Relative velocity/Boat & River Question

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Homework Statement


280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?
uqT7ydn.png


Homework Equations


[itex]V_x = Vcosθ[/itex]
[itex]V_y = Vsinθ[/itex]
[itex]x = V_0 t[/itex] (starting points set to 0 for x and y, no acceleration involved)
[itex]V_{BS} = V_{BW} + V_{WS}[/itex]

The Attempt at a Solution


y component of velocity unaffected by current, so [itex]V_{BS} = V_{BW} + V_{WS}[/itex] will be relevent for x component only.
[itex]V_x = Vcosθ = 2.7 cosθ[/itex]
[itex]V_y = Vsinθ = 2.7 sinθ[/itex]
[itex]V_{BS} = V_{BW} + V_{WS} = (2.7 cosθ) - 1.35[/itex]

[itex]y = V_0 t = 280 = (2.7 sinθ)t[/itex]
[itex]t = 280/(2.7 sin)[/itex]

[itex]x = V_0 t = 120 = ((2.7 cosθ) - 1.35) t[/itex]
[itex]t = 120/((2.7 cosθ)-1.35)[/itex]

time it takes to go across river (y component) equals time to travel upstream (x component)
[itex]t = t = 280/(2.7 sin) = 120/((2.7 cosθ)-1.35)[/itex]
[itex]280((2.7 cosθ) - 1.35) = 120 (2.7 sinθ)[/itex]
[itex](756 cosθ) - 378 = (324 sinθ)[/itex]
[itex](14 cosθ) - 7 = (6 sinθ)[/itex]
[itex]7((2 cosθ) - 1) = (6 sinθ)[/itex]
[itex](2 cosθ) - 1 = (6/7) sinθ[/itex]
[itex]((2 cosθ) - 1)/sinθ = 6/7[/itex]
[itex](2(cosθ / sinθ)) - (1/sinθ) = 6/7[/itex]
[itex]2 cotθ - cscθ = 6/7[/itex]

I then plotted this using a graphing calculator and found where [itex]y = 2 cot(x) - csc(x)[/itex] crossed [itex]y = 6/7[/itex] and found θ = 39.5°. This answer does seem to check, but I have a feeling I'm doing something wrong, and that I should be able to solve for θ without using a graphing calculator. I'm not sure if the law of sines or law of cosines will come into play, or if I'm missing a certain trigonometric identity. Any insight would be greatly appreciated. Thanks.
 
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Answers and Replies

  • #2
ehild
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Your equation 14cos(θ)-6sin(θ)=7 is correct.
You can proceed by substituting cosθ=sqrt(1-sin2θ) and solve a quadratic equation.

The other method is to rewrite the equation as A(cosβcosθ-sinβsinθ)=Acos(β+θ)=7, where Acosβ=14 and Asinβ=6: tanβ=3/7 and A2=142+62.

ehild
 
  • #3
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Thanks. I used the quadratic equation way and got the same answer as my previous attempt.
 
  • #4
ehild
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Splendid. :smile:

ehild
 
  • #5
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I have a question for this. Doesn't going upstream mean going against the water flow?
 
  • #6
ehild
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Yes, it is against the water flow.

ehild
 
  • #7
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Yes, it is against the water flow.

ehild
In the picture wouldn't the 120m distance be the whole horizontal distance? I am bit confused on that.
 
  • #8
ehild
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120 m is the distance along the river where the boat is due.

ehild
 
  • #9
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120 m is the distance along the river where the boat is due.

ehild
Dalkiel had this: "280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?"

In the bolded part wouldn't it make more sense if it did not say upstream and just " destination 120m" or something like that? Because the water is coming from the right side.
 
  • #10
ehild
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120 m alone could mean both upstream and downstream, that is to the right from the starting point or to the left.

ehild
 
  • #11
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120 m alone could mean both upstream and downstream, that is to the right from the starting point or to the left.

ehild
Oh so even though the destination will be at an angle from starting point it still means upstream or downstream. Thanks!
 
  • #12
ehild
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That 120 m is the distance on the other side of the river from the point opposite to the starting point.

ehild
 
  • #13
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That 120 m is the distance on the other side of the river from the point opposite to the starting point.

ehild
so 280 is horizontal distance then
 
  • #14
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so 280 is horizontal distance then
280 is the width of the river. if there was no current, and the boat wanted to go straight across the river, it would have to travel exactly 280 m. In my (horribly drawn) picture, 280 is the vertical distance, or the y component of the displacement.
 
  • #15
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The answer that was found (39.5°) is correct, but there was actually another way to solve this, using the law of sines. I'll try to draw another picture to describe it when I get home.
 
  • #16
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The answer that was found (39.5°) is correct, but there was actually another way to solve this, using the law of sines. I'll try to draw another picture to describe it when I get home.
Thanks! I was mostly confused with the wording of the question.
 
  • #17
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Sorry about the wording, I basically just paraphrased the problem to the basic components.

Here's the solution that was explained to me, using another horribly drawn ms paint picture as a guide.

yomTMYR.png


The yellow line is straight across the river, green line is where we want to go, brown is where we have to aim, dark blue is the velocity of the water. Since we know the length of two sides of the right triangle, we can find the angles. tanβ = 120/280. so β=23.2° and γ=66.8°. Since we know γ, we can find σ= 113.2°. Then use law of sines to find 2.7/sin(113.2) = 1.35/sinα. α=27.4. 90-β-α=θ=39.4°. Which matches all previous attempts.
I personally didn't care for this method because I'm mixing angles involved with distance with those using velocity. My initial intuition would be to not mix the two, but this seemed to work out.
 
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  • #18
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Sorry about the wording, I basically just paraphrased the problem to the basic components.

Here's the solution that was explained to me, using another horribly drawn ms paint picture as a guide.

yomTMYR.png


The yellow line is straight across the river, green line is where we want to go, brown is where we have to aim, dark blue is the velocity of the water. Since we know the length of two sides of the right triangle, we can find the angles. tanβ = 120/280. so β=23.2° and γ=66.8°. Since we know γ, we can find σ= 113.2°. Then use law of sines to find 2.7/sin(113.2) = 1.35/sinα. α=27.4. 90-β-α=θ=39.4°. Which matches all previous attempts.
I personally didn't care for this method because I'm mixing angles involved with distance with those using velocity. My initial intuition would be to not mix the two, but this seemed to work out.
Oh I see and thanks for posting this up. Do you mind posting up the whole question? :shy:
 
  • #19
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Oh I see and thanks for posting this up. Do you mind posting up the whole question? :shy:
Sure, but it's actually a combination of two questions. One is direct from a textbook, and then and then there's my professor's part.
Original from text (problem #70): A boat, whose speed in still water is 2.70 m/s, must cross a 280-m-wide river and arrive at a point 120 m upstream from where it starts. To do so, the pilot must head the boat at a 45.0° upstream angle. What is the speed of the river's current?

Professor's question: In prob 70, suppose the river current is 1.35 m/s and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be in that case, in order to reach the same destination?
 

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