# Spaceship Coasting Problem

1. Apr 29, 2014

### meesa

1. The problem statement, all variables and given/known data

The position of a spaceship is $(3 + t, 2 + ln(t), 7 - \frac{4}{t^2 + 1})$ and the coordinates of the space station are (6, 4, 9). The captain wants the spaceship to coast into the space station. When should the engines be turned off?

2. Relevant equations

$r' = (1, \frac{1}{t}, \frac{8t}{(t^2 + 1)^2})$

3. The attempt at a solution

Every solution I've found online, but none of them are complete, and I have no idea what I'm missing. So would somebody please just work out the entire problem? PLEASE? All the help online is very vague and the most I've gotten from it is that I need the derivative, but I can't figure out how to relate that and the original together to get something.

Thanks.

Last edited: Apr 29, 2014
2. Apr 29, 2014

### Curious3141

What is the acceleration when the engine is turned off? What happens to the velocity?

Check your y-coordinate derivative. It's wrong.

Sorry, no one is allowed to give full solutions to HW problems here. It's in the rules.

Last edited: Apr 29, 2014
3. Apr 29, 2014

### meesa

The acceleration would be zero. We're in mystical space where everything is perfect with lovely numbers.

Fixed the y derivative.

I've seen full solutions here before.

The answer is 1 according to the back of the book.

4. Apr 29, 2014

### Curious3141

Those were in contravention of the rules if they were solutions to homework problems. You can PM a mentor or admin if you have an issue with the policy.

I can guide you through the general approach if you answer the questions I asked in my previous post (acceleration, velocity).

5. Apr 29, 2014

### meesa

Just saw that and edited my post right before yours came though. r' is the velocity.

A general approach will work, if you can do it without being too abstract. Appreciate your time.

6. Apr 29, 2014

### Curious3141

What happens to the acceleration when the engines are cut?

What happens to the velocity?

7. Apr 29, 2014

### meesa

The velocity would be constant, since acceleration is zero. What I wrote in my first post is the exact question.

8. Apr 29, 2014

### Curious3141

OK, what you have is a vector equation for displacement.

s(t) = sx(t) i + sy(t) j + sz(t) k

where i, j and k are the usual orthogonal unit vectors for 3-D. The individual expressions for the displacements along each axis are given in the original question.

Differentiate s(t) wrt t to find v(t) in the same form. You've already done the work here, just put it in vector form.

Now let the time when the engines are cut be T.

The displacement at at that time will be s(T).

The velocity at time T will be v(T).

Thereafter, the spaceship will move at the constant velocity v(T).

The displacement at any time t' (t'>T) will be given by:

(t' - T)v(T) + s(T) = s(t')

You want to equate s(t') to 6i + 4j + 9k and solve for T. You will have 3 equations in two variables, and if they can be solved to give consistent solutions, you have an answer.

EDIT: Confirmed that T = 1 is a consistent and valid solution. Now try it yourself.