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 = 7.89 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis.
magnitude    	  

direction    	  ° counterclockwise from the +x-axis


To solve for the average velocity of the Moon over the interval from \( t = 0 \) to \( t = 7.89 \) days, we need to compute both the magnitude and direction of the average velocity vector.

### 1. **Position functions**
The position of the Moon as a function of time is given by:
\[
x(t) = r \cos(\omega t)
\]
\[
y(t) = r \sin(\omega t)
\]
where:
- \( r = 3.84 \times 10^8 \, \text{m} \) (the orbital radius),
- \( \omega = 2.46 \times 10^{-6} \, \text{radians/s} \) (angular velocity).

### 2. **Average velocity formula**
The average velocity is given by:
\[
\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t}
\]
where \( \Delta \vec{r} = \vec{r}(t_{\text{final}}) - \vec{r}(t_{\text{initial}}) \) is the change in position over time.

Since we are asked for the average velocity from \( t = 0 \) to \( t = 7.89 \, \text{days} \), the total time interval is:
\[
\Delta t = 7.89 \, \text{days} = 7.89 \times 24 \times 3600 \, \text{seconds} = 681696 \, \text{seconds}.
\]

### 3. **Position at initial and final times**
At \( t = 0 \):
\[
x(0) = r \cos(0) = r = 3.84 \times 10^8 \, \text{m},
\]
\[
y(0) = r \sin(0) = 0.
\]

At \( t = 7.89 \, \text{days} \):
\[
x(7.89) = r \cos(\omega \cdot 681696),
\]
\[
y(7.89) = r \sin(\omega \cdot 681696).
\]

Let's calculate these values.At \( t = 0 \):
- \( x(0) = 3.84 \times 10^8 \, \text{m} \)
- \( y(0) = 0 \, \text{m} \)

At \( t = 7.89 \, \text{days} \):
- \( x(7.89) \approx -4.07 \times 10^7 \, \text{m} \)
- \( y(7.89) \approx 3.82 \times 10^8 \, \text{m} \)

### 4. **Change in position (\(\Delta r\))**
The change in position is:
\[
\Delta x = x_{\text{final}} - x_{\text{initial}} = -4.07 \times 10^7 \, \text{m} - 3.84 \times 10^8 \, \text{m}
\]
\[
\Delta y = y_{\text{final}} - y_{\text{initial}} = 3.82 \times 10^8 \, \text{m} - 0 \, \text{m}
\]

Now let's calculate the magnitude and direction of the average velocity:
\[
\text{Magnitude} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t}
\]
\[
\text{Direction} = \text{atan2}(\Delta y, \Delta x)
\]

I'll compute these next.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, I can walk you through how to calculate the magnitude and direction of the average velocity step by step. Let me know if you'd like to proceed that way.

Here are some related questions you might want to explore:
1. How does angular velocity relate to orbital velocity?
2. What is the formula for centripetal force in circular motion?
3. How does changing the orbital radius affect the Moon's velocity?
4. What is the Moon's average speed during its orbit?
5. How can the direction of motion in circular paths be described using vectors?

**Tip:** When working with circular motion, breaking motion into components (like \(x(t)\) and \(y(t)\)) often helps solve complex problems.

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 x 10^8 m and 𝜔 = 2.46 x 10^-6 radians/s. What is the average velocity of the Moon measured over the interval from t = 0 to t = 7.89 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis.

Explore how to calculate the average velocity of the Moon in its circular orbit around the Earth over a period of 7.89 days. Using angular velocity, trigonometric functions, and vector calculations, this problem illustrates key concepts in circular motion and vector magnitude. Suitable for high school physics and trigonometry students.

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