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) = 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 = 3.92 days? Find its magnitude and direction, given as an angle measured counterclockwise from the positive x-axis.

Answer 1

Answer: Average Velocity = - 643.42 i + 512.66 j m/s

Magnitude = 822.7 m/s

Direction = 141.45°

Explanation:

r = 3.84 x 10^8 m

w = 2.46 x 10^-6 rad/s

Formula for Average velocity = displacement / time

at t = 0

x(0) = r

y(0) = 0

at t = 8.45 days

= 8.45 x 24 x 3600 s =730080 sec

w t = 2.46 x 10^-6 x 730080 = 1.80 rad Or 102.90°

xf = r cos(w t) = - 0.2233r

yf = r sin(w t) = 0.9747r

Displacement = (xf - x0)i + (yf - y0)j = -1.2233r i + 0.9747r j

<v> = dispalcement / t = (-1.2233r i + 0.9747r j ) / (730080 s )

= - 643.42 i + 512.66 j m/s

Magnitude

= sqrt(643.42^2 + 512.66^2)

= 822.7 m/s

Direction

= 180 - tan^-1(512.66 / 643.42)

= 141.45°

Answer 2

Answer:

Explanation:

Given

x(t) = r cos(ωt)

y(t) = r sin(ωt)

where r = 3.84×10⁸ m and ω = 2.46×10-⁶radians/s

Average velocity = Δx /Δt

At t = 0s

x(0) = 3.84×10⁸×cos(2.46×10-⁶×0)

x(0) = 3.84×10⁸×1 = 3.84×10⁸m

At t = 3.92days = 3.92×86400s = 3.39×10⁵s

x(3.39×10⁵) = 3.84×10⁸×cos(2.46×10-⁶×3.39×10⁵) = –2.58×10⁸m

Δx = x(3.39×10⁵) – x(0) = –2.58×10⁸–3.84×10⁸m = –6.42×10⁸m

Δt = 3.39×10⁵ – 0 = 3.39×10⁵

vx = –6.42×10⁸/3.39×10⁵ = –1894m/s

At t = 0s

y(0) = 3.84×10⁸×sin(2.46×10-⁶×0)

y(0) = 3.84×10⁸×0= 0m

At t = 3.92days = 3.92×86400s = 3.39×10⁵s

y(3.39×10⁵) = 3.84×10⁸×sin(2.46×10-⁶×3.39×10⁵) = 2.84×10⁸m

Δy = y(3.39×10⁵) – y(0) = 2.84×10⁸ – 0 = 2.84×10⁸m

Δt = 3.39×10⁵ – 0 = 3.39×10⁵

vy = 2.84×10⁸/3.39×10⁵= 838m/s

vy = 838m/s

v = √(vx² + vy²) = √((-1894)² + 838²)

v = 2071m/s

θ = tan-¹(vy/vx) = tan-¹(838/1894) = -23.9°