The Moon orbits the Earth in an approximately circular path. The position of the moon as a function of time is given by:
x(t)	 = 	r cos(𝜔t)
y(t)	 = 	r sin(𝜔t)
where r = 3.84  108 m and 𝜔 = 2.46  10-6 radians/s. What is the average velocity of the Moon measured over the interval from t = 0 to t = 6.9 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis.
magnitude    	  m/s
direction    	  ° counterclockwise from the +x-axis

The position of the Moon as a function of time is given by the parametric equations:

\[
x(t) = r \cos(\omega t)
\]
\[
y(t) = r \sin(\omega t)
\]

Where:
- \( r = 3.84 \times 10^8 \, \text{m} \)
- \( \omega = 2.46 \times 10^{-6} \, \text{rad/s} \)

### Step 1: Convert the time interval to seconds

The given time interval is \( t = 6.9 \, \text{days} \). We first need to convert this into seconds:
\[
t = 6.9 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour}
\]

### Step 2: Determine the positions at \( t = 0 \) and \( t = 6.9 \) days

At \( t = 0 \), the position of the Moon is:
\[
x(0) = r \cos(0) = r = 3.84 \times 10^8 \, \text{m}
\]
\[
y(0) = r \sin(0) = 0
\]

At \( t = 6.9 \, \text{days} \), the position of the Moon is:
\[
x(6.9) = r \cos(\omega t)
\]
\[
y(6.9) = r \sin(\omega t)
\]
Substitute \( \omega t = 2.46 \times 10^{-6} \times t \).

### Step 3: Calculate the displacement vector

The displacement vector over the time interval is:
\[
\Delta x = x(6.9) - x(0)
\]
\[
\Delta y = y(6.9) - y(0)
\]

### Step 4: Compute the average velocity

The average velocity magnitude is given by:
\[
v_{\text{avg}} = \frac{\sqrt{(\Delta x)^2 + (\Delta y)^2}}{\Delta t}
\]

Where \( \Delta t = 6.9 \, \text{days} \) converted to seconds.

### Step 5: Direction of the average velocity

The direction is the angle \( \theta \) of the displacement vector relative to the positive x-axis:
\[
\theta = \tan^{-1}\left(\frac{\Delta y}{\Delta x}\right)
\]

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Here are 5 relative questions to expand this concept:
1. How does the angular velocity \( \omega \) affect the Moon's orbital period?
2. What is the formula for instantaneous velocity in circular motion?
3. How can we calculate the centripetal acceleration of the Moon given the parameters?
4. What role does the gravitational force play in maintaining the Moon's orbit?
5. How can we estimate the time it takes for the Moon to complete one full orbit?

**Tip:** The direction of velocity in circular motion is always tangent to the orbit at any given point.

The Moon orbits the Earth in an approximately circular path. The position of the moon as a function of time is given by: x(t) = r cos(𝜔t) y(t) = r sin(𝜔t) where r = 3.84 * 10^8 m and 𝜔 = 2.46 * 10^-6 radians/s. What is the average velocity of the Moon measured over the interval from t = 0 to t = 6.9 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis.

This problem explores the average velocity of the Moon in its orbit around the Earth over a 6.9-day period. Using trigonometric and circular motion formulas, we calculate both the magnitude and direction of the Moon's velocity, applying key concepts like angular velocity and displacement. Perfect for advanced high school or undergraduate physics students.

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