Solving 2-D Motion Vectors and 2-D Kinematics

Question

A lunar probe is moving 1,670 m/s at a 73.0* degree angle. It needs to land on the moon 3.71x10^8m away in a 46.0* direction in 8.64x10^4s. What is the direction of the acceleration that the engine must produce?

Answer 1

Assumptions I made with this problem: -y (down/South of vertical axis), +x (right/East of horizontal axis), -x (left/West of horizontal axis), +y (up/North of vertical axis): directions that are given are (polar coordinates) Δy is 314° and v_i=287° (polar coordinates assume +x = 0° and counterclockwise is positive).

Step 1: establish the main equation needed: You are given displacement, initial velocity, and time. You are looking for acceleration. The equation is: Δy_x=v_iΔt+.5aΔt²

Step 2: Rearrange the equation for acceleration:

a=(2Δy-2v_iΔt)/Δt²

Step 3: This is a 2d problem. Therefore, define the 2d given Δy and v_i into its 1d components. Time crosses the dimensions.

a_x=(2*3.71E8*cos314°-2*1670*8.64E4*cos287°)/((8.64E4)²) = + 0.0577 m/s²

a_y=(2*3.71E8*sin314°-2*1670*8.64E4*sin287°)/((8.64E4)²) = – 0.0345 m/s²

Step 4: establish the quadrant: Given the signs of a_x and a_y, we can predict the resultant acceleration vector to be located in (+, -) quadrant IV. If magnitude of the acceleration were needed, you would Pythagorean Theorem (using this phrase as a verb) the two components of acceleration a_x and a_y.

Step 5: However, since the problem is looking only for direction, then the equation is, using right-triangle trig:

Θ_R=tan^-1(.0345/.0577) = 30.878º S (-y) of E (+x) or polar coordinate would be 329.122º

Note: Again, this problem required making an assumption regarding whether the angles given respected the vertical or horizontal axis. I assumed the latter.