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[Uni Physics w/ calc] Boat crossing a river

  • #1
The problem is:

A 230-m-wide river has a uniform flow speed of 1.4 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.6 m/s with respect to the water. There is a clearing on the north bank 63 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

What I have been trying is, multiplying 4.6 by the costheta and sintheta (theta being the angle the boats velocity points toward from west towards north) to get the components of the velocity and then subtracting 1.4 from the x component because of the rivers current. Then I get position information by integrating the velocity equations. I set the x component of the position equal to -63 (I made west positive and east negative) and then solve for t. Then I plug that t into the y component of the position and try to solve for theta. I end up with 230costheta+63sintheta=70. I tried to solve it and quickly realized it is pretty much impossible to find theta from that. So if somebody could figure out what I need to change in how I'm trying to do this problem that would be great, I have been stuck on this problem for 3 days. Last problem in my homework set. Thank you!
 

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  • #2
haruspex
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The problem is:

A 230-m-wide river has a uniform flow speed of 1.4 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.6 m/s with respect to the water. There is a clearing on the north bank 63 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

What I have been trying is, multiplying 4.6 by the costheta and sintheta (theta being the angle the boats velocity points toward from west towards north) to get the components of the velocity and then subtracting 1.4 from the x component because of the rivers current. Then I get position information by integrating the velocity equations. I set the x component of the position equal to -63 (I made west positive and east negative) and then solve for t. Then I plug that t into the y component of the position and try to solve for theta. I end up with 230costheta+63sintheta=70. I tried to solve it and quickly realized it is pretty much impossible to find theta from that. So if somebody could figure out what I need to change in how I'm trying to do this problem that would be great, I have been stuck on this problem for 3 days. Last problem in my homework set. Thank you!
There are two components of position, giving you two equations sharing the same time value.
 
  • #3
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26
The resultant of vector addition should be the path taken by the boat.
 
  • #4
The problem is:

A 230-m-wide river has a uniform flow speed of 1.4 m/s through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of 4.6 m/s with respect to the water. There is a clearing on the north bank 63 m upstream from a point directly opposite the clearing on the south bank. (a) At what angle, measured relative to the direction of flow of the river, must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

What I have been trying is, multiplying 4.6 by the costheta and sintheta (theta being the angle the boats velocity points toward from west towards north) to get the components of the velocity and then subtracting 1.4 from the x component because of the rivers current. Then I get position information by integrating the velocity equations. I set the x component of the position equal to -63 (I made west positive and east negative) and then solve for t. Then I plug that t into the y component of the position and try to solve for theta. I end up with 230costheta+63sintheta=70. I tried to solve it and quickly realized it is pretty much impossible to find theta from that. So if somebody could figure out what I need to change in how I'm trying to do this problem that would be great, I have been stuck on this problem for 3 days. Last problem in my homework set. Thank you!
Can someone PLEASE just actually sit down and try to solve this problem and then explain to me the steps they took. I've been looking for help on many website and everyone does the same thing. They read the question, think for what seems to be 3 seconds, then reply something like "add the vectors". That's not helpful. I need someone who really knows their stuff to solve the problem and then explain to me how they went about doing it. I know I'm complaining about the help I have gotten, but it's useless. I went to a TA's office hours and he had 15 people in his office and by the time it would have been my turn the office hours would have ended. I'm desperate.
 
  • #5
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Haha. Ok ok. Calm down.

Let's break it into 2 parts, alright?
1) First, the river is flowing at 1.4 m/s. This means that if the boat is left in the river on it's own, it will flow at this speed.by its own.
2) Secondly, the boat by itself has a speed of 4.6 m/s, which means that in still water, it will travel with this speed.

Now upstream means in the direction opposite to the direction of flow of water. So what you basically have to do is-
Travel 230 m from South to North.
Travel 63 m from East to West against the current (against 1.4 m/s).
Now these two journeys need to take the same time. This will give you your two vectors and the angle between them.

Got a fair idea?
 

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  • #6
Nathanael
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My approach for part (a) doesn't involve time. I would write an equation that says "the ratio of (east-west component of boat's velocity minus velocity of the river) to (south-north component of boat's velocity) equals the ratio of east-west distance (63 meters) to south-north distance (230 meters)"
 
  • #7
Haha. Ok ok. Calm down.

Let's break it into 2 parts, alright?
1) First, the river is flowing at 1.4 m/s. This means that if the boat is left in the river on it's own, it will flow at this speed.by its own.
2) Secondly, the boat by itself has a speed of 4.6 m/s, which means that in still water, it will travel with this speed.

Now upstream means in the direction opposite to the direction of flow of water. So what you basically have to do is-
Travel 230 m from South to North.
Travel 63 m from East to West against the current (against 1.4 m/s).
Now these two journeys need to take the same time. This will give you your two vectors and the angle between them.

Got a fair idea?
Hi thank you so much for this response, it's coming together. Although, in the question it wants me to find the angle relative to the flow of the water. In your picture can you point out which angle that is? In your picture is the angle the question is asking for the one that creates a less than 90 degree angle or the one that creates an over 90 degree angle with the desired path (the arrow pointing right and up) of the boat?
 
  • #8
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Hi thank you so much for this response, it's coming together. Although, in the question it wants me to find the angle relative to the flow of the water. In your picture can you point out which angle that is? In your picture is the angle the question is asking for the one that creates a less than 90 degree angle or the one that creates an over 90 degree angle with the desired path (the arrow pointing right and up) of the boat?
Glad to help :)
Come on, try it now! Read the question once again and see which angle they are asking for.
 
  • #9
My approach for part (a) doesn't involve time. I would write an equation that says "the ratio of (east-west component of boat's velocity minus velocity of the river) to (south-north component of boat's velocity) equals the ratio of east-west distance (63 meters) to south-north distance (230 meters)"
When I tried to do that I end up with 1058costheta-322=289.8sintheta I don't know how to solve for theta from that. Thanks for the response though! I'm thinking maybe I can somehow get the cos or the sin in terms of the other using some other relationship?
 
  • #10
Glad to help :)
Come on, try it now! Read the question once again and see which angle they are asking for.
It says the angle relative to the flow of the river does that mean the angle that points toward the flow or against it?
 
  • #11
Nathanael
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When I tried to do that I end up with 1058costheta-322=289.8sintheta I don't know how to solve for theta from that. Thanks for the response though! I'm thinking maybe I can somehow get the cos or the sin in terms of the other using some other relationship?
You just need one more constraint. That comes from the nature of sine and cosine; [itex]\sin^2\theta+\cos^2\theta=1[/itex]
 
  • #12
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It says the angle relative to the flow of the river does that mean the angle that points toward the flow or against it?
Its the angle BETWEEN the direction of flow of the river and the direction of travel of the boat.
 
  • #13
haruspex
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Can someone PLEASE just actually sit down and try to solve this problem and then explain to me the steps they took.
That's not how it works. You are supposed to post the detailed steps you took, then others comment. Unfortunately you only posted a description of what you did. That makes it much harder to figure out exactly where you are stuck.
 
  • #14
You just need one more constraint. That comes from the nature of sine and cosine; [itex]\sin^2\theta+\cos^2\theta=1[/itex]
Okay so I tried some stuff and couldn't get anywhere. Could you please tell me what the constraint is?
 
  • #15
Nathanael
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Okay so I tried some stuff and couldn't get anywhere. Could you please tell me what the constraint is?
Sorry if the way I said it confused you; [itex]\sin^2\theta+\cos^2\theta=1[/itex] is the constraint. You can use it to get a quadratic equation for either cosθ or sinθ.
 
  • #16
Sorry if the way I said it confused you; [itex]\sin^2\theta+\cos^2\theta=1[/itex] is the constraint. You can use it to get a quadratic equation for either cosθ or sinθ.
So I solved for sin theta and then plugged it into the identity your referencing. Then I used wolfram alpha to find the zeros of the quadratic which were costheta = 0.0305627 and costheta = 0.535652. Now, the issue is I have 2 submissions left on this question before I'm locked out. Does anybody think either of these thetas are correct and if so, which one?
 
  • #17
Nathanael
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So I solved for sin theta and then plugged it into the identity your referencing. Then I used wolfram alpha to find the zeros of the quadratic which were costheta = 0.0305627 and costheta = 0.535652. Now, the issue is I have 2 submissions left on this question before I'm locked out. Does anybody think either of these thetas are correct and if so, which one?
Take the arccos to find the angles. Only one of them is reasonable. Which one?
 
  • #18
Sorry if the way I said it confused you; [itex]\sin^2\theta+\cos^2\theta=1[/itex] is the constraint. You can use it to get a quadratic equation for either cosθ or sinθ.
Here is a more detailed account of what I did.
So I have 1058costheta-322=289.8sintheta from earlier. I take this and solve for sintheta. Then, I plug what sintheta equals into sinsquared+cossquared=1. I substitute x for costheta. Then I plug the quadratic into wolfram alpha. It says x=0.535652 and x=0.0305627
 
  • #19
Take the arccos to find the angles. Only one of them is reasonable. Which one?
Why is only one of them reasonable? One of them equals 57. something and the other is 88. someth
 
  • #20
Nathanael
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Why is only one of them reasonable? One of them equals 57. something and the other is 88. someth
Well, what if the river was perfectly still. Then what angle would you be traveling?
 
  • #21
Well, what if the river was perfectly still. Then what angle would you be traveling?
Okay so this is crazy. I used the 50 seomthing degree angle and using that I got part B. Then, when enter part A it tells me the angle is wrong. I think I have the angle that the boat is traveling in but not the angle the boat is pointed in.
 
  • #22
WELP BOYS. I tried a final answer, and wrong. My father was telling me that the angle is the angle that goes into the current and that that's what "relative to the water means". GUESS WHAT? I had the right angle. He was just wrong. It was the angle -180. FMLMFMLFMFLMFLMFLFLMFLFMFLMFLFMFLMFLFMLFMLFMFLMFLMLFMLF 3 days. 3 days I have been trying. Well thanks everyone. Gonna go die now
 
  • #23
Nathanael
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I had the right angle. He was just wrong. It was the angle -180.
Perhaps you can message your instructor and tell him/her you misunderstood what "relative to the flow of water" meant and maybe get partial credit.

Anyway, the understanding is what's important; grades are secondary (imo). As long as you understand the problem, the 3 days was not wasted.
 

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