Rescue Boat to Save Little Child

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Rescue Boat to Save Little Child

Postby QED-1 » Mon Jul 01, 2013 1:59 am UTC

Hi all, I need help to solve a homework question. I hope you guys won't mind and I just need a hint (or two). Question is:

A child in danger of drowning in a river is being carried downstream by a current that flows uniformly with a speed of 2.0 m/s. The child is 200 m from the shore and 1500 m upstream of the boat dock from which the rescue team sets out. If their boat speed is 8.0 m/s with respect to the water, at what angle from the shore should the pilot leave the shore to go directly to the child?

I think of this as a river flowing north/south with the boat dock 1500 m south of the child on the same bank that the question says is 200 m from the child. The rescue boat is running north approaching the child who is flowing downstream (south) with the shoreline being the y-axis and the breadth of the river being the x-axis. The child moves so the theta is increasing as time passes but I'm not sure how to solve for it.

I was thinking solve for the time when they are equal with r=d/t but that only got me with the fact that for every unit of distance the child travels, the boat does 4x more. Am I right by saying that the resultant vector created by the velocity of the boat is 8? And the x-component is 8*sin(theta) and y-component is 8*cos(theta)? But now I'm thinking, what if instead of 8*cos(theta) it's actually 8*cos(theta)-2? That way the child moving is subtracted from the moving boat.

I'm not sure what to do, could anyone do me a favor and help me out?
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Re: Rescue Boat to Save Little Child

Postby jedelmania » Tue Jul 02, 2013 8:55 am UTC

The child moves so the theta is increasing as time passes but I'm not sure how to solve for it.


And the x-component is 8*sin(theta) and y-component is 8*cos(theta)


What are you calling theta? Is it the angle between the boat and the child (which changes with time) or the angle at which the boat is moving (which doesn't)? The first statement implies the former and the second statement implies the latter
But now I'm thinking, what if instead of 8*cos(theta) it's actually 8*cos(theta)-2? That way the child moving is subtracted from the moving boat.


Spoiler:
Why would you want to subtract the child moving from the boat? I'm not saying you are definately wrong, but please articulate the reasoning behind this



If you want a little clue:
Spoiler:
I think you are right that this is a trigonometry type question and you are putting the cos and sin functions in the right place.
Where are the boat/child now? Where will they be in 1, 2 or 3 seconds?
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Re: Rescue Boat to Save Little Child

Postby scribbler » Tue Jul 02, 2013 11:19 am UTC

As an alternative to jedelmania's method, I would suggest working off this:

QED-1 wrote:I was thinking solve for the time when they are equal with r=d/t but that only got me with the fact that for every unit of distance the child travels, the boat does 4x more.


But I think you mean the boat travels 4 units for every unit of distance the child travels (your phrasing implies the boat is moving at 10 m/s). So, what can you say about the intersection point?

Hint:
Spoiler:
Call the intersection point (x,y).
Derive expressions for the distance the boat travels and the child travels and put them into an equation.


Further hint:
Spoiler:
The intersection point is on the child's path. This leaves with you with only 1 variable in the above equation.


Finding the intersection point tells you how far north the boat must travel, allowing you to find the angle with basic trig.
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Re: Rescue Boat to Save Little Child

Postby brandbarth » Tue Jul 02, 2013 2:05 pm UTC

stromversetzung.jpg


Both the child and the boat are moved downstream the same distance in the same time (black arrows = "Stromversetzung" - I don't know an english word for that :( ). So if you aim directly for the child's position starting position at t0 (red arrow), you will reach the child at tx (blue arrow), no matter how fast the boat is. As long as it's significantly faster than the current.

And, because it's only asked for the angle, the rest is basic trigonometry.
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Re: Rescue Boat to Save Little Child

Postby dudiobugtron » Wed Jul 03, 2013 2:23 am UTC

brandbarth wrote:As long as it's significantly faster than the current.

It doesn't have to be much faster than the current at all. As long as it is moving relative to the water, then you'll eventually reach the child.
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Re: Rescue Boat to Save Little Child

Postby QED-1 » Wed Jul 03, 2013 3:06 am UTC

Ah, didn't realize the thread went through. We figured it out, and this is what we came out with:

Let's use brandbarth's picture (thanks for the time to do that!). We found the angle which the boat would have left at if the child was holding on to a rock. That angle would be tan^-1(200/1500)=7.595 deg. Now that we have that angle, we can use it for a velocity graph. y-component would be +8*sin(7.595) = +1.057 m/s and x-component would be -8*cos(7.595) = -7.920 m/s. Again, that would be for the child holding on to a rock and using brandbarth's picture. Note that the y-component is + and the x-component is - this is because this relative to the shore. However, the child is moving, and to the right, so the child's velocity is +2m/s. This is added to the x-component velocity to take the moving child into account. Since the river flows uniformly we assume the child only has a x velocity and so the y-component for velocity is untouched. Now you solve for the angle:

tan^-1(+1.057/(-7.920+2))=-10.1 which is correct except for the fact that it's supposed to be a positive angle because we just want the angle.

I really appreciate the help brandbarth, scribbler, and jedelmania! What really messed with my mind was the "respect to the water" which comes up in other questions as " velocity with relative to X" and then they are solved in different ways in the book with the hypotenuse being the velocity and sometimes a component. I guess that would be in the physics forum. Thanks again, and I hope I was clear enough.
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Re: Rescue Boat to Save Little Child

Postby Derek » Wed Jul 03, 2013 4:01 pm UTC

The current effects the boat and the child equally, so you can ignore it. Just solve the problem as if there is no current.

(black arrows = "Stromversetzung" - I don't know an english word for that )

Google Translate gives "current dislocation", so maybe current vector? Or current velocity?
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Re: Rescue Boat to Save Little Child

Postby Dinosaur! » Thu Jul 11, 2013 6:49 am UTC

Derek wrote:The current effects the boat and the child equally, so you can ignore it. Just solve the problem as if there is no current.

(black arrows = "Stromversetzung" - I don't know an english word for that )

Google Translate gives "current dislocation", so maybe current vector? Or current velocity?


Typing 'current' into an english-german dictionary gives a number of definitions, one of which is strom. Translated back to English, we can see that strom means "river" (we get our word "stream from it").[1]

From this, I'd feel safe inferring that a better translation for Stromversetzung would be something like "stream's dislocation." Base on that interpretation, I'd say that the black arrows are displacement vectors for the river's current, the red arrow is the displacement vector for where one should aim, and the blue arrow is their sum, ie, where one ends up reaching the child

[1]from wordreference.c0m; but I cannot post links.
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Re: Rescue Boat to Save Little Child

Postby eSOANEM » Fri Jul 12, 2013 8:43 pm UTC

My google search for Stromversetzung just gave drift which is a perfectly sensible translation for this context.
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Re: Rescue Boat to Save Little Child

Postby korona » Mon Jul 15, 2013 9:59 pm UTC

"Strom" means "flow" in this context (it can also mean electric current) and "versetzung" means "displacement", which makes sense in this context. I never heard the word "Stromversetzung" before, I don't think that it is a common german word.
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