A space station is located at (0, 0, 0). A rocket travels along the path FG) = (

Question

A space station is located at (0, 0, 0). A rocket travels along the path FG) = (Int-2) it(2+2)/t (3-9)k where t is measured in appropriate units of time. a) If it shuts off its engines at t = 1 , and thus "releases" from the above path, will the rocket coast into the space station? b) Use 3D Calc Plotter to graph F(t) for .25 s s5. Have it place the velocity and acceleration vectors on the graph at the point corresponding to t-1. Does the graph support your conclusion in part a)? Print out your graph to turn in. Make sure to label the velocity and acceleration vectors on the graph. c) Compute aT and aN when t=1 d) Is the curvature of i(t) larger at 1 or2?

Explanation / Answer

r(t) = (lnt - 2)i + (2/t + 2)j + (3t - 9)k
Space station is at (0,0,0)

We need the velocity so v(t) = 1/t - 2/t^2 + 3

Then we use:
r(t) + mv(t) = 0i + 0j + 0k [which is the location of the space station, right]
=> (lnt - 2 + m/t)i + (2/t + 2 - 2m/t^2)j + (3t - 9 + 3m)k = 0i + 0j + 0k

We need the slope at t = 1 hours to figure out where it will coast to, so solve for m:
=> lnt - 2 + m/t = 0
=> m = 2t - tlnt
=> With t=1, this becomes => m = 2(1) - (1)ln(1) = 2
So the slope is 2.

Now (from above) we have:
2/t + 2 - 2m/t^2 = 0

We plug in the known t and m values.
=> 2/(1) + 2 - 2(2)/(1)^2 = 0
=> 0 = 0

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