A particle moves in a plane such that its coordinates changes with time as `x = at` and `y = bt`, where a and b are constants. Find the position vector of the particle and its direction at any time t.

Options

A. `(a) hati + (bt) hatj`
B. `(at) hati + (b) hatj`
C. `(at) hati + (bt) hatj`
D. `(a) hati + (b) hatj`

Explanation

To solve the problem, we need to find the position vector of the particle and its direction at any time \( t \).

### Step-by-Step Solution:

1. **Identify the given equations**:
The coordinates of the particle are given as:
\[ x = at \]
\[ y = bt \]
where \( a \) and \( b \) are constants.

2. **Write the position vector**:
The position vector \( \mathbf{r} \) of the particle in a two-dimensional plane can be expressed in terms of its coordinates \( x \) and \( y \):
\[ \mathbf{r} = x \hat{i} + y \hat{j} \]
Substituting the expressions for \( x \) and \( y \):
\[ \mathbf{r} = (at) \hat{i} + (bt) \hat{j} \]
Thus, the position vector becomes:
\[ \mathbf{r} = at \hat{i} + bt \hat{j} \]

3. **Determine the direction of the particle**:
The direction of the particle can be described by the angle \( \theta \) it makes with the x-axis. This angle can be found using the tangent function:
\[ \tan(\theta) = \frac{y}{x} \]
Substituting the expressions for \( x \) and \( y \):
\[ \tan(\theta) = \frac{bt}{at} = \frac{b}{a} \]

4. **Find the angle \( \theta \)**:
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]

### Final Answer:
- The position vector of the particle at any time \( t \) is:
\[ \mathbf{r} = at \hat{i} + bt \hat{j} \]
- The direction of the particle at any time \( t \) is given by the angle:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]