A particle moves in x-y plane and at time t is at the point `(t^2, t^3 -2 t),` then which of the following is/are correct?

Options

A. At `t = 0,` particle is moving parallel to y-axis
B. At `t = 0,` direction of velocity and acceleration are perpendicular
C. At `t = sqrt(2/3)`, particle is moving parallel to x-axis
D. At `t=0,` particle is at rest

Explanation

To solve the problem, we need to analyze the motion of the particle given its position in the x-y plane as a function of time \( t \). The position of the particle is given by the coordinates:

\[ x(t) = t^2 \]
\[ y(t) = t^3 - 2t \]

### Step 1: Find the velocity components
The velocity components in the x and y directions can be found by taking the derivative of the position functions with respect to time \( t \).

\[ v_x = \frac{dx}{dt} = \frac{d(t^2)}{dt} = 2t \]
\[ v_y = \frac{dy}{dt} = \frac{d(t^3 - 2t)}{dt} = 3t^2 - 2 \]

### Step 2: Find the acceleration components
The acceleration components can be found by taking the derivative of the velocity components with respect to time \( t \).

\[ a_x = \frac{dv_x}{dt} = \frac{d(2t)}{dt} = 2 \]
\[ a_y = \frac{dv_y}{dt} = \frac{d(3t^2 - 2)}{dt} = 6t \]

### Step 3: Analyze the situation at \( t = 0 \)
Now, let's evaluate the position, velocity, and acceleration at \( t = 0 \).

- Position:
\[ x(0) = 0^2 = 0 \]
\[ y(0) = 0^3 - 2(0) = 0 \]

- Velocity:
\[ v_x(0) = 2(0) = 0 \]
\[ v_y(0) = 3(0)^2 - 2 = -2 \]

- Acceleration:
\[ a_x(0) = 2 \]
\[ a_y(0) = 6(0) = 0 \]

### Step 4: Check the statements
1. **At \( t = 0 \), the particle is moving parallel to the y-axis.**
- True, because \( v_x(0) = 0 \) and \( v_y(0) \neq 0 \).

2. **At \( t = 0 \), the direction of velocity and acceleration are perpendicular.**
- True, because \( v_y(0) \) is non-zero and \( a_x(0) \) is non-zero while \( a_y(0) = 0 \).

3. **At \( t = \sqrt{\frac{2}{3}} \), the particle is moving parallel to the x-axis.**
- True, because we can check that \( v_y(\sqrt{\frac{2}{3}}) = 0 \).

4. **At \( t = 0 \), the particle is at rest.**
- False, because the particle has a velocity component in the y-direction.

### Conclusion
The correct statements are:
1. True
2. True
3. True
4. False

### Final Answer
The correct options are A, B, and C.

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