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Position Vector Problem

  • Thread starter bodensee9
  • Start date
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1. Homework Statement
The position function of a spaceship is

r(t) = (3+t)i + (2+ln t)j + (7 - 4/(t^2+1)k and the coordinates of the space station are (6,4,9). If the spaceship were to "coast" into the space station, when should the engines be turned off?


2. Homework Equations

The relevant equations are r' = velocity, and r'' = acceleration.

3. The Attempt at a Solution

I am really not sure how to go about solving this problem. So, this means that r" is zero but then there is still velocity? So, if r' is i + 1/tj + 8t/(t^2+1)^2k. r" = -1/t^2 + (8(t^2+1)^4 - 16t^2(t^2+1))/(t^2+1)^4. But then I am not sure what to do after that. Thanks!!
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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1. Homework Statement
The position function of a spaceship is

r(t) = (3+t)i + (2+ln t)j + (7 - 4/(t^2+1)k and the coordinates of the space station are (6,4,9). If the spaceship were to "coast" into the space station, when should the engines be turned off?
Am I correct that that r(t) is the position of the space craft with the engines on? And with the engines off it will continue in that direction?


2. Homework Equations

The relevant equations are r' = velocity, and r'' = acceleration.

3. The Attempt at a Solution

I am really not sure how to go about solving this problem. So, this means that r" is zero but then there is still velocity? So, if r' is i + 1/tj + 8t/(t^2+1)^2k. r" = -1/t^2 + (8(t^2+1)^4 - 16t^2(t^2+1))/(t^2+1)^4. But then I am not sure what to do after that. Thanks!!
Yes, with the engines turned of the space ship will still have non zero velocity. Look at the velocity vector for each t (the derivative of r) as well as the vector from r(t) directly to the space station. When those vectors are parallel, turn off the engines will cause the space ship to continue to the space station.
 
178
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So do you mean that if the vector formed by subtracting r(t) from (6,4,9) needs to be parallel to r'? So this means that <3-t, 2-ln t, 2+4/t^2+1> needs to be a multiple of <i, 1/tj, 8t/(t^2+1)^2k?

But how do you go about solving for that, because you have 3-t must be a multiple of i, and 2-ln t must be a multiple of 1/t? So does that mean you try to make ln t go away? Thanks.
 

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