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Kinematics - Boat over a constant River.

  • Thread starter Exuro89
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  • #1
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Homework Statement


A 400m wide river flows from west to east at 30 m/min. Your boat moves at 100m.min relative to the water no matter which direction you point. To cross the river, you start from a dock at point A on the south bank. There is a boat landing directly opposite at point B on the north bank, and also one at point C, 75m downstream from B.

(a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled?

(b)If you initially aim your boat directly toward point C and do not change the bearing relative to the shore, where on the north shore will you land?

(c) To reach point C:(i)at what bearing must you aim you boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) what is the speed of your boat as measured by an observer standing on the river bank?

[PLAIN]http://i.min.us/jcg2eG.png [Broken]

Homework Equations


X = X_0+V_0*t
X^2 + Y^2 = R^2
cos(theta)=opp/hypo
sin(theta)=adj/hypo


The Attempt at a Solution


I'm stuck at C but I'll show you what I did for parts A and B to make sure that I'm attempting the problem correctly.

(a)What I did for the first part was find the time it take to go across the river. I found this by using
Y = Y_0 + V_0y*t
400m = 0m + 100m/min * t
400m/100m/min = t
t = 4min

Because the river is the only x component affecting the boat, I can take that and multiply that by 4 minutes to get the distance on the north shore.

X = X_0 + V_0x*t
X = 0 + 30m/min*4min
X = 120m

The second part of A asks the distance traveled. Because I have both X and Y distance I can use the pythagorean theorem to find the distance traveled.

X^2 + Y^2 = R^2
120^2 + 400^2 = R^2
R = 417.6m

(b) So for this question the boat is initially going towards point C, so I can find the degree right of the y axis because I have both Y length 400, and X length 75.

tan(theta) = 75/400
theta = tan^-1(75/400)
theta = 10.6 degrees

I can take theta and find both components of V_0

V_0y = 100cos(theta)
V_0y = 98.3m/min

V_0x = 100sin(theta)
V_0x = 18.4m/min

After this I can find how long it take to get across

Y = Y_0 + V_0y*t
400m = 0 + 98.3m/min*t
t = 4.07min

With the time I can multiply it by the x component(along with the added river's x component) to get the x distance

X = X_0 + V_0x*t
X = 0 + (18.4m/min+30m/min)*4.07min
X = 197m

So by pointing towards point C initially you would go 197m east of point C.

(c)Okay I'm having trouble with this. I'm not quite sure how I would find the bearing. I tried this, as I don't know either theta or time, i figured I could figure out 2 equations for time and then equal them to each other, but Im stuck.

Y = Y_0 + V_0y*t X = X_0 + V_0x*t
400m = 0 + 100m/mincos(theta)*t 75m = 0 + (100m/minsin(theta)+30m/min)*t

I solved for t, made them equal, but can't get theta out. Am I doing this wrong, what should I try, and are the earlier answers correct?

Thanks.
 
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Answers and Replies

  • #2
2,685
20
Sorry, didn't read your post first so ended up typing help tips to the lot.

OK, for part C:

You need to calculate the angle required to give the horizontal component of your boats velocity equal to 30m/min. That will then cancel out the river's flow and allow the boat to travel in a straight line.

You'll also have the vertical velocity from the above so you can get the time to cross.

The distance travelled in this case will simply be the width of the river.

So from all this you will have:
1. Bearing of the boat (angle required).
2. Time taken to cross
3. Distance travelled
4. Speed relative to an observer (assuming that's oncoming speed then that will be the vertical component used for time taken).
 
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  • #3
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I'm actually trying to reach point C, not point B. To reach point B I understand that I would need an x component of -30m/min to cancel out the river, but I don't know what I would need to get to point C. I know that it will have distances of y component 400m and x component 75m so I could find the distance it takes, but I'm not sure that helps me, seeing as my velocity would be going somewhere north-westish. I don't know how to solve for theta.
 
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  • #4
2,685
20
I'm actually trying to reach point C, not point B. To reach point A I understand that I would need an x component of -30m/min to cancel out the river, but I don't know what I would need to do to get to point C.
Bugger, thought it was Part C not point C.

Hang on I'll be back to answer it shortly.
 
  • #5
2,685
20
Sorry for the delay. OK, you're on the right track.

So I'll step through it as you did:

X = X0 + V0SIN(θ)t
Y = Y0 + V0COS(θ)t

t = X / (X0 + V0SIN(θ))
t = Y / (Y0 + V0COS(θ))

Rearrange:

X / (X0 + V0SIN(θ)) = Y / (Y0 + V0COS(θ))

Rearrange:

X / Y = (X0 + V0SIN(θ)) / (Y0 + V0SIN(θ))

Now would be a good time to sub in your values:

75 / 400 = (0 + 100SIN(θ)) / (0 + 100COS(θ)) = SIN(θ) / COS(θ) = 0.75

Now, you need to remember trig identities. Specifically, SIN(θ) / COS(θ) = TAN(θ)

So you're left with TAN(θ) = 0.75.

Can you take it from here?
 
  • #6
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I think you did 75/100 which is .75 and not 75/400 which is .1875.

I'm still trying to grasp this. Theta would be 10.6 degrees but wouldn't that only work if the river wasn't flowing? Don't I need to add the river's 30m/min or is that included? The way it's set up now looks like the boat is traveling 10.6 degrees east of the y axis which would mean it's going to be traveling much farther than point C with the added 30m/min.

EDIT: I think I got it. Give me a sec to conceptualize this out.
 
  • #7
2,685
20
I think you did 75/100 which is .75 and not 75/400 which is .1875.
Sorry, correct it's 400.
I'm still trying to grasp this. Theta would be 10.6 degrees but wouldn't that only work if the river wasn't flowing? Don't I need to add the river's 30m/min or is that included? The way it's set up now looks like the boat is traveling 10.6 degrees east of the y axis which would mean it's going to be traveling much farther than point C with the added 30m/min.

EDIT: I think I got it. Give me a sec to conceptualize this out.
You're right, I forgot to add the rivers flow to the X calcs.
 
  • #8
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Okay so what would we change to the equation to add in river flow?

Would it be 75/400 = (100sin(theta) + 30)/100cos(theta) ?
 
  • #9
2,685
20
Right in your post before you deleted it, you used 18.4 in the time check section which is wrong, it should be 11.57.

18.4 is the value you need but I can't get it for the life of me. You need the boat to move backwards at 18.4 and then it works.
 
  • #10
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How did you get 11.57? I found 18.4 by using theta = 10.6 then finding the x component. I then assumed that that included the river's flow and that if I subtracted 30 I could have the boat's actual component which would be -11.6 which sounds right as the boat would need to be going against the water.

EDIT: It should be -11.57 yeah? That would be the boat's x component and then tied with the river's velocity it would make the boat travel at 18.4m/min? So this IS on track then? THe only reason I deleted that post was because you said something was wrong so I figured everything I was doing was incorrect :P.
 
  • #11
2,685
20
How did you get 11.57? I found 18.4 by using theta = 10.6 then finding the x component. I then assumed that that included the river's flow and that if I subtracted 30 I could have the boat's actual component which would be -11.6 which sounds right as the boat would need to be going against the water.

EDIT: It should be -11.57 yeah?
Yeah.

Which means the boat would be aiming up river but moving backwards with a speed of 11.57. This only covers 40m though, you need a speed of ~-18.5 to get it to C.
 
  • #12
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And that makes sense. The boat going straight across went some 120ish meters out, so the boat would need to point in a negative direction to get to C. That would mean that the 11.6m/min the boat is travelling west BUT the 30m/min river causes it to go 18.4m/min East which means it would make it to point C correct?
 
  • #13
2,685
20
And that makes sense. The boat going straight across went some 120ish meters out, so the boat would need to point in a negative direction to get to C. That would mean that the 11.6m/min the boat is travelling west BUT the 30m/min river causes it to go 18.4m/min East which means it would make it to point C correct?
No.

If the boat points 10.6 degrees from North towards West, it would have 18.4 heading west. Subtract this from 30 gives you 11.6 heading east. This means after the 4 minutes to cross the river it will have travelled ~44m east. Not enough.
 
  • #14
34
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Okay I think I see what you're saying. Ugh this is a tough question....
 
  • #15
verty
Homework Helper
2,164
198
Hiya. This is a VERY difficult question, I think there are no shortcuts. Let the boat start at (0,0), then it crosses the river in a straight line and ends up at (75,400).

Using the line equation, we have:

y - 0 = m(x - 0), y = mx

y - 400 = m(x - 75), y = mx + 400 - 75m

400 = 75m
m = 400/75

y = (400/75)x

The boat stays on this line until it reaches point C. Now, where is the boat at time t = 1 minute? Let θ be the angle with the river bank. It is at:

( 100 cos(θ) + 30, 100 sin(θ) )

And this point must be on that line:

100 sin(θ) = (400/75) (100 cos(θ) + 30)
75 sin(θ) = 400 cos(θ) + 120

This can be solved but the algebra is horrendous. It is safer doing it by computer. According to the computer, the solution of:

[tex]a sin(x) + b cos(x) = c[/tex]

is

[tex]x = 2 tan^{-1}(\frac{ a \pm \sqrt{a^2 + b^2 - c^2} }{b+c}).[/tex]

You could use this formula to find the angle.
 
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  • #16
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"100 sin(θ) = (400/75) (100 cos(θ) + 30)"

Shouldn't this be 100sin(θ) = (75/400)100cos(θ) - 30? 100sin(θ) Would be the x component no?

When I was first doing this I got it down to this but I had no idea how to get θ by itself and that's pretty much why I posted here. It seems like this question is much harder than I initially thought(I normally "miss" things that cause me to have trouble)
 
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  • #17
verty
Homework Helper
2,164
198
"100 sin(θ) = (400/75) (100 cos(θ) + 30)"

Shouldn't this be 100cos(θ) = (400/75)(100sin(θ) + 30)? 100sin(θ) Would be the x component no?
I chose to let θ be the angle to the river bank.
 
  • #18
34
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Think I'm off in my math. I did as you said and got

theta = 6.53

V_0y = 99.35

V_0x = -11.06+30

t = 4.02

When I check for the X length I'm a little high, at 76.26m instead of 75, so maybe I'll trying that again with more precision.

So, is there no easier way to do this? This is my first physics class and so far everything else has been far easier than this question. i don't know how I would come up with that equation algebraically.
 
  • #19
verty
Homework Helper
2,164
198
I think it was a typo, it should have said B, not C. Anyway, you found the same angle I did. Good job.
 
  • #20
34
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Yeah, had it been B this wouldn't be much of an issue. So the angle is going to be somewhere between 6.5 and 6.6 degrees. This was such a pain. Hopefully this type of question isn't on the exam. I'm doing the questions out of an e-book so maybe it's a typo on there.
 
  • #21
34
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Well I found out what I needed to do to solve for theta. The Law of Sines made it easy to work it out. After that it was a piece of cake.
 
  • #22
2,685
20
Ah, well done.
 

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