A particle moves in the xy plane and at time t is at the point `(t^(2), t^(3)-2t)`. Then :-

Options

A. At `t=2//3 s`, directions of velocity and acceleration are perpendicular
B. At `t=0`, directions of velocity and acceleration are perpendicular
C. At `t=sqrt(2/3) s`, particle is moving parallel to x-axis
D. Acceleration of the particle when it is at point `(4, 4)` is `2hat(i)+24 hat(j)`

Explanation

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

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

### Step 1: Determine the velocity vector
The velocity vector \( \vec{v}(t) \) is the derivative of the position vector \( \vec{r}(t) \) with respect to time \( t \).

\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(t^2) \hat{i} + \frac{d}{dt}(t^3 - 2t) \hat{j} \]

Calculating the derivatives:

\[ \vec{v}(t) = (2t) \hat{i} + (3t^2 - 2) \hat{j} \]

### Step 2: Determine the acceleration vector
The acceleration vector \( \vec{a}(t) \) is the derivative of the velocity vector \( \vec{v}(t) \) with respect to time \( t \).

\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(2t) \hat{i} + \frac{d}{dt}(3t^2 - 2) \hat{j} \]

Calculating the derivatives:

\[ \vec{a}(t) = (2) \hat{i} + (6t) \hat{j} \]

### Step 3: Check if velocity and acceleration are perpendicular at \( t = \frac{2}{3} \)
To check if the velocity and acceleration vectors are perpendicular, we need to compute the dot product \( \vec{a}(t) \cdot \vec{v}(t) \) and set it to zero.

\[ \vec{a}(t) \cdot \vec{v}(t) = (2) \cdot (2t) + (6t) \cdot (3t^2 - 2) \]

Calculating the dot product:

\[ = 4t + 18t^3 - 12t = 18t^3 - 8t \]

Setting the dot product to zero:

\[ 18t^3 - 8t = 0 \]

Factoring out \( t \):

\[ t(18t^2 - 8) = 0 \]

This gives us \( t = 0 \) or \( 18t^2 - 8 = 0 \). Solving for \( t \):

\[ 18t^2 = 8 \implies t^2 = \frac{8}{18} = \frac{4}{9} \implies t = \frac{2}{3} \]

### Step 4: Verify the condition at \( t = \frac{2}{3} \)
Now we substitute \( t = \frac{2}{3} \) into the expression for the dot product:

\[ \vec{a}\left(\frac{2}{3}\right) \cdot \vec{v}\left(\frac{2}{3}\right) = 18\left(\frac{2}{3}\right)^3 - 8\left(\frac{2}{3}\right) \]

Calculating:

\[ = 18 \cdot \frac{8}{27} - \frac{16}{3} = \frac{144}{27} - \frac{144}{27} = 0 \]

Since the dot product is zero, the velocity and acceleration are perpendicular at \( t = \frac{2}{3} \).

### Step 5: Determine the slope of the particle's path
To find when the particle is moving parallel to the x-axis, we need to set the slope \( \frac{dy}{dx} = 0 \).

Calculating \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \):

\[ \frac{dy}{dt} = 3t^2 - 2, \quad \frac{dx}{dt} = 2t \]

Setting \( \frac{dy}{dx} = 0 \):

\[ \frac{dy}{dx} = \frac{3t^2 - 2}{2t} = 0 \]

This gives us:

\[ 3t^2 - 2 = 0 \implies t^2 = \frac{2}{3} \implies t = \sqrt{\frac{2}{3}} \]

### Conclusion
The particle is moving parallel to the x-axis when \( t = \sqrt{\frac{2}{3}} \).