The position of a particle is given by `vec(r) = 3that(i) - 4t^(2)hat(j) + 5hat(k).` Then the magnitude of the velocity of the particle at t = 2 s is

Options

A. `sqrt(265)m//s`
B. `sqrt(276)m//s`
C. `sqrt(246)m//s`
D. `sqrt(255)m//s`

Answer

To find the magnitude of the velocity of the particle at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Find the velocity vector
The velocity vector \( \vec{v} \) is the derivative of the position vector \( \vec{r} \) with respect to time \( t \). The position vector is given as:
\[ \vec{r}(t) = 3t \hat{i} - 4t^2 \hat{j} + 5 \hat{k} \]

To find the velocity, we differentiate \( \vec{r}(t) \):
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t \hat{i} - 4t^2 \hat{j} + 5 \hat{k}) \]
Calculating the derivative:
\[ \vec{v}(t) = 3 \hat{i} - 8t \hat{j} + 0 \hat{k} \]
Thus,
\[ \vec{v}(t) = 3 \hat{i} - 8t \hat{j} \]

### Step 2: Substitute \( t = 2 \) seconds
Now we substitute \( t = 2 \) seconds into the velocity vector:
\[ \vec{v}(2) = 3 \hat{i} - 8(2) \hat{j} \]
Calculating this gives:
\[ \vec{v}(2) = 3 \hat{i} - 16 \hat{j} \]

### Step 3: Calculate the magnitude of the velocity vector
The magnitude of the velocity vector \( \vec{v} \) is given by:
\[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2} \]
In our case, \( v_x = 3 \), \( v_y = -16 \), and \( v_z = 0 \). Thus,
\[ |\vec{v}(2)| = \sqrt{(3)^2 + (-16)^2 + (0)^2} \]
Calculating the squares:
\[ |\vec{v}(2)| = \sqrt{9 + 256 + 0} = \sqrt{265} \]

### Step 4: Final result
Therefore, the magnitude of the velocity of the particle at \( t = 2 \) seconds is:
\[ |\vec{v}(2)| = \sqrt{265} \text{ units} \]

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