Two particles leave the origin at the same time. The first particle travels along the y-axis and its distance, in meters, from the origin at time t seconds is Y (t) = 3t. The second particle travels along the x-axis and its distance from the origin, in meters, at time t seconds is X(t) = 4t. a. What is the distance between the two particles at time t? b. At what rate is the distance between the two particles changing at time t?

We have two particles traveling along the coordinate axes:

- Particle 1 moves along the \( y \)-axis with a position given by \( Y(t) = 3t \).
- Particle 2 moves along the \( x \)-axis with a position given by \( X(t) = 4t \).

### Part (a): Distance Between the Particles at Time \( t \)
At any time \( t \), the particles are located at points \( (0, 3t) \) and \( (4t, 0) \). The distance between them is the straight-line distance, which can be found using the Pythagorean theorem.

The distance \( D(t) \) between the two particles is given by:

\[
D(t) = \sqrt{(X(t) - 0)^2 + (Y(t) - 0)^2}
\]
Substitute \( X(t) = 4t \) and \( Y(t) = 3t \):

\[
D(t) = \sqrt{(4t)^2 + (3t)^2}
= \sqrt{16t^2 + 9t^2}
= \sqrt{25t^2}
= 5t
\]

Thus, the distance between the two particles at time \( t \) is:

\[
D(t) = 5t
\]

### Part (b): Rate of Change of Distance Between the Particles at Time \( t \)
To find how the distance between the particles changes over time, we compute the derivative of \( D(t) \) with respect to \( t \).

\[
\frac{dD(t)}{dt} = \frac{d}{dt} \left( 5t \right) = 5
\]

Thus, the rate of change of the distance between the two particles is constant and equal to:

\[
\frac{dD(t)}{dt} = 5 \, \text{meters per second}
\]

### Summary:
- **Part (a):** The distance between the two particles at time \( t \) is \( 5t \) meters.
- **Part (b):** The rate at which the distance between the particles is changing is \( 5 \, \text{m/s} \).

Would you like more details on any part? Here are some related questions:

1. What is the physical interpretation of the rate of change of distance between two particles?
2. How would the result change if one of the particles moved in a different direction, such as diagonally?
3. How do you find the velocity of each particle at a given time?
4. What if the particles were accelerating instead of moving at constant velocities?
5. How does this problem relate to real-world physics scenarios?

**Tip:** When dealing with problems involving distance and rates of change, always check if you can apply the Pythagorean theorem to simplify the calculations.

Two particles leave the origin at the same time. The first particle travels along the y-axis and its distance, in meters, from the origin at time t seconds is Y(t) = 3t. The second particle travels along the x-axis and its distance from the origin, in meters, at time t seconds is X(t) = 4t. a. What is the distance between the two particles at time t? b. At what rate is the distance between the two particles changing at time t?

Learn how to calculate the distance between two particles traveling along the x and y axes using the Pythagorean theorem and find the rate of change of that distance over time. This problem explores kinematics, algebra, and basic calculus concepts. Suitable for students in Grades 10-12.

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