Calculating Velocity & Acceleration of the Moon

[planke]

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 one orbit is 29.5 days.) To approximate the instantaneous acceleration of the Moon at t = 0, calculate the magnitude and direction of the average acceleration during the following two time intervals.

(a) between t = 0 and t = 0.400 days
______ m/s
______ degrees (counterclockwise from the +x axis)

(b) between t = 0 and t = 0.0040 days
_____ m/s
_____ degrees (counterclockwise from the +x axis)

Homework Equations

aav = delta v / delta t
ax = delta vx / delta t
ay = delta vy / delta t
magnitude = sqroot (ay^2 + ax^2)
theta = tan-1 (ay / ax)

The Attempt at a Solution

I got the magnitudes of both at 0.00233 m/s. I am having trouble finding the directions though.

I just used ax = delta vx / delta t and ay = delta vy / delta t to find the values of ax and ay, then I subbed these numbers into this eqn: magnitude = sqroot (ay^2 + ax^2). My answer for the direction in part (a) was 2.43558 degrees... which is wrong. Can anyone think of a better way to do this problem?

 

[ideasrule]

Why did you calculate acceleration? You're supposed to calculate velocity. Theta would be the inverse tan of Vy/Vx.

 

[kerimek]

I would recommend using the formula for calculating the instantaneous acceleration at a specific time, which is a(t) = v(t) / t. In this case, we can use the given velocity equation, varrowbold(t), to calculate the velocity at t = 0. Then, we can plug this value into the formula for acceleration to find the magnitude and direction.

(a) At t = 0, the velocity of the Moon is v = 945 m/s. Therefore, the instantaneous acceleration at t = 0 is a(0) = 945 m/s / (0.4 days * 24 hours/day * 3600 seconds/hour) = 0.00233 m/s^2. The direction of this acceleration can be found using the formula theta = tan^-1 (ay / ax). In this case, ay = -vωcos(ωt) = -945 m/s * 2.46x10^-6 rad/s * cos(0) = 0 m/s^2, and ax = vωsin(ωt) = 945 m/s * 2.46x10^-6 rad/s * sin(0) = 0 m/s^2. Therefore, theta = tan^-1 (0/0) is undefined.

(b) Similarly, at t = 0, the velocity of the Moon is v = 945 m/s. Therefore, the instantaneous acceleration at t = 0 is a(0) = 945 m/s / (0.004 days * 24 hours/day * 3600 seconds/hour) = 0.00233 m/s^2. The direction of this acceleration can be found using the formula theta = tan^-1 (ay / ax). In this case, ay = -vωcos(ωt) = -945 m/s * 2.46x10^-6 rad/s * cos(0) = 0 m/s^2, and ax = vωsin(ωt) = 945 m/s * 2.46x10^-6 rad/s * sin(0) = 0 m/s^2. Therefore, theta = tan^-1 (0/0) is undefined.

Overall, the direction of the average acceleration during these time intervals is undefined due to the fact that the velocity and acceleration equations are both sinusoidal functions, and at t = 0, the acceleration is equal