The position of a particle is given by
`r = 3t hati +2t^(2) hatj +8 hatk`
where, t is in seconds and the coefficients have the proper units for r to be in metres.
(i) Find v (t) and a(t) of the particles.
(ii) Find the magnitude and direction of v(t) and a(t) at `t = 1s`.

Options

A. `(2.0hat(i) + 4.0that(j))` , `a=4.0ms^(-2)` `5.0ms^(-1)` , `53^(@)` with x-axis`
B. `(3.0hat(i) + 4.0that(j))` , `a=4.0ms^(-2)` `5.0ms^(-1)` , `53^(@)` with x-axis`
C. `(3.0hat(i) + 4.0that(j))` , `a=2 ms^(-2)` `5.0ms^(-1)` , `53^(@)` with x-axis`
D. `(3.0hat(i) + 4.0that(j))` , `a=4.0ms^(-2)` `5.0ms^(-1)` , `45^(@)` with x-axis`

Explanation

To solve the problem step-by-step, we will first find the velocity and acceleration functions of the particle, and then we will calculate their magnitudes and directions at \( t = 1 \) second.

### Step 1: Find the velocity \( v(t) \)

The position vector of the particle is given by:
\[ \mathbf{r}(t) = 3t \hat{i} + 2t^2 \hat{j} + 8 \hat{k} \]

To find the velocity \( v(t) \), we differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \):
\[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(3t \hat{i} + 2t^2 \hat{j} + 8 \hat{k}) \]

Calculating the derivatives:
- The derivative of \( 3t \) with respect to \( t \) is \( 3 \).
- The derivative of \( 2t^2 \) with respect to \( t \) is \( 4t \).
- The derivative of \( 8 \) (a constant) with respect to \( t \) is \( 0 \).

Thus, the velocity function is:
\[ \mathbf{v}(t) = 3 \hat{i} + 4t \hat{j} + 0 \hat{k} = 3 \hat{i} + 4t \hat{j} \]

### Step 2: Find the acceleration \( a(t) \)

To find the acceleration \( a(t) \), we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \):
\[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(3 \hat{i} + 4t \hat{j}) \]

Calculating the derivatives:
- The derivative of \( 3 \) (a constant) with respect to \( t \) is \( 0 \).
- The derivative of \( 4t \) with respect to \( t \) is \( 4 \).

Thus, the acceleration function is:
\[ \mathbf{a}(t) = 0 \hat{i} + 4 \hat{j} + 0 \hat{k} = 0 \hat{i} + 4 \hat{j} \]

### Step 3: Evaluate \( v(t) \) and \( a(t) \) at \( t = 1 \) second

Now we will evaluate the velocity and acceleration at \( t = 1 \) second.

1. **Velocity at \( t = 1 \)**:
\[ \mathbf{v}(1) = 3 \hat{i} + 4(1) \hat{j} = 3 \hat{i} + 4 \hat{j} \]

2. **Acceleration at \( t = 1 \)**:
\[ \mathbf{a}(1) = 0 \hat{i} + 4 \hat{j} \]

### Step 4: Find the magnitude and direction of \( v(1) \) and \( a(1) \)

1. **Magnitude of velocity**:
\[ |\mathbf{v}(1)| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s} \]

2. **Direction of velocity**:
The direction can be found using the angle \( \theta \) with respect to the x-axis:
\[ \tan(\theta) = \frac{4}{3} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ \]

3. **Magnitude of acceleration**:
\[ |\mathbf{a}(1)| = 4 \, \text{m/s}^2 \]

4. **Direction of acceleration**:
The direction of acceleration is along the \( \hat{j} \) axis, which is simply \( 90^\circ \) from the x-axis.

### Final Results
- Velocity at \( t = 1 \) second: \( 5 \, \text{m/s} \) at an angle of \( 53^\circ \) from the x-axis.
- Acceleration at \( t = 1 \) second: \( 4 \, \text{m/s}^2 \) in the \( \hat{j} \) direction.