A particle moves along a parabolic parth `y = 9x^(2)` in such a way that the x-component of velocity remains constant and has a value `1/3 ms ^(-1)` . The acceleration of the particle is -

Options

A. `1/3overset^lms^(-2)`
B. `3overset^jms^(-2)`
C. `2/3overset^jms^(-2)`
D. `2overset^jms^(-2)`

Answer

To solve the problem, we need to find the acceleration of a particle moving along the parabolic path given by the equation \( y = 9x^2 \), where the x-component of velocity is constant at \( \frac{1}{3} \, \text{m/s} \).

### Step-by-Step Solution:

1. **Identify the given information:**
- The path of the particle is described by the equation \( y = 9x^2 \).
- The x-component of velocity, \( v_x = \frac{1}{3} \, \text{m/s} \).
- Since \( v_x \) is constant, the acceleration in the x-direction, \( a_x = 0 \).

2. **Differentiate the path equation to find the y-component of velocity:**
- We start with the equation \( y = 9x^2 \).
- Differentiate \( y \) with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]
- First, find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = 18x \]
- Now, substitute \( \frac{dx}{dt} = v_x = \frac{1}{3} \):
\[ \frac{dy}{dt} = 18x \cdot \frac{1}{3} = 6x \]

3. **Differentiate again to find the y-component of acceleration:**
- Differentiate \( \frac{dy}{dt} \) with respect to time \( t \):
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(6x) = 6 \frac{dx}{dt} \]
- Since \( \frac{dx}{dt} = v_x = \frac{1}{3} \):
\[ \frac{d^2y}{dt^2} = 6 \cdot \frac{1}{3} = 2 \, \text{m/s}^2 \]

4. **Determine the total acceleration of the particle:**
- The total acceleration \( \vec{a} \) has components:
- \( a_x = 0 \)
- \( a_y = 2 \, \text{m/s}^2 \)
- Therefore, the acceleration vector is:
\[ \vec{a} = 0 \hat{i} + 2 \hat{j} = 2 \hat{j} \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the particle is \( 2 \, \text{m/s}^2 \) in the positive y-direction.

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