material particle is chasing another one and both of them are moving on the same straight line. After they pass a particular point, their velocities v vary with timet as shown in the figure. When will the chase end?

To determine when the chase between two material particles will end, we need to analyze their velocities over time and how their positions change as they move along the same straight line. The chase will conclude when both particles are at the same position, meaning the distance covered by both particles becomes equal at some point in time.

Understanding Velocity and Position

First, let's clarify the relationship between velocity, time, and position. The position of an object can be calculated by integrating its velocity over time. If we denote the velocities of the two particles as \( v_1(t) \) and \( v_2(t) \), their positions \( x_1(t) \) and \( x_2(t) \) can be expressed as:

• Particle 1: \( x_1(t) = x_{1_0} + \int v_1(t) \, dt \)
• Particle 2: \( x_2(t) = x_{2_0} + \int v_2(t) \, dt \)

Here, \( x_{1_0} \) and \( x_{2_0} \) are the initial positions of the two particles at the starting point of the chase. If we assume they start from the same point, we can set \( x_{1_0} = x_{2_0} \).

Finding the Time of Intersection

To find when the chase ends, we need to set the positions equal to each other:

Condition for the chase to end: \( x_1(t) = x_2(t) \)

This leads us to the equation:

\( \int v_1(t) \, dt = \int v_2(t) \, dt \)

To solve this, we can differentiate both sides with respect to time, which gives us:

\( v_1(t) = v_2(t) \)

However, this condition alone does not guarantee that the chase will end, as it only indicates that their velocities are equal at that moment. We must also consider their initial positions and how far each has traveled over time.

Analyzing the Velocity Graph

If you have a graph showing the velocities of both particles over time, you can identify the points where the velocities intersect. These points indicate when the particles are moving at the same speed. However, to determine when the chase ends, you need to check their positions at these times.

For example, if you find that:

• At time \( t_1 \), \( v_1(t_1) = v_2(t_1) \) but \( x_1(t_1) < x_2(t_1) \)
• At time \( t_2 \), \( v_1(t_2) = v_2(t_2) \) and \( x_1(t_2) = x_2(t_2) \)

Then the chase ends at \( t_2 \) because that is when both particles occupy the same position.

Example Scenario

Let’s consider a practical example. Suppose Particle 1 has a velocity function defined as \( v_1(t) = 2t \) and Particle 2 has \( v_2(t) = 3t - 1 \). We can find their positions:

• For Particle 1: \( x_1(t) = \int 2t \, dt = t^2 + C_1 \)
• For Particle 2: \( x_2(t) = \int (3t - 1) \, dt = \frac{3}{2}t^2 - t + C_2 \)

Assuming both start from the same position (let's say \( C_1 = C_2 = 0 \)), we can set their position equations equal to find the time when they meet:

\( t^2 = \frac{3}{2}t^2 - t \)

Solving this equation will give us the time at which the chase ends. By rearranging and simplifying, we can find the specific time value.

Final Thoughts

In summary, to determine when the chase ends, analyze the velocity functions, integrate them to find position functions, and then solve for the time when both positions are equal. This approach will give you a clear understanding of the dynamics involved in the chase between the two particles.