Question

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)-cos(wt) y(t) = r sin(ωt) where r 3.84 x 108 m and - 2.46 x 10-5 radians/s. What is the average velocity of the Moon measured over the interval from t to t 3.02 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis. 822.7 magnitude How is average velocity calculated using initial and final time? Given the times and functions for position, can you find initial and final positions? Since average velocity is a vector quantity, how do you find its components, and from there its magnitude? Make sure your calculator is in radian mode. m/s 76.2 direction Knowing the components of the average velocity, can you use trigonometric functions to find its direction?0 counterclockwise from the +x-axis

Answer 1

At t = 0
X₁ = rcos(w*0) = r
Y₁ = rSin(w*0) = 0
At t = 3.02 days
= 3.02*24*3600 = 260928 s
X₂ = rCos*(2.46*10^-6*260928) = 0.801r
Y₂ = rSin(2.46*10^-6*260928) = 0.599r
Now displacement is given by
D = (X₂ - X₁) i +(Y₂ - Y₁) j
= (0.801r - r) i +(0.599r -0) j = -0.199r + 0.599 r j
now we know that the average velocity is given by
V= Displacement/time
V= (-0.199r + 0.599 r j )/(260928 )
Putting the value of r in above equation (r) = 3.84*10⁸ m
V = - 292.862 i + 881.53 j
Now the magnitude of the velocity = [(-292.862)² +(881.53)²]^1/2 = 928.904 m/s
Now the direction
tan $\theta$ = 881.53/(-292.862)
$\theta$ = 71.622 from -x axis
From + x axis
= 180 - $\theta$ = 108.38 degree