A particle move according to the law a=-ky find the velocity as a function of distance y where v zero is veloctiy

To find the velocity of a particle as a function of distance \( y \) given that its acceleration \( a \) is proportional to the negative of its position, we start with the relationship between acceleration, velocity, and position. The acceleration \( a \) can be expressed as the derivative of velocity \( v \) with respect to time \( t \), and we know that \( a = -ky \), where \( k \) is a positive constant. Let's break this down step by step.

Understanding the Relationship

We know that acceleration \( a \) can be defined as:

• \( a = \frac{dv}{dt} \) (the rate of change of velocity with respect to time)
• \( a = v \frac{dv}{dy} \) (using the chain rule, where \( \frac{dv}{dy} \) is the rate of change of velocity with respect to position)

By equating these two expressions for acceleration, we have:

\( v \frac{dv}{dy} = -ky \)

Separating Variables

Next, we can separate the variables to isolate \( v \) and \( y \). Rearranging the equation gives us:

\( v \, dv = -ky \, dy \)

Integrating Both Sides

Now we can integrate both sides. The left side integrates with respect to \( v \), and the right side integrates with respect to \( y \):

• Left side: \( \int v \, dv = \frac{v^2}{2} + C_1 \)
• Right side: \( \int -ky \, dy = -\frac{ky^2}{2} + C_2 \)

Combining these results, we have:

\( \frac{v^2}{2} = -\frac{ky^2}{2} + C \)

where \( C = C_2 - C_1 \) is a constant of integration.

Solving for Velocity

To express \( v \) in terms of \( y \), we can rearrange the equation:

\( v^2 = -ky^2 + 2C \)

Taking the square root gives us:

\( v = \sqrt{2C - ky^2} \)

Determining the Constant

To find the constant \( C \), we can use the initial condition provided in the problem. If we know the initial velocity \( v_0 \) when \( y = 0 \), we can substitute these values into our equation:

\( v_0 = \sqrt{2C} \)

From this, we can solve for \( C \):

\( C = \frac{v_0^2}{2} \)

Final Expression for Velocity

Substituting \( C \) back into our equation for \( v \), we get:

\( v = \sqrt{v_0^2 - ky^2} \)

This equation shows how the velocity of the particle changes as a function of the distance \( y \). The velocity decreases as \( y \) increases, reflecting the negative acceleration proportional to the position. This relationship is typical in systems where a restoring force acts against the motion, such as in harmonic oscillators.