How close will the two ships get?

[thomasrules]

Homework Statement

Ship A, sailing due east at 8 km/h, sights ship B 5km to the southeast when ship B is sailing due north at 6km/h. How close to each other will the two ships get?

Homework Equations

The Attempt at a Solution

I drew the picture. Ship a distance is 8t and B is 6t

Use the a^2+b^2=c^2

 

[Dick]

Seems to me there are two coordinates of interest here. Hint.

 

[thomasrules]

sorry I'm lost

 

[HallsofIvy]

Homework Statement

Ship A, sailing due east at 8 km/h, sights ship B 5km to the southeast when ship B is sailing due north at 6km/h. How close to each other will the two ships get?

Homework Equations

The Attempt at a Solution

I drew the picture. Ship a distance is 8t and B is 6t

Use the a^2+b^2=c^2

"Ship a distance is 8t and B is 6t" distance from what? Set up a coordinate system, say with A initially at the origin. What are the coordinates of ship A at time t? What are the initial coordinates of ship B? What are the coordinates of ship A at time t? What is the (square of the) distance between those points as a function of t?

 

[thomasrules]

ok $$D^2=(-3.53+6t^2)+8t^2 dD/dt=(200t-42.36)/whatever\\0=200t-42.36\\t=0.212\\$$
I plugged that in the distance formula and got a wrong answer

 

[Dick]

Follow HallsOfIvy's suggest and write down the position of the two ships in xy coordinates as a function of time first. It's difficult to tell where you went wrong from what you post.

 

[thomasrules]

yes I have a drawing. I used the sin law to find the distance of the 2 sides.

Coordinates (3.53,0) and (3.53,-3.53)

By the way how do you put spaces in latex coding I tried \\ it didnt work

 

[arildno]

Eeh??

Do you even know what coordinates are??

Considered as a function of time, what is ship A's position measured from an origin lying where A was, and sighted B somewhere at t=0?
And, with the same choice of origin, what is B's position as a function of time?

 

[Dick]

At what time? Where is the time dependence? (Not sure about your latex question, sorry).

 

[thomasrules]

yes those are the coordinates.from the origin (0,0) using pythagorean theorem,

i use 5sin45=a=3.53, b=3.53

 

[arildno]

1. What choice have you made of positive axes?
2. What was A's position at t=0?
3. What is A's position as a function of time?
4. What is B's position as a function of time?

 

[thomasrules]

A's position at t=0 is (0,0)
A's position as a function of time is (3.53-8t,0)
B's position as a function of time is (3.53,-3.53-6t)

 

[arildno]

Is (3.53-8*0,0)=(0,0)?? ?

 

[thomasrules]

ah crap...
A is (8t,0)

 

[thomasrules]

**** or is it
A(8t,0)
B(3.53,-3.53+6t)

My guess is that's right if not I quit school

 

[Dick]

You can stay in school! Now what is D^2 as a function of t?

 

[thomasrules]

D^2=(-3.53+6t-0)^2+(3.53-8t)^2

I did derivative. Set it to 0. Got 0.7

Thats not the answer

I'm suppost to plug time into the D^2 and get the distance correct?

 

[Dick]

You seem to be trying to do the right thing. What is the derivative of D^2?

 

[thomasrules]

I got it to be:

After the expansion and derivative

dD/dt=200t-98.84/(the rest here doesn't matter cause I'm setting other side to zero)

 

[Dick]

All seems ok. What is the answer supposed to be?

 

[thomasrules]

The ships get 0.707km to each other

btw thanks for all your help so far

 

[Dick]

You're welcome. But that answer only differs from 0.7 because you are rounding off differently.

 

[thomasrules]

NO! but we only have found the time man not the distance

 

[Dick]

I got it to be:

After the expansion and derivative

dD/dt=200t-98.84/(the rest here doesn't matter cause I'm setting other side to zero)

You are saying your solution for the TIME is 0.7? That's not what this says.

 

[thomasrules]

OMG YOU ARE CORRECT!

Thanks man. I'm slow. What's your paypal how much do I owe you.


And now I have another similar one I'm struggling with it :S

 

[Dick]

Good luck!

 

[HallsofIvy]

Because you (cleverly) chose to make A's initial position the origin of your coordinate system, A's position at t= 0 is, indeed, (0,0). Since A is moving due east (and you chose to make the positive x-axis that direction) at 8 km/h, A's position at time t hours is (8t, 0). Yep, that's what you got!

Initially, B was "5 km to the south east" so B's initial position is $(5\frac{\sqrt{2}}{2}, 5\frac{\sqrt{2}}{2}$ ($sin(45^o)= \frac{\sqrt{2}}{2})$. Since B is moving straight North at 6 km/h, B's position at time t hours must be $(5\frac{\sqrt{2}}{2}, 5\frac{\sqrt{2}}{2}+ 6t)$). If $5\frac{\sqrt{2}}{2}= 3.53$(and it is) you are correct. Glad you don't have to quit school!

Now what is the distance between those points? (Hint: since distance is always positive, minimizing distance is the same as mininizing distance squared- so you can ignore the square root in the distance formula.)

 

[thomasrules]

Yea I got it thanks Halls of Ivy...

I get to stay in school :D