The Motion of a Spaceship in Space

[TheSpaceGuy]

Homework Statement

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.

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.

 

[Quinzio]

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.

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

Part b) has me really confused.

Why ?

 

[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?

 

[Quinzio]

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

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.

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

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.

2. I don't understand how it can hit the beacon. [/B]

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.

 

[mujin]

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.

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.

 

[Quinzio]

Please write down A.
B will be just a step beyond.

 

[mujin]

Please write down A.
B will be just a step beyond.

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.

 

[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.

 

[Quinzio]

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

 

[steven80612]

I don't 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?

 

[steven80612]

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

 

[Quinzio]

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

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

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

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

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.

 

[Quinzio]

I don't 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?

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.

 

[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.

 

[TheSpaceGuy]

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.

That was an awesome example! It really helped me see the picture clearer.

 

[Quinzio]

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.

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.