A particle moves along the parabolic path `x = y^2 + 2y + 2` in such a way that Y-component of velocity vector remains `5ms^(-1)` during the motion. The magnitude of the acceleration of the particle is

Options

A. `50 ms^(-2)`
B. `100 ms^(-2)`
C. `10(sqrt2) ms^(-2)`
D. `0.1 ms^(-2)`

Explanation

To solve the problem, we need to find the magnitude of the acceleration of a particle moving along the parabolic path defined by the equation \( x = y^2 + 2y + 2 \), with the Y-component of the velocity remaining constant at \( 5 \, \text{m/s} \).

### Step-by-Step Solution:

1. **Understand the given equation**:
The path of the particle is given by:
\[ x = y^2 + 2y + 2 \]
This is a parabolic equation in terms of \( y \).

2. **Differentiate the equation with respect to time**:
To find the relationship between the velocities in the x and y directions, we differentiate both sides with respect to time \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(y^2 + 2y + 2) \]
Using the chain rule, we get:
\[ \frac{dx}{dt} = 2y \frac{dy}{dt} + 2 \frac{dy}{dt} \]
Simplifying this, we can factor out \( \frac{dy}{dt} \):
\[ \frac{dx}{dt} = (2y + 2) \frac{dy}{dt} \]

3. **Substitute the known value of \( \frac{dy}{dt} \)**:
We know that the Y-component of the velocity \( \frac{dy}{dt} = 5 \, \text{m/s} \). Substituting this into the equation gives:
\[ \frac{dx}{dt} = (2y + 2) \cdot 5 \]
Thus:
\[ \frac{dx}{dt} = 10y + 10 \]

4. **Determine the acceleration in the x-direction**:
To find the acceleration, we need to differentiate \( \frac{dx}{dt} \) with respect to time:
\[ \frac{d^2x}{dt^2} = \frac{d}{dt}(10y + 10) \]
Again applying the chain rule:
\[ \frac{d^2x}{dt^2} = 10 \frac{dy}{dt} \]
Substituting \( \frac{dy}{dt} = 5 \, \text{m/s} \):
\[ \frac{d^2x}{dt^2} = 10 \cdot 5 = 50 \, \text{m/s}^2 \]

5. **Calculate the total acceleration**:
Since the Y-component of the velocity is constant, the Y-component of acceleration is zero. Therefore, the total acceleration of the particle is purely in the X-direction:
\[ a = \sqrt{(a_x)^2 + (a_y)^2} = \sqrt{(50)^2 + (0)^2} = 50 \, \text{m/s}^2 \]

### Final Answer:
The magnitude of the acceleration of the particle is \( 50 \, \text{m/s}^2 \).

---