The position of a particle is given by `vecr = 3t hati + 2t^(2) hatj + 5hatk`, where t is in seconds and the coefficients have the proper units for `vecr` to be in meters. The direction of velocity of the particle at `t = 1` s is

Options

A. `53^(@)` with x-axis
B. `37^(@)` with x-axis
C. `30^(@)` with y-axis
D. `60^(@)` with y-axis

Explanation

To find the direction of the velocity of the particle at \( t = 1 \) second, we will follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given by:
\[ \vec{r} = 3t \hat{i} + 2t^2 \hat{j} + 5 \hat{k} \]

### Step 2: Differentiate the position vector to find the velocity vector
The velocity vector \( \vec{v} \) is the time derivative of the position vector \( \vec{r} \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3t \hat{i} + 2t^2 \hat{j} + 5 \hat{k}) \]
Differentiating each component:
- The derivative of \( 3t \hat{i} \) is \( 3 \hat{i} \)
- The derivative of \( 2t^2 \hat{j} \) is \( 4t \hat{j} \)
- The derivative of \( 5 \hat{k} \) is \( 0 \hat{k} \)

Thus, the velocity vector is:
\[ \vec{v} = 3 \hat{i} + 4t \hat{j} \]

### Step 3: Substitute \( t = 1 \) second into the velocity vector
Now, we will substitute \( t = 1 \) into the velocity vector:
\[ \vec{v} = 3 \hat{i} + 4(1) \hat{j} = 3 \hat{i} + 4 \hat{j} \]

### Step 4: Calculate the angle with respect to the x-axis
To find the angle \( \theta \) that the velocity vector makes with the x-axis, we use the tangent function:
\[ \tan \theta = \frac{v_y}{v_x} = \frac{4}{3} \]
Now we can find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 5: Calculate \( \theta \)
Using a calculator or trigonometric tables:
\[ \theta \approx 53^\circ \]

### Conclusion
The direction of the velocity of the particle at \( t = 1 \) second is \( 53^\circ \) with respect to the x-axis.

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