A particle moves uniformly with speed v along a parabolic path `y = kx^(2)`, where k is a positive constant. Find the acceleration of the particle at the point x = 0.

Answer

To find the acceleration of a particle moving uniformly along a parabolic path given by the equation \( y = kx^2 \), where \( k \) is a positive constant, we can follow these steps:

### Step 1: Understand the Path
The path of the particle is defined by the equation \( y = kx^2 \). This is a parabolic curve that opens upwards.

### Step 2: Differentiate the Path Equation
To find the velocity components of the particle, we need to differentiate the path equation with respect to time \( t \).

1. Differentiate \( y = kx^2 \):
\[ \frac{dy}{dt} = \frac{d(kx^2)}{dt} = 2kx \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is the velocity in the x-direction, denoted as \( v_x \).

### Step 3: Find the Velocity Components
At any point on the path, the velocity in the y-direction \( v_y \) is given by:
\[ v_y = 2kx \cdot v_x \]
where \( v_x \) is the speed of the particle in the x-direction.

### Step 4: Differentiate Again to Find Acceleration
Next, we differentiate \( v_y \) with respect to time \( t \) to find the acceleration in the y-direction \( a_y \):
\[ a_y = \frac{d(v_y)}{dt} = \frac{d(2kx \cdot v_x)}{dt} \]
Using the product rule:
\[ a_y = 2k \left( \frac{dx}{dt} \cdot v_x + x \cdot \frac{dv_x}{dt} \right) = 2k(v_x \cdot v_x + x \cdot a_x) \]
where \( a_x \) is the acceleration in the x-direction.

### Step 5: Evaluate at \( x = 0 \)
At the point \( x = 0 \):
- The term \( x \cdot a_x \) becomes \( 0 \).
- The velocity \( v_x \) remains \( v \) (the constant speed of the particle).

Thus, the acceleration in the y-direction simplifies to:
\[ a_y = 2k \cdot v^2 \]

### Conclusion
The acceleration of the particle at the point \( x = 0 \) is:
\[ \boxed{2kv^2} \]

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