A particle moves in the plane xy with constant acceleration 'a' directed along the negative y-axis. The equation of motion of the particle has the form `y= px - qx^2` where p and q are positive constants. Find the velocity of the particle at the origin of co-ordinates.

Explanation

To find the velocity of the particle at the origin of coordinates given the equation of motion \( y = px - qx^2 \), we can follow these steps:

### Step 1: Understand the motion
The equation \( y = px - qx^2 \) describes the trajectory of the particle in the xy-plane. The acceleration is directed along the negative y-axis, which indicates that the particle is experiencing a downward acceleration.

### Step 2: Identify parameters
From the equation of motion, we can identify:
- The coefficient \( p \) corresponds to the slope of the trajectory at the origin.
- The coefficient \( q \) relates to the curvature of the trajectory.

### Step 3: Compare with the general projectile motion equation
The general equation for projectile motion is given by:
\[ y = x \tan(\theta) - \frac{g x^2}{2u^2(1 + \tan^2(\theta))} \]
where:
- \( g \) is the acceleration due to gravity (which we can replace with \( a \) since the acceleration is constant).
- \( u \) is the initial velocity.
- \( \theta \) is the angle of projection.

### Step 4: Match coefficients
By comparing the coefficients from both equations, we can write:
1. \( p = \tan(\theta) \)
2. \( q = \frac{a}{2u^2(1 + \tan^2(\theta))} \)

### Step 5: Express \( u \) in terms of \( p \) and \( q \)
From the second equation, we can rearrange to find \( u^2 \):
\[ u^2 = \frac{a}{2q(1 + \tan^2(\theta))} \]
Substituting \( \tan(\theta) = p \):
\[ u^2 = \frac{a}{2q(1 + p^2)} \]

### Step 6: Solve for \( u \)
Taking the square root to find \( u \):
\[ u = \sqrt{\frac{a}{2q(1 + p^2)}} \]

### Step 7: Conclusion
Thus, the velocity of the particle at the origin of coordinates is given by:
\[ u = \sqrt{\frac{a}{2q(1 + p^2)}} \]