A particle starts from point A moves along a straight line path with an acceleration given by a = p - qx where p, q are connected and x is distance from point A . The particle stops at point B. The maximum velocity of the particle is

Options

A. `p/q`
B. `p/sqrt(q)`
C. `q/p`
D. `sqrt(q/p)`

Explanation

To find the maximum velocity of the particle moving from point A to point B with the given acceleration \( a = p - qx \), we can follow these steps:

### Step 1: Understand the relationship between acceleration and maximum velocity
The particle will reach its maximum velocity when the acceleration becomes zero. Therefore, we set the acceleration equation to zero:

\[ 0 = p - qx \]

### Step 2: Solve for \( x \)
From the equation \( 0 = p - qx \), we can rearrange it to find \( x \):

\[ qx = p \implies x = \frac{p}{q} \]

### Step 3: Express acceleration in terms of velocity
We know that acceleration \( a \) can also be expressed in terms of velocity \( v \) and position \( x \):

\[ a = v \frac{dv}{dx} \]

### Step 4: Substitute the expression for acceleration
Now we can substitute the expression for acceleration into our equation:

\[ v \frac{dv}{dx} = p - qx \]

### Step 5: Rearrange and integrate
Rearranging gives us:

\[ v \, dv = (p - qx) \, dx \]

Now, we can integrate both sides. The left side integrates to:

\[ \int v \, dv = \frac{v^2}{2} \]

The right side integrates as follows:

\[ \int (p - qx) \, dx = px - \frac{qx^2}{2} \]

### Step 6: Combine results from integration
Combining the results from both sides, we have:

\[ \frac{v^2}{2} = px - \frac{qx^2}{2} \]

### Step 7: Solve for \( v^2 \)
Multiplying through by 2 gives:

\[ v^2 = 2px - qx^2 \]

### Step 8: Substitute \( x = \frac{p}{q} \)
Now, we substitute \( x = \frac{p}{q} \) into the equation for \( v^2 \):

\[ v^2 = 2p\left(\frac{p}{q}\right) - q\left(\frac{p}{q}\right)^2 \]

This simplifies to:

\[ v^2 = \frac{2p^2}{q} - \frac{qp^2}{q^2} \]

### Step 9: Simplify the expression
Combining the terms gives us:

\[ v^2 = \frac{2p^2}{q} - \frac{p^2}{q} = \frac{p^2}{q} \]

### Step 10: Find \( v \)
Taking the square root gives us the maximum velocity:

\[ v = \sqrt{\frac{p^2}{q}} = \frac{p}{\sqrt{q}} \]

### Conclusion
Thus, the maximum velocity of the particle is:

\[ \boxed{\frac{p}{\sqrt{q}}} \]