# Hitting a Target Question

1. Sep 26, 2010

### TheSpaceGuy

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

A spaceship maneuvering in space, far from any gravitational influences, is executing a predetermined acceleration program which yields a position vector r(t) for the ship, relative to a small space beacon at the origin given by

r(t) = (t - 2)i + j(t - 3)^2 + k(t - 4)^3

a) Suppose that the captain shuts down the engines at time t = a. Find the subsequent motion of the ship.

b) Show that if t = a is chosen appropriately then the ship will hit the beacon.

3. The attempt at a solution

For Part a) I was thinking about finding the derivative of r(t) at t=a since r(t) is a position vector.

Part b) has me really confused.

2. Sep 26, 2010

### Quinzio

Yes, the idea is good but write down the equation of the motion

Why ?

3. Sep 27, 2010

### TheSpaceGuy

1. Won't the derivative of r(t) be the equation of motion?

2. I don't understand how it can hit the beacon. I have it moving away from the beacon?

4. Sep 27, 2010

### Quinzio

It will be a part of.
In real world imagine you have to jump on a slowly running train.
Your brain automatically processes two informations: the speed of the train and the relative position of the train. The speed itself is not enough. If this example confuses you, forget it.

Starts by remembering the motion equation e.g. for the train. In 3d it's more complex, but in the end it's the same soup.

Are you sure sure ??? Have you visualized the motion of the ship as if you were watching a movie ? I don't think, so trust the equations.

Imagine the pilot doesn't have a "window" to look ahead, but only a little hole. He sits in the middle of the ship, he looks in the hole. Now can you "see" which are the condition of hitting the beacon ?
First solve A, you'll better understand B.

5. Sep 27, 2010

### mujin

I figured out A, but stuck at B. Can you explain a little more? I really can't think of what to do in order to solve the question.

6. Sep 27, 2010

### Quinzio

B will be just a step beyond.

7. Sep 27, 2010

### mujin

what do you mean by a step beyond? I have the equation of the tangent line (L(t)=r(a)+r'(a)t) as the answer for A (is this right?). then, to show it will hit the beacon, what exactly should I look for?

I originally thought that part B means the beacon is on the position vector r(t), so if I set r(t) equal to the linear equation, then I will find another intersection point besides t=a, and that point is where the beacon is located. but obviously, I was wrong.
and the statement of the question says that the beacon is at the origin, this confuses me, too.

8. Sep 27, 2010

### Quinzio

Ok, so the equation of the tangent in t is

$$\frac{x-(t-2)}{f_x(t)}= \frac{y - (t-3)^2}{f_y(t)}= \frac{z - (t-4)^3}{f_z(t)}$$

Find the derivatives.
The beacon is in (0,0,0)
Replace x, y, z with zeros and solve for t.
That's it.

9. Sep 27, 2010

### Quinzio

I don't read you anymore, hope it's clear...

10. Sep 28, 2010

### steven80612

I dont understand how you get that equation of tangent line.
I thought its L(t) = r(t) + tr'(t)
and it should yield to
L(a) = (a-2)i+(a-3)^2j+(a-4)^3k + t( i + 2(a-3)j + 3(a-4)^2k)

is this correct?

11. Sep 28, 2010

### steven80612

in addition how would you solve for B? can you explain in detail?

12. Sep 28, 2010

### Quinzio

$$\frac{x-(t-2)}{f_x(t)}= \frac{y - (t-3)^2}{f_y(t)}= \frac{z - (t-4)^3}{f_z(t)}$$

$$f_x = 1$$
$$f_y = 2(t-3)$$
$$f_x = 3(t-4)^2$$

$$x-(t-2)= \frac{y - (t-3)^2}{2(t-3)}= \frac{z - (t-4)^3}{3(t-4)^2}$$

In the origin $$O(0,0,0)$$

$$t-2= \frac{(t-3)^2}{2(t-3)}= \frac{(t-4)^3}{3(t-4)^2}$$

$$t-2= \frac{t-3}{2}= \frac{t-4}{3}$$

Which is true only for $$t=1$$.

So the istant searched is t=1.
If in t=1 you switch off the engine, the ship will hit the target in the origin, the beacon.

Last edited: Sep 28, 2010
13. Sep 28, 2010

### Quinzio

I can see what you mean but at least it should be corrected lilke this:

$$L(t_0+\Delta t) = r(t_0) + \Delta t\ r'(t_0)$$

and find solution for the void vector (0).
It 's the same.

14. Sep 28, 2010

### zedleppelin17

I have a question. How did you get:
$$\frac{x-(t-2)}{f_x(t)}= \frac{y - (t-3)^2}{f_y(t)}= \frac{z - (t-4)^3}{f_z(t)}$$

From:
$$L(t_0+\Delta t) = r(t_0) + \Delta t\ r'(t_0)$$

I understand what you did to solve part B at the Origin, but this confused me a bit.

15. Sep 28, 2010

### TheSpaceGuy

16. Sep 28, 2010

### Quinzio

I didn't solve B to get A.
Part A is the general equation of a straight line.
But using B you must get something equivalent.

B says that we search a scalar $$\Delta t$$, for which
$$r(t_0) + \Delta t\nabla(t_0) = 0$$

That is if the vector position and it's speed $$\nabla(t_0)$$ are parallel in some point, we have a solution.