Relative motion/ Vector Question

[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 I am not sure where to start..i could do trial and error..but there has to be a simpler way..

 

[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
[/color]

 

[quantumdude]

JamesR,

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

I'm cracking down on the rules.

 

[quantumdude]

im guessing it has something to do with a max-min problem..but I am not sure where to start..i could do trial and error..

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?

 

[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 I've ever done

 

[quantumdude]

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?

 

[jtbell]

Have you studied any calculus yet, in particular derivatives?

 

[suspenc3]

Yes I have..forgot that ppart..oops.its been a long time since I've 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

 

[quantumdude]

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

I will now restore JamesR's post.

 

[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)]

 

[quantumdude]

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

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}}

 

[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 I am not sure if this is correct

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

 

[quantumdude]

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

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

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.