A particle moves along the parabolic path `y=ax^(2)` in such a The accleration of the particle is:

Options

A. `2a^(2)chatj`
B. `2ac^(2)hatj`
C. `achatk`
D. `a^(2)c^(2)hatk`

Explanation

To find the acceleration of a particle moving along the parabolic path given by \( y = ax^2 \), we will follow these steps:

### Step 1: Establish the relationship between \( x \) and \( t \)

Given that the initial velocity in the x-direction \( u_x = c \), we can express the relationship between \( x \) and \( t \) as follows:

\[ \frac{dx}{dt} = c \]

Integrating both sides, we have:

\[ \int dx = \int c \, dt \]

This gives us:

\[ x = ct + C \]

Assuming the particle starts from the origin (when \( t = 0, x = 0 \)), we find \( C = 0 \). Thus, the relationship becomes:

\[ x = ct \]

### Step 2: Substitute \( x \) into the equation for \( y \)

Now we substitute \( x = ct \) into the equation for \( y \):

\[ y = a x^2 = a (ct)^2 = a c^2 t^2 \]

### Step 3: Find the velocity in the y-direction

To find the velocity in the y-direction, we differentiate \( y \) with respect to \( t \):

\[ \frac{dy}{dt} = \frac{d}{dt}(a c^2 t^2) = a c^2 \cdot 2t = 2a c^2 t \]

### Step 4: Find the acceleration in the y-direction

Next, we differentiate the velocity in the y-direction to find the acceleration:

\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(2a c^2 t) = 2a c^2 \]

### Step 5: Express the acceleration vector

Since the particle is only moving in the y-direction (the x-direction does not have any acceleration), we can express the acceleration vector as:

\[ \vec{a} = 0 \hat{i} + 2a c^2 \hat{j} \]

Thus, the acceleration of the particle is:

\[ \vec{a} = 2a c^2 \hat{j} \]

### Final Answer

The acceleration of the particle is \( 2a c^2 \hat{j} \).

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