The position vector of a moving particle at't' sec is given by `vecr = 3hati + 4t^(2)hatj - t^(3)hatk` . Its displacement during an interval of t = Is to 3 sec is

Options

A. `hatj - hatk`
B. `3hati + 4hatj - hatk`
C. `9hati + 36hatj - 27 hatk`
D. `32 hatj - 26 hatk`

Answer

To find the displacement of a moving particle during the interval from \( t = 1 \) second to \( t = 3 \) seconds, we will follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given by:
\[ \vec{r}(t) = 3\hat{i} + 4t^2\hat{j} - t^3\hat{k} \]

### Step 2: Calculate the position vector at \( t = 1 \) second
Substituting \( t = 1 \) into the position vector:
\[ \vec{r}(1) = 3\hat{i} + 4(1^2)\hat{j} - (1^3)\hat{k} \]
Calculating this gives:
\[ \vec{r}(1) = 3\hat{i} + 4\hat{j} - 1\hat{k} = 3\hat{i} + 4\hat{j} - \hat{k} \]

### Step 3: Calculate the position vector at \( t = 3 \) seconds
Now substituting \( t = 3 \) into the position vector:
\[ \vec{r}(3) = 3\hat{i} + 4(3^2)\hat{j} - (3^3)\hat{k} \]
Calculating this gives:
\[ \vec{r}(3) = 3\hat{i} + 4(9)\hat{j} - 27\hat{k} = 3\hat{i} + 36\hat{j} - 27\hat{k} \]

### Step 4: Calculate the displacement
The displacement \( \Delta \vec{r} \) during the interval from \( t = 1 \) to \( t = 3 \) seconds is given by:
\[ \Delta \vec{r} = \vec{r}(3) - \vec{r}(1) \]
Substituting the values we calculated:
\[ \Delta \vec{r} = (3\hat{i} + 36\hat{j} - 27\hat{k}) - (3\hat{i} + 4\hat{j} - \hat{k}) \]
Now simplifying this:
\[ \Delta \vec{r} = (3\hat{i} - 3\hat{i}) + (36\hat{j} - 4\hat{j}) + (-27\hat{k} + \hat{k}) \]
\[ \Delta \vec{r} = 0\hat{i} + 32\hat{j} - 26\hat{k} \]
Thus, the displacement is:
\[ \Delta \vec{r} = 32\hat{j} - 26\hat{k} \]

### Final Answer
The displacement during the interval from \( t = 1 \) second to \( t = 3 \) seconds is:
\[ \Delta \vec{r} = 32\hat{j} - 26\hat{k} \]

---