Find the magnitude and direction of the Moon\'s average velocity between t = 0 a

Question

Find the magnitude and direction of the Moon's average velocity between t = 0 and t = 5.51 days.

t converted to seconds = 5.51 * 24 * 60 *60 = 476064 seconds

Using the equation and information given:

r(ti) = 3.84 108 m

r(tf) = 3.84 108 m [cos (2.46 10-6 radians/s * 476064 s) + sin (2.46 10-6 radians/s * 476064 s)]

There is something wrong with my calculations because I've gotten a few different answers and all of them are wrong: 16.3 m/s, 250 m/s, 1.65 m/s

Please help!

Relative to the center of the Earth, the position of the Moon can be approximated by (t) = r [cos (wt) + sin (wt) ], where r = 3.84 108 m and w = 2.46 10^-6 radians/s. Find the magnitude and direction of the Moon's average velocity between t = 0 and t = 5.51 days. t converted to seconds = 5.51 * 24 * 60 *60 = 476064 seconds Using the equation and information given: r(ti) = 3.84 108 m r(tf) = 3.84 108 m [cos (2.46 10-6 radians/s * 476064 s) + sin (2.46 10-6 radians/s * 476064 s)] r(tf) = 3.84 108 m + 7.84 106 m vav = [ r(tf) - r(ti) ] / t vav = [ 3.84 108 m + 7.84 106 m - 3.84 108 m ] / 476064 seconds = ? There is something wrong with my calculations because I've gotten a few different answers and all of them are wrong: 16.3 m/s, 250 m/s, 1.65 m/s

Explanation / Answer

In my opinion, the way to do this is by differentiating the given function to get a formula for velocity. Your way (i.e.) finding distance travelled and dividing by time will give you distance travelled RELATIVE to earth. But remember that velocity is displacement per unit time not distance per unit time.

Thus, we get veloicty by differentiating the function for displacement w.r.t time. This will be:

-rsin(t) (i^) + rcos(t) (j^).

Now find the average value of this function between t1 and t2. I remember from calculus there is a method to calculate average value of a function over time but I can't remember how to do it. Will get back if I can recall it.

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