A problem from science fiction - intersecting curves

[Emspak]

1. Homework Statement [/b]

A spaceship is traveling on a curved path, f(t) = (t, t²)

(We'll assume that the path isn't affected by gravity, this is a math problem, not physics :-) )

It has to release a pod to intersect a space station that has an orbit described by the following:

g(t) = (4+cos([itex]\frac{πt}{8})[/itex], 8-sin([itex]\frac{πt}{8}[/itex]))

The pod will travel at a tangent from the curved path f(t) and I want to know at what time s I should release it to get to the space station and how long it will take.

The Attempt at a Solution

OK, so I know that since I am releasing the pod at time s, my first vector to add will be (s, s²). And those numbers will be constant.

I know that f'(t) = (1, 2t)

So that means that a pod will be zooming off the curve in a path described by (s+t, s²+2t). (correct?)

So to see if the pod will hit the space station and when, I need

(s+t, s²+2t) = (4+cos([itex]\frac{π(s+t)}{8})[/itex], 8-sin([itex]\frac{π(s+t)}{8}[/itex]))But after that I am stuck, and I feel like there is some simple step I am missing. I know that the time s of release can't be more than s=3 and less than 0 because if s=0 the line is horizontal and never hits the circle described, and if s≥3 it will miss the circle full stop. So there are some constraints here. I am almost there, I feel like.

 

[Simon Bridge]

So that means that a pod will be zooming off the curve in a path described by (s+t, s2+2t). (correct?)

Not quite.

For the spacecraft , v(t)=(1,2t)
At time t=s, the speed of the spacecraft , and thus the pod, will be... ?
... and thus, the position-time equation of the pod will be... ?

 

[Emspak]

A time t=s the speed of the spacecraft is f'(s) or (1,2s) so that's the speed of the pod.

The position of the pod would have to be (s+t, s²+(2s)t) after that, yes?

So you'd want (s+t, s²+(2s)t) = (4+cos[itex]\frac{(s+t)π}{8}[/itex], 8-sin[itex]\frac{(s+t)π}{8}[/itex])

That's still pretty ugly when you want to solve it, no?

 

[Simon Bridge]

It's usually nastier than that because the pod will be traveling along a conic section as well.
And then - real orbits are seldom that tidy.

You want to solve for s - you'll find there are lots of solutions from the restraints you have specified.

 

[Emspak]

OK, but when I try to do that I get s = 4-t+cos([itex]\frac{(t+s)π}{8}[/itex])

And I still have a lot of possibilities for both s and t. I do know that -1<s+t-4<1, b/c of the cosine function. And I know that s>0, and t>0 so for this to work 3< s+t < 5. Am I getting it right so far?

Further, s²+2st = 8-sin([itex]\frac{(t+s)π}{8}[/itex])

which gets me s(1+2t) = 8-sin([itex]\frac{(t+s)π}{8}[/itex])

or 8-s(1+2t)=sin([itex]\frac{(t+s)π}{8}[/itex])

since -1<8-s(1+2t) <1 7<s(1+2t)<9, yes?

 

[Simon Bridge]

I still have a lot of possibilities for both s and t

Yes you do - why would you expect any different?

You are using t as the elapsed time between release and capture.
You should find an equation for this time in terms of the release time and the speed of the pod.