General

[quote=iPhone5]if it was that easy, i would not be here[/quote]

Wow, you're right. The question is definitely poorly worded or the answer is wrong. Using the sine law, I get the wrong velocity (according to the answer) and when I manipulate the vectors and put them together, I get too many unknown angles (cause if we add some vectors together, you also need to add unknown angles to it to complete the triangle).

Cosine Law is no help and simply subtracting the resultant vector from the current vector to find the vector of the boat doesn't help either. Ask your teacher and definitely post once you get the solution, I'm curious as hell now. The best velocity I got for the boat was 7.something knots.

[quote=NoobCake]That clearly says what to do... use the sine law (a/sinA = b/sinB = c/sinC) first then solve for velocity[/quote]

if it was that easy, i would not be here

[quote=iPhone5]here, right from the work sheet.

http://gyazo.com/c9b025f9ffb5e385a0c15ae41f7fc103[/quote]

That clearly says what to do... use the sine law (a/sinA = b/sinB = c/sinC) first then solve for velocity

[quote=Aranmurder]you want a calculus solution or an algebra one[/quote]

alebra

you want a calculus solution or an algebra one

Yeah, I'm getting 12 knots too now. There's something that I'm doing wrong, but I'm not sure what it is atm, sorry.

We have some sub physics teacher and he gets his homework from other teachers online websites LOL. QQ i doubt he even looks at em.

[quote=iPhone5]I was thinking the samething, but you helped a lot and that matters. Thanks! Why am i getting 12 knots tho QQ T.T SIGN[/quote]

I'll take a look at it again in a minute

EDIT: Oh, this problem says what speed and angle does it need to be going at initially to be going at 23 degrees north of east once the current is factored in. The problems we always did last year were always find what the resultant vector is given this speed and bearing of the boat and the current. I see why we were getting wrong numbers early. Brb 1 sec, reworking this problem

[quote=iPhone5]I was thinking the samething, but you helped a lot and that matters. Thanks! Why am i getting 12 knots tho QQ T.T SIGN[/quote]

cos = adj / hyp
sin = opp / hyp

make sure your signs are correct for the direction of each component.

remember - 23 deg N of E is the path you want the boat to take. that's not the direction it will be headed.

[quote=UAPaladin]@xXMCheifXx: oh, my bad then[/quote]

I was thinking the samething, but you helped a lot and that matters. Thanks! Why am i getting 12 knots tho QQ T.T SIGN

@xXMCheifXx: oh, my bad then

[quote=xXMCheifXx]you're reading the question wrong. the boat wants to travel ALONG the 23 deg N or E path. since there's a current pulling it down it'll have to point more north to make up for that.

that answer in his SS is correct.[/quote]

Can you please help me out then sign

[quote=UAPaladin]That answer is wrong, either the textbook writer or your teacher who made the problem screwed up. There's no way that the resultant angle can be 43 degrees when initially you had an angle at 23 degrees and there is a force pulling it down. At this point, if the problem's that important, I'd say email your teacher.[/quote]

you're reading the question wrong. the boat wants to travel ALONG the 23 deg N or E path. since there's a current pulling it down it'll have to point more north to make up for that.

that answer in his SS is correct.

[quote=iPhone5]here, right from the work sheet.

http://gyazo.com/c9b025f9ffb5e385a0c15ae41f7fc103[/quote]

That answer is wrong, either the textbook writer or your teacher who made the problem screwed up. There's no way that the resultant angle can be 43 degrees when initially you had an angle at 23 degrees and there is a force pulling it down. At this point, if the problem's that important, I'd say email your teacher.

[quote=UAPaladin]The answer should be 10.19 at 2.47 degrees north of east. Are you sure that you're looking at the answer for the right problem?[/quote]

here, right from the work sheet.

http://gyazo.com/c9b025f9ffb5e385a0c15ae41f7fc103

[quote=iPhone5]I never know when to over complicate it or not to over complicate it. The physics questions have such horrible word choices. On some of the questions I did, this wording essentially meant the boat was resisting the current, in this one it just folows alongside the current. Thanks for your help m8.[/quote]

The answer should be 10.19 at 2.47 degrees north of east. Are you sure that you're looking at the answer for the right problem?

[quote=UAPaladin]The boat doesn't actively try to push back against the current in this problem, it's just doing what the problem simply says. It's moving at a constant speed in the same direction the whole time and doesn't try to make up for what the current's doing to it, and the problem is asking for what the current does to it basically. You're overcomplicating the problem, the boat only does what the problem says it does.

EDIT: I'm just solving it for you on paper right now and I'll SS it.[/quote]

I never know when to over complicate it or not to over complicate it. The physics questions have such horrible word choices. On some of the questions I did, this wording essentially meant the boat was resisting the current, in this one it just folows alongside the current. Thanks for your help m8.

[quote=iPhone5]Why? should I not have to do what I am doing in order to figure out the heading so the boat can go in its desired path despite the current? I am adding in the y direction because the current "subtracts" velocity; the adding will cancel the effect of the current. Do you get what I am trying to say?[/quote]

The boat doesn't actively try to push back against the current in this problem, it's just doing what the problem simply says. It's moving at a constant speed in the same direction the whole time and doesn't try to make up for what the current's doing to it, and the problem is asking for what the current does to it basically. You're overcomplicating the problem, the boat only does what the problem says it does.

EDIT: I'm just solving it for you on paper right now and I'll SS it.

[quote=UAPaladin]That's wrong. Add the values in the x direction and subtract them in the y-direction. Both the boat and the current are going east, so they're heading the same way in the x-direction so you're supposed to add them. The boat's y-direction is north though, while the current's is south, which are opposite directions, so you're supposed to subtract them.[/quote]

Why? should I not have to do what I am doing in order to figure out the heading so the boat can go in its desired path despite the current? I am adding in the y direction because the current "subtracts" velocity; the adding will cancel the effect of the current. Do you get what I am trying to say?

edit: just tried it your way, I got 10.1 knots, that is not the correct answer. The correct answer is 8.7 knots. Help >,>

[quote=iPhone5]I added the values in the y direction and subtracted in the x direction. I got the heading to be 12 knots still. I really dont know anymore. I tried again... x.x ty[/quote]

That's wrong. Add the values in the x direction and subtract them in the y-direction. Both the boat and the current are going east, so they're heading the same way in the x-direction so you're supposed to add them. The boat's y-direction is north though, while the current's is south, which are opposite directions, so you're supposed to subtract them.

[quote=UAPaladin]@iPhone5: You did something wrong then. Did you subtract the values in the y-direction because they were moving opposite of each other? That would be your problem if you didn't.[/quote]

I added the values in the y direction and subtracted in the x direction. I got the heading to be 12 knots still. I really dont know anymore. I tried again... x.x ty

@iPhone5: You did something wrong then. Did you subtract the values in the y-direction because they were moving opposite of each other? That would be your problem if you didn't.

ty i will try again

[quote=iPhone5]did that got 12 knots, answer is 8.7 knots[/quote]

you shouldn't be getting anything over 10.5 knots.. since that's the maximum boat speed.
it's going to be less than 10.5 because a part of the current will be pushing it backwards.

i'll trade you the answer for the non-zero Riemann tensors for the flat space FRWJ metric

[quote=UAPaladin]Break the 2 parts into 2 different triangles. The boat and it's heading and speed, and the current and its heading and speed. Get a vertical and horizontal component of each triangle (previously, you only had the hypotenuse and the angle, but with that you can use some trig to get the x and y components), and sum the 2 triangles together. You'll have to add the horizontal components of the triangle together because both the boat and the current are going east, but you have to subtract the vertical components, because the boat's heading north, but the current's pushing south. Build a new triangle with your new horizontal and vertical components, solve for the hypotenuse to get the chart speed, then solve for the angle formed between the horizontal component and the hypotenuse to get its heading.

I'd solve it for you, but I'm a bit too lazy to atm.[/quote]

did that got 12 knots, answer is 8.7 knots

Break the 2 parts into 2 different triangles. The boat and it's heading and speed, and the current and its heading and speed. Get a vertical and horizontal component of each triangle (previously, you only had the hypotenuse and the angle, but with that you can use some trig to get the x and y components), and sum the 2 triangles together. You'll have to add the horizontal components of the triangle together because both the boat and the current are going east, but you have to subtract the vertical components, because the boat's heading north, but the current's pushing south. Build a new triangle with your new horizontal and vertical components, solve for the hypotenuse to get the chart speed, then solve for the angle formed between the horizontal component and the hypotenuse to get its heading. As long as you do this correctly, you'll get the right answer.

I'd solve it for you, but I'm a bit too lazy to atm.

[quote=xXMCheifXx]do it yourself. you don't learn anything if you have other people do it for you.[/quote]

If Ive been stuck in it for days, i need some help other than draw a diagram. I drew 10 so far. x.x

[quote=iPhone5]can someone solve or give a more detailed procedure. thanks[/quote]

do it yourself. you don't learn anything if you have other people do it for you.

can someone solve or give a more detailed procedure. thanks

you can also shift the axes to make the boat wanted to travel "due north" instead of 23 deg N of E. this makes the current 15 deg S of E.

just remember to switch back to the original coordinates at the end of your calculation since that's what the answer needs to be in (just a shift by 23 deg)

I already tried doing this for the past hour. I need help. I drew pictures and everything. I get the velocity of the heading to be 12 knots, but its wrong.

draw a pic with the vectors and speeds. It'll become clearer and you'll see its a lot easier than you think