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How wide is the river?
I'm trying to solve this problem. http://home.att.net/~numericana/answ...onal.htm#ferry I understand most of it except for one thing. With the equation method of solving, how is it valid to divide the two equations? I know you are trying to eliminate the speed variable but I have never heard of that method (dividing equations) of solving before. Can someone please explain why that works?
And for the part below in blue text, I don't understand this part Quote:
I have spent over two hours on this and I just can't understand. Please help. |
Re: How wide is the river?
Two ferry boats start at the same instant on opposite sides of the river. One is faster than the other. They cross at a point 720 yards from the left shore on their way to their respective destinations, where each one spends 10 minutes to change passengers before the return trip. They meet again at a point 400 yards from the right shore.
How wide is the river? I tried to draw a picture but it wouldn't load...essentially what happens is...and this might be totally wrong since I am not feeilng up to reading that whole solution thing. Boat #1 is the one on the left side of the river, and boat #2 is on the right side, so at the beginning, boat #1 travels 720 feet until it meets boat #2. Boat #2 goes "E" yards from the right side, before it first meets Boat #1. It then travels 720 feet further across the river. The length of the river is for now, r = 720 + e. Distance = velocity * time So, for now, in regards to their first meeting, 720 = v1 * T(a), and e = v2 * T(a). That time is the same becuase they took the same amount of time to go those respective distances. So then, Boat #2 travels 720 more yards at v2 in some more time, and thus 720 + e = v2 * (t(a) + t(b)). Looking at Boat #1, it goes that E distance at v1, at what we'll call time c. OK, so then, 720 + e = v1 * (t(a) + t(c)). Using the transitive property, we find that... (v2)(TA) + (v2)(Tb) = (V1)(ta) + (V1)(Tc) or (V2)(TA + TB) = (V1)(TA + TC) That's how far I am right now, I am deciding how to add the next factor...I hope it kinda helps... |
Re: How wide is the river?
Thanks for replying but I couldn't follow you. However, I did find this that explains is pretty well, I think. http://mathcentral.uregina.ca/QQ/dat...9.00/gil1.html
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Re: How wide is the river?
Edit: SORRY! I tried, it seems it's not followable. Maybe I just think funny.
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So then, before I add the 10 minute thingy...Let's look at what happens with Boat #1 again. |
Re: How wide is the river?
it is a perfectly acceptable method.
originally we would use it to see if something is true. say you have 2 equations with the same variables in both, dividing through allows you to see if its true,, as you should get 1=1 or something similar |
Re: How wide is the river?
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Consider that an equation means that the values on each side of the equals sign are, well, equal. It's perfectly valid to do the same arithmetic operation to each side of an equation, e.g. add 5, or multiply by 12, or subtract x, or divide by a. Start with the equation a / u = (w-a) / v. You can accept multiplying both sides by u, right? You can also divide both sides by something more complicated, such as (w+b) / u. Here's the fun part: if you know that (w+b) / u = (2w-b) / v, you can divide the left side of the original equation by the left side of the new equation, while dividing the other side of the original equation by the other side of the new one. You're still dividing both sides by the same thing. |
Re: How wide is the river?
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Re: How wide is the river?
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Since the original problem had nonzero values for both a and b, we know that we're not dividing by zero. |
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