# Relative motion/ Vector Question

1. Oct 17, 2005

### suspenc3

Please..give It A Try

"Ship A is located 4 km north and 2.5 km east of ship B. Ship A has a velocity of 22km/h towards the south and ship B has a velociy of 40 km/h in a direction of 37(degrees) north of east.
A)what is the velocity of A relative to B
B)write an expression for the position of A relative to B as a function of t, where t is equal to 0.
***C) at what time is the separation between the ships least?
Ive completed A and B..but i have no idea where to start for C..Can anyone help?

wellllllllll;
first i found the point of A relative to B=
...ans = [32km/h(t) + 2.5 km]i + [2.0km/h(t) + 4.0km]j
im guessing it has something to do with a max-min problem..but im not sure where to start..i could do trial and error..but there has to be a simpler way..

Last edited: Oct 17, 2005
2. Oct 17, 2005

### James R

If you've done (B), then you can find the distance between A and B as a function of time by taking the magnitude of the vector expression in (B).

For example, if your answer for (B) is:

$$r(t) = x(t)i + y(t)j$$

then

$$|r(t)| = \sqrt{x(t)^2 + y(t)^2}$$

To find the minimum separation you need

$$\frac{d(|r(t)|)}{dt} = 0$$

and solve for t. Then find |r| using the t value.

(Hint: since $|r|$ is a minimum when $|r|^2$ is a mimimum, you can alternatively solve $d(|r|^2)/dt = 0$ for t, which will be easier.)

Does that help?

Post restored.

--TM

Last edited by a moderator: Oct 17, 2005
3. Oct 17, 2005

### Tom Mattson

Staff Emeritus
JamesR,

I've "soft deleted" your post. I'll restore it when suspenc3 shows some work.

I'm cracking down on the rules.

4. Oct 17, 2005

### Tom Mattson

Staff Emeritus
You're right about it being a max-min problem, but you're wrong about needing to do it by trial and error.

How do you normally deal with max-min problems?

5. Oct 17, 2005

### suspenc3

find the quantity were trying to minimize
find the variables involved
find a formula relating the quantity and the variables

This question is unlike most max min problems ive ever done

Last edited: Oct 17, 2005
6. Oct 17, 2005

### Tom Mattson

Staff Emeritus
Have you taken Calculus I yet? The reason I'm asking is that you are forgetting a crucial step, and if you have had Calc I it should be obvious what that step is. Simply setting up a relationship isn't enough.

Do you know what I'm getting at?

7. Oct 17, 2005

### Staff: Mentor

Have you studied any calculus yet, in particular derivatives?

8. Oct 17, 2005

### suspenc3

Yes I have..forgot that ppart..oops.its been a long time since ive dealt with min problems

so..
d= [32km/h(t)]i + [4km/h(t)]j..Take the derivative with respect to d?

or am i just way off

Last edited: Oct 17, 2005
9. Oct 17, 2005

### Tom Mattson

Staff Emeritus
You're getting there. But you want to take the derivative with respect to t, not d.

I will now restore JamesR's post.

10. Oct 17, 2005

### suspenc3

yes this helps..but i still am a little fuzzy

r(t) = 32(t)i + 4(t)j...

0 = (radical of)[2(32t) + 2(4t)]

11. Oct 17, 2005

### Tom Mattson

Staff Emeritus
No, you have to take the derivative first, and then set it equal to zero. In order to take the derivative you have to use the Chain Rule:

$$\frac{d}{dx}[f(x)]^{\frac{1}{2}}=\frac{1}{2}\frac{f'(x)}{[f(x)]^{1/2}}$$

12. Oct 17, 2005

### suspenc3

R1(t) = {1/2[x(t)^2 + y(t)^2]^-1/2}{x(2t) + y(2t}

Ive been out of school for a few years..so im not sure if this is correct

...0=72/18t..I must have done something wrong here!

Last edited: Oct 17, 2005
13. Oct 18, 2005

### Tom Mattson

Staff Emeritus
I've started to work the problem myself, and I see that your starting point is wrong. It is close to being correct though.

You appear to have added the position vectors for A and B, when you should have subtracted them. The position of A relative to B is $\vec{r}_{AB}=\vec{r}_A-\vec{r}_B$.

I'd like for you to do the following:

1.) Clean up your expression for $\vec{r}_{AB}$, and then post it here.
2.) Obtain the expression for $|\vec{r}_{AB}|$, the distance between A and B, and post it here.

Then, we can get to work on taking that derivative.

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