The position of a particle is given by  `vecr = 3.01t hati +2. 0 t^2 hatj +5.0 hatk` where t is in seconds and the coefficients have the proper units for `vecr` to be in metres. What is the magnitude and direction of velocity of the particle at t = 1 s? .

Options

A. `5 ms^(-1) , tan(4/3)` with x - axis.
B. `5 ms^(-1) , tan(3/4)` with x - axis.
C. `4 ms^(-1) , tan(3/4)` with x - axis.
D. `4 ms^(-1) , tan(4/3)` with x - axis.

Answer

To solve the problem, we need to find the velocity of the particle at \( t = 1 \) second given the position vector:

\[ \vec{r} = 3.01t \hat{i} + 2.0t^2 \hat{j} + 5.0 \hat{k} \]

### Step 1: Differentiate the Position Vector
The velocity vector \( \vec{v} \) is the time derivative of the position vector \( \vec{r} \):

\[ \vec{v} = \frac{d\vec{r}}{dt} \]

Differentiating each component of \( \vec{r} \):

1. For the \( \hat{i} \) component:
\[ \frac{d}{dt}(3.01t) = 3.01 \]

2. For the \( \hat{j} \) component:
\[ \frac{d}{dt}(2.0t^2) = 4.0t \]

3. For the \( \hat{k} \) component (constant):
\[ \frac{d}{dt}(5.0) = 0 \]

Thus, the velocity vector is:

\[ \vec{v} = 3.01 \hat{i} + 4.0t \hat{j} + 0 \hat{k} \]

### Step 2: Substitute \( t = 1 \) second
Now, we substitute \( t = 1 \) into the velocity vector:

\[ \vec{v} = 3.01 \hat{i} + 4.0(1) \hat{j} + 0 \hat{k} = 3.01 \hat{i} + 4.0 \hat{j} \]

### Step 3: Calculate the Magnitude of the Velocity
The magnitude of the velocity vector \( \vec{v} \) is given by:

\[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2} \]

Substituting the values:

\[ |\vec{v}| = \sqrt{(3.01)^2 + (4.0)^2} \]

Calculating:

\[ |\vec{v}| = \sqrt{9.0601 + 16} = \sqrt{25.0601} \approx 5.0 \, \text{m/s} \]

### Step 4: Calculate the Direction of the Velocity
The direction of the velocity can be described by the angle \( \theta \) it makes with the x-axis, which can be found using:

\[ \tan \theta = \frac{v_y}{v_x} \]

Substituting the values:

\[ \tan \theta = \frac{4.0}{3.01} \]

Calculating \( \theta \):

\[ \theta = \tan^{-1}\left(\frac{4.0}{3.01}\right) \approx \tan^{-1}(1.327) \approx 53.1^\circ \]

### Final Answer
The magnitude of the velocity of the particle at \( t = 1 \) second is approximately \( 5.0 \, \text{m/s} \) and the direction is approximately \( 53.1^\circ \) above the positive x-axis.

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