A point mass performs straight line motion along positive x-axis At t=0 point mass is at point `A(x_(1),0)` it moves such that its velocity is given by `v=(a)/(x)`, where a is positive constant and x is the x-coordinate of position vector mass at a certain time t. Find the time required to move from A to B `(x_(2),0)`

Options

A. `(x_(2)^(2)-x_(1)^(2))/(2a)`
B. `(x_(2)^(2)-x_(1)^(2))/(a)`
C. `(2x_(2)^(2)-x_(1)^(2))/(2a)`
D. `(2x_(2)^(2)-x_(1)^(2))/(a)`

Explanation

To solve the problem step by step, we will analyze the motion of the point mass and derive the time required to move from point A to point B.

### Step 1: Understand the given information
We know that:
- The point mass starts at position \( A(x_1, 0) \) at time \( t = 0 \).
- The velocity of the mass is given by the equation \( v = \frac{a}{x} \), where \( a \) is a positive constant and \( x \) is the position of the mass.

### Step 2: Relate velocity to position and time
The velocity \( v \) can also be expressed as the derivative of position with respect to time:
\[ v = \frac{dx}{dt} \]
Setting the two expressions for velocity equal gives us:
\[ \frac{dx}{dt} = \frac{a}{x} \]

### Step 3: Rearranging the equation
We can rearrange this equation to separate variables:
\[ x \, dx = a \, dt \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side will be integrated with respect to \( x \) from \( x_1 \) to \( x_2 \), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_{x_1}^{x_2} x \, dx = \int_{0}^{t} a \, dt \]

### Step 5: Perform the integration
The left side integrates to:
\[ \frac{x^2}{2} \bigg|_{x_1}^{x_2} = \frac{x_2^2}{2} - \frac{x_1^2}{2} \]
The right side integrates to:
\[ a t \bigg|_{0}^{t} = a t - 0 = a t \]

### Step 6: Set the integrals equal to each other
Now we can set the results of the integrals equal to each other:
\[ \frac{x_2^2}{2} - \frac{x_1^2}{2} = a t \]

### Step 7: Solve for time \( t \)
Rearranging the equation gives:
\[ \frac{x_2^2 - x_1^2}{2} = a t \]
Thus, solving for \( t \):
\[ t = \frac{x_2^2 - x_1^2}{2a} \]

### Final Result
The time required to move from point A to point B is:
\[ t = \frac{x_2^2 - x_1^2}{2a} \]

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