A particle moves along the parabola `y^2=2ax` in such a way that its projection on y-axis has a constant velocity. Show that its projection on the x-axis moves with constant acceleration

Options

A. constant velocity
B. constant acceleration
C. variable velocity
D. variable acceleration

Answer

To solve the problem step by step, we will analyze the motion of a particle along the parabola defined by the equation \( y^2 = 2ax \). We will show that if the projection of the particle on the y-axis has a constant velocity, then its projection on the x-axis moves with constant acceleration.

### Step 1: Understand the given equation
The equation of the parabola is given by:
\[ y^2 = 2ax \]
This relates the coordinates \(x\) and \(y\) of the particle.

### Step 2: Differentiate the equation with respect to time \(t\)
To find the relationship between the velocities in the x and y directions, we differentiate both sides of the equation with respect to time \(t\):
\[ \frac{d}{dt}(y^2) = \frac{d}{dt}(2ax) \]
Using the chain rule, we get:
\[ 2y \frac{dy}{dt} = 2a \frac{dx}{dt} \]
Dividing both sides by 2, we have:
\[ y \frac{dy}{dt} = a \frac{dx}{dt} \]

### Step 3: Express \(\frac{dy}{dt}\) in terms of \(\frac{dx}{dt}\)
From the equation derived, we can express \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} = \frac{a}{y} \frac{dx}{dt} \]

### Step 4: Differentiate again with respect to time \(t\)
Next, we differentiate \(\frac{dy}{dt}\) with respect to time \(t\) to find the acceleration:
\[ \frac{d^2y}{dt^2} = \frac{d}{dt}\left(\frac{a}{y} \frac{dx}{dt}\right) \]
Using the product rule, we get:
\[ \frac{d^2y}{dt^2} = \frac{a}{y} \frac{d^2x}{dt^2} + \frac{dx}{dt} \frac{d}{dt}\left(\frac{a}{y}\right) \]
Now, we need to differentiate \(\frac{a}{y}\):
\[ \frac{d}{dt}\left(\frac{a}{y}\right) = -\frac{a}{y^2} \frac{dy}{dt} \]
Substituting this back into our equation gives:
\[ \frac{d^2y}{dt^2} = \frac{a}{y} \frac{d^2x}{dt^2} - \frac{a}{y^2} \frac{dy}{dt} \frac{dx}{dt} \]

### Step 5: Use the condition of constant velocity in the y-direction
Since it is given that the projection on the y-axis has a constant velocity, we have:
\[ \frac{dy}{dt} = k \quad (\text{constant}) \]
Thus, \(\frac{d^2y}{dt^2} = 0\). Substituting this into our equation gives:
\[ 0 = \frac{a}{y} \frac{d^2x}{dt^2} - \frac{a}{y^2} k \frac{dx}{dt} \]

### Step 6: Rearranging the equation
Rearranging the above equation leads to:
\[ \frac{a}{y} \frac{d^2x}{dt^2} = \frac{a}{y^2} k \frac{dx}{dt} \]
Multiplying through by \(y\) gives:
\[ \frac{d^2x}{dt^2} = \frac{k}{y} \frac{dx}{dt} \]

### Step 7: Show that \(\frac{d^2x}{dt^2}\) is constant
Since \(k\) is constant and \(y\) is a function of \(x\) (from the parabola equation), we can express \(y\) in terms of \(x\):
\[ y = \sqrt{2ax} \]
Substituting this into the equation shows that \(\frac{d^2x}{dt^2}\) is proportional to a constant, thus indicating that the acceleration in the x-direction is constant.

### Conclusion
We have shown that if the projection of the particle on the y-axis moves with constant velocity, then the projection on the x-axis moves with constant acceleration.