I had completely spaced converting days into seconds. However, what I ended up doing was:
differentiate v(t) = v( -sin(wt) + cos(wt)) which is
-v(w) (cos(wt) + sin(wt)) you have the acceleration
v(w) = 2.32 x 10^-3 meter/sec^2
At t = 0 days, the acceleration is - 2.32 x 10^-3 meters/sec^2...
Oh thank you much! I just tried that equation and my response "differs from the correct answer by orders of magnitude". I'm not too sure what that means... I feel like i must be off by 10^x in my answer but not too sure how to go about figuring that out?? I got 945 when t=2 and 1.89 when t=.002...
Hi thanks for the help. I was thinking since the moon is moving around us, that it would be a centripetal acceleration problem? But maybe that's where I was going wrong. So I was using that formula in relation to vectors...
a = -lim (v/2*theta)2vsin(theta)ey as lim of theta approaches zero =...
Homework Statement
1. Homework Statement
The velocity of the Moon relative to the center of the Earth can be approximated by varrowbold(t) = v [−sin (ωt) xhatbold + cos (ωt) yhatbold], where v = 945 m/s and ω = 2.46 multiplied by 10−6 radians/s. (The time required for the Moon to complete...