View Full Version : How wide is the river?

sanddrag

04-10-2005, 07:44 PM

I'm trying to solve this problem. http://home.att.net/~numericana/answer/recreational.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 The boat which had travelled 720 yd at the first meeting has therefore travelled three times that (2160 yd) at the second meeting. Can someone represent this graphically for me?

I have spent over two hours on this and I just can't understand. Please help.

Eugenia Gabrielov

04-10-2005, 08:09 PM

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...

sanddrag

04-10-2005, 08:14 PM

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/database/QQ.09.00/gil1.html

Eugenia Gabrielov

04-10-2005, 08:19 PM

Edit: SORRY! I tried, it seems it's not followable. Maybe I just think funny.

So then, before I add the 10 minute thingy...Let's look at what happens with Boat #1 again.

To review, it goes 720 yards at velocity 1 or v1. it goes 720 + e yards (the whole river) in v1 (TA + TC). Then it stops for 10 minutes. Then it makes it another 400 yards before it meets up with Boat #2. So the equation goes as such. But what would happen if we kinda...ruled out those 10 minutes for one second? Let's say TF is the time it takes to make those 400 yards

Distance = velocity * time

1120 + E = v1 (TA + TC + TF), and then we'll have to factor in those 10 minutes in a moment.

So what happens to Boat #2?

It makes it E yards. Then it makes it 720 yards. Then it makes it G yards. Forgetting the 10 extra loadage minutes for a second, let's say...

720 + E + G = v2 (TA + TB + TG)

So what exactly is G? G is R - 400, cause it's the distance traveled discluding 400 yards of river.. G is 720 + E - 400 which is equal to 320 + E. Therefore,

720 + E + 320 + E = 1040 + 2E

1040 + 2E = v2 ( TA + TB + TG), later factoring in 10 minutes of passage time, and...

1120 + E = v1 (TA + TC + TF), again, I'm still not sure precisely how to factor those next 10 minutes in.

Denman

04-11-2005, 03:34 AM

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

Alan Anderson

04-11-2005, 10:13 AM

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?

Why it works is easy.

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.

Greg Ross

04-11-2005, 10:44 AM

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.

You have to be careful that you're not dividing by zero, though. (Which would happen in the above example if w = -b or if 2w = b.) Otherwise you might convince yourself that 1 = 2 (http://www.chiefdelphi.com/forums/showpost.php?p=231527&postcount=60). (See my response too. (http://www.chiefdelphi.com/forums/showpost.php?p=231566&postcount=62))

Alan Anderson

04-11-2005, 01:50 PM

You have to be careful that you're not dividing by zero, though. (Which would happen in the above example if w = -b or if 2w = b.)

Good point. However, since the above example had (w+b) / u = (2w-b) / v, dividing by zero would require both w = -b and 2w = b. That's only possible if w and b are both equal to zero.

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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