i} + 3t \hat{j} + 9 \hat{k} \]

2. **Calculate the Initial Position Vector**:
At \( t = 0 \) seconds, we substitute \( t = 0 \) into the position vector:
\[ \vec{r}(0) = 2(0)^2 \hat{i} + 3(0) \hat{j} + 9 \hat{k} = 0 \hat{i} + 0 \hat{j} + 9 \hat{k} = 9 \hat{k} \]

3. **Calculate the Position Vector at \( t = 2 \) seconds**:
Now, we substitute \( t = 2 \) seconds into the position vector:
\[ \vec{r}(2) = 2(2)^2 \hat{i} + 3(2) \hat{j} + 9 \hat{k} = 2(4) \hat{i} + 6 \hat{j} + 9 \hat{k} = 8 \hat{i} + 6 \hat{j} + 9 \hat{k} \]

4. **Find the Displacement Vector**:
The displacement vector \( \vec{D} \) is given by:
\[ \vec{D} = \vec{r}(2) - \vec{r}(0) = (8 \hat{i} + 6 \hat{j} + 9 \hat{k}) - (0 \hat{i} + 0 \hat{j} + 9 \hat{k}) \]
Simplifying this, we get:
\[ \vec{D} = 8 \hat{i} + 6 \hat{j} + 9 \hat{k} - 9 \hat{k} = 8 \hat{i} + 6 \hat{j} \]

5. **Calculate the Magnitude of the Displacement Vector**:
The magnitude of the displacement vector \( |\vec{D}| \) is calculated using the formula:
\[ |\vec{D}| = \sqrt{(8)^2 + (6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ m} \]

### Final Answer:
The magnitude of the displacement in the first two seconds is \( 10 \) meters.

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