the velocity v and displacement r of a body are related as v²=kr, where k is constant. what will be trhe velocity after 1 second/ (given that displacement is zero at t=0)

To find the velocity of a body after 1 second, given the relationship \( v^2 = kr \) and that the displacement \( r \) is zero at \( t = 0 \), we need to analyze the situation step by step.

Understanding the Relationship

The equation \( v^2 = kr \) indicates that the square of the velocity \( v \) is directly proportional to the displacement \( r \). Here, \( k \) is a constant that relates the two quantities. When \( r = 0 \), we can see that \( v^2 \) must also be zero, which implies that \( v = 0 \) at that moment.

Initial Conditions

At time \( t = 0 \), since the displacement \( r \) is zero, the velocity \( v \) is also zero. This means that the body starts from rest. As time progresses, the displacement will change, and so will the velocity.

Finding Velocity Over Time

To determine how the velocity changes after 1 second, we need to consider how displacement evolves over time. Since we have a relationship between \( v \) and \( r \), we can express \( v \) in terms of \( r \) and then relate it to time.

Using Calculus

We know that velocity is the derivative of displacement with respect to time, or \( v = \frac{dr}{dt} \). From the equation \( v^2 = kr \), we can express \( v \) as:

• \( v = \sqrt{kr} \)

Substituting this into the derivative gives us:

• \( \frac{dr}{dt} = \sqrt{kr} \)

Solving the Differential Equation

This is a separable differential equation. We can rearrange it as follows:

• \( \frac{dr}{\sqrt{r}} = \sqrt{k} \, dt \)

Integrating both sides, we get:

• \( 2\sqrt{r} = \sqrt{k} t + C \)

Where \( C \) is the constant of integration. To find \( C \), we use the initial condition \( r(0) = 0 \):

• \( 2\sqrt{0} = \sqrt{k} \cdot 0 + C \)
• Thus, \( C = 0 \).

This simplifies our equation to:

• \( 2\sqrt{r} = \sqrt{k} t \)

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

• \( r = \frac{k t^2}{4} \)

Calculating Velocity at \( t = 1 \) Second

Now, we can substitute \( t = 1 \) second into our equation for \( r \):

• \( r(1) = \frac{k \cdot 1^2}{4} = \frac{k}{4} \)

Next, we can find the velocity at this displacement:

• \( v = \sqrt{kr} = \sqrt{k \cdot \frac{k}{4}} = \sqrt{\frac{k^2}{4}} = \frac{k}{2} \)

Final Result

Therefore, the velocity of the body after 1 second is \( \frac{k}{2} \). This shows how the relationship between velocity and displacement evolves over time, starting from rest and increasing as the body moves away from its initial position.