# Homework Help: Calc III : Spaceship reaching a space station at constant velocity.

1. Jan 22, 2012

### mimzy

1. The problem statement, all variables and given/known data
The Position Function of a spaceship is
r(t) = (5t+9)i+(2t2-1)j+t2k
and the coordinates of a space station are P(54,89,45). The captain wants the spaceship to reach the space station in constant velocity. When should the engines be turned off?

2. Relevant equations
r'(t) = v(t)= <5,4t,2t>

and

r''(t) = v'(t)= a(t) <0,4,2>

3. The attempt at a solution
I have tried to use the dot product of a(t) * [P-v(t)] but that doesn't work :l I also thought of finding the angle of the tangent vector that shots towards P in order to find the time but I end up confusing myself afterwards D;

2. Jan 22, 2012

### Dick

Your second idea is the right one. You want the vector between the ship and the station to be parallel to the velocity of the ship with the velocity vector pointing at the station. Then you can shut the engine off and then just cruise in. So you want to solve [P-r(t)]=k*v(t) with a value of k>0.

3. Jan 23, 2012

### Joffan

What kind of gravity field is the ship operating in? ;-)

OK ignore that. But it does seem to me that the x ordinate implies a definite value of t that we need to arrive at. I haven't pursued the rest of the equations to find out if there is one time we can turn off acceleration in both y and z directions, or whether they require separate treatment.

4. Jan 23, 2012

### Dick

You are over complicating it. This question is rigged to give you a simple solution for a value of t to set the acceleration to zero.

5. Jan 23, 2012

### Joffan

Well, Dick, I feel it's good practice to work the two axes out separately, and we are in the Calculus+ section. But now I've actually done the working, I see what you mean - but only once I had the same quadratics in both y & z cases.

For mimzy: I set a tc for the cutoff time and recast the equations using that (and your velocity information). If I'd had to, I would have had a ta for arrival as well, but that wasn't necessary in this case.