Position vector of a particle is expressed as function of time by equation `vec(r)=2t^(2)+(3t-1) hat(j) +5hat(k)`. Where r is in meters and t is in seconds.

Options

A. It always moves in a plane that is parallel to the x-y plane.
B. At the instant t=0 s, it is observed at point (0m, -1m, 5m), moving with velocity 3 m/s in the positive y-direction.
C. Its acceleration vector is uniform.
D. It is an example of three dimensional motion.

Answer

To solve the problem, we need to analyze the given position vector of the particle, which is expressed as a function of time:

\[ \vec{r}(t) = 2t^2 \hat{i} + (3t - 1) \hat{j} + 5 \hat{k} \]

Where \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) are the unit vectors in the x, y, and z directions respectively.

### Step 1: Identify the components of the position vector
From the position vector, we can identify the components as follows:
- \( x(t) = 2t^2 \)
- \( y(t) = 3t - 1 \)
- \( z(t) = 5 \)

### Step 2: Determine if the motion is confined to a plane
To check if the motion is confined to a plane, we can analyze the z-component. Since \( z(t) = 5 \) is a constant, it indicates that the motion occurs in a plane where \( z = 5 \). Therefore, the motion is confined to the plane parallel to the XY-plane.

### Step 3: Analyze the trajectory in the XY-plane
Next, we can express the relationship between \( x \) and \( y \) by eliminating \( t \):
1. From \( y(t) = 3t - 1 \), we can express \( t \) as:
\[ t = \frac{y + 1}{3} \]
2. Substitute \( t \) into \( x(t) \):
\[ x = 2\left(\frac{y + 1}{3}\right)^2 = 2 \cdot \frac{(y + 1)^2}{9} = \frac{2(y + 1)^2}{9} \]
This equation shows that the trajectory in the XY-plane is a parabola.

### Step 4: Calculate the velocity vector
The velocity vector \( \vec{v}(t) \) is the derivative of the position vector with respect to time:
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(2t^2 \hat{i} + (3t - 1) \hat{j} + 5 \hat{k}) = 4t \hat{i} + 3 \hat{j} + 0 \hat{k} \]

### Step 5: Evaluate the velocity at \( t = 0 \)
Substituting \( t = 0 \):
\[ \vec{v}(0) = 4(0) \hat{i} + 3 \hat{j} + 0 \hat{k} = 0 \hat{i} + 3 \hat{j} + 0 \hat{k} = 3 \hat{j} \]
This indicates that the particle is moving in the positive y-direction at \( 3 \, \text{m/s} \).

### Step 6: Calculate the acceleration vector
The acceleration vector \( \vec{a}(t) \) is the derivative of the velocity vector:
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(4t \hat{i} + 3 \hat{j} + 0 \hat{k}) = 4 \hat{i} + 0 \hat{j} + 0 \hat{k} \]
This indicates that the acceleration is constant and equal to \( 4 \hat{i} \).

### Step 7: Determine if the acceleration is uniform
Since the acceleration vector is constant, we conclude that the acceleration is uniform.

### Step 8: Conclusion about the motion
Based on the analysis:
- The motion is confined to a plane parallel to the XY-plane.
- The velocity at \( t = 0 \) is \( 3 \, \text{m/s} \) in the positive y-direction.
- The acceleration vector is uniform.

### Final Answer
1. The particle always moves in a plane that is parallel to the XY-plane. (Correct)
2. At \( t = 0 \), the position is \( (0, -1, 5) \) and the velocity is \( 3 \hat{j} \) (Correct).
3. The acceleration vector is uniform (Correct).
4. The motion is not an example of three-dimensional motion; it is confined to a two-dimensional plane. (Incorrect)