Math Problem Statement

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(πœ”t) y(t) = r sin(πœ”t) where r = 3.84 108 m and πœ” = 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 Β° counterclockwise from the +x-axis

Solution

To solve for the average velocity of the Moon over the interval from t=0t = 0 to t=7.89t = 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)=rcos⁑(Ο‰t)x(t) = r \cos(\omega t) y(t)=rsin⁑(Ο‰t)y(t) = r \sin(\omega t) where:

  • r=3.84Γ—108 mr = 3.84 \times 10^8 \, \text{m} (the orbital radius),
  • Ο‰=2.46Γ—10βˆ’6 radians/s\omega = 2.46 \times 10^{-6} \, \text{radians/s} (angular velocity).

2. Average velocity formula

The average velocity is given by: vβƒ—avg=Ξ”rβƒ—Ξ”t\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t} where Ξ”rβƒ—=rβƒ—(tfinal)βˆ’rβƒ—(tinitial)\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=0t = 0 to t=7.89 dayst = 7.89 \, \text{days}, the total time interval is: Ξ”t=7.89 days=7.89Γ—24Γ—3600 seconds=681696 seconds.\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=0t = 0: x(0)=rcos⁑(0)=r=3.84Γ—108 m,x(0) = r \cos(0) = r = 3.84 \times 10^8 \, \text{m}, y(0)=rsin⁑(0)=0.y(0) = r \sin(0) = 0.

At t=7.89 dayst = 7.89 \, \text{days}: x(7.89)=rcos⁑(Ο‰β‹…681696),x(7.89) = r \cos(\omega \cdot 681696), y(7.89)=rsin⁑(Ο‰β‹…681696).y(7.89) = r \sin(\omega \cdot 681696).

Let's calculate these values.At t=0t = 0:

  • x(0)=3.84Γ—108 mx(0) = 3.84 \times 10^8 \, \text{m}
  • y(0)=0 my(0) = 0 \, \text{m}

At t=7.89 dayst = 7.89 \, \text{days}:

  • x(7.89)β‰ˆβˆ’4.07Γ—107 mx(7.89) \approx -4.07 \times 10^7 \, \text{m}
  • y(7.89)β‰ˆ3.82Γ—108 my(7.89) \approx 3.82 \times 10^8 \, \text{m}

4. Change in position (Ξ”r\Delta r)

The change in position is: Ξ”x=xfinalβˆ’xinitial=βˆ’4.07Γ—107 mβˆ’3.84Γ—108 m\Delta x = x_{\text{final}} - x_{\text{initial}} = -4.07 \times 10^7 \, \text{m} - 3.84 \times 10^8 \, \text{m} Ξ”y=yfinalβˆ’yinitial=3.82Γ—108 mβˆ’0 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: Magnitude=Ξ”x2+Ξ”y2Ξ”t\text{Magnitude} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} Direction=atan2(Ξ”y,Ξ”x)\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)x(t) and y(t)y(t)) often helps solve complex problems.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Circular Motion
Angular Velocity
Vector Calculations

Formulas

x(t) = r cos(πœ”t)
y(t) = r sin(πœ”t)
Average Velocity: v_avg = Ξ”r/Ξ”t
Magnitude of Average Velocity: |v_avg| = sqrt(Ξ”x^2 + Ξ”y^2)/Ξ”t
Direction of Average Velocity: ΞΈ = atan2(Ξ”y, Ξ”x)

Theorems

Trigonometric functions for circular motion
Pythagorean theorem for magnitude of vectors

Suitable Grade Level

Grades 11-12