Question

I really need help with this!!!!! The speed of the current in a river is 6 mph. A ferry operator who works that part of the river is looking to buy a new boat for his business. Every day, his route takes him 22.5 miles against the current and back to his dock, and he needs to make the trip a total of 9 hours. He has a boat in mind but he can only test it on the lake with no current. How fast must the boat go on the lake in order to for it to serve the ferry operators needs?

Answer 1

Let \(s\) be the speed of the boat, and \(c\) be the speed of the current. When the boat goes with the current, its speed is \(s+c\). When the boat goes against the current, its speed is \(s-c\). The wording is a bit confusing, but I interpret it as his route takes him 22.5 miles with the current and 22.5 miles against the current, taking a total of 9 hours.
The speed of the current, \(c\), is 6 miles/hour.
From the formula \(d = st\) where \(d\) is distance, \(s\) is speed, and \(t\) is time, we can determine that the time to go a given distance at a given speed is \[d=st\]\[t = \frac{d}{s}\]

Trip with the current:\[t_{with} = \frac{22.5 \text{ miles}}{s+6 \text{ miles/hr}} = \frac{22.5}{s+6} \text{ hr}\]
Trip against the current:\[t_{against} = \frac{22.5 \text{ miles}}{s-6 \text{ miles/hr}} = \frac{22.5}{s-6} \text{ hr}\]
Both trips together: \[t_{with}+t_{against} = 9 \text{ hr}\]Final equation to solve, with units removed for clarity\[\frac{22.5}{s+6} + \frac{22.5}{s-6} = 9\]Solve that equation for \(s\) and you've got the speed (in miles/hr) the boat needs to maintain in still water to be able to meet the requirements. You'll get two solutions, but it should be obvious that one of them does not apply to this problem!

Answer 2

@whpalmer4 Okay. I've got what you said here, I'm just not completely sure what to do next. Every time I tried before it cancelled out.

Answer 3

Well, let's do something to make the writing a bit easier for now: we'll substitute \(a=22.5\). Any objection?

Answer 4

That gives us \[\frac{a}{s+6}+\frac{a}{s-6}=0\]

Answer 5

We'll plug the \(22.5\) back in when we are done and ready for the final answer.

Answer 6

Do you agree we need to form a common denominator for those two fractions in order to proceed with the solution?

Answer 7

okay.

Answer 8

Yes, I do.

Answer 9

I made a mistake while typing and hit the 0 instead of the 9. It won't work to cross-multiply here.

Answer 10

Okay. So should we go back to your original thought?

Answer 11

Okay, let's try that again!

\[\frac{a}{s+6}+\frac{a}{s-6}=9\]We need to make a common denominator. Do you know how to do that?

Answer 12

I'm not completely sure so I can't say I do.

Answer 13

Okay, no problem. We really have two choices: we can make a common denominator, or we can "clear the fractions" out of the problem.

When "clearing the fractions" we multiply each term of the equation by the product of all of the denominators. An example:

\[\frac{1}{2} + \frac{1}{3} = x\]We clear the fractions by multiplying by 2*3:\[2*3*\frac{1}{2} + 2*3*\frac{1}{3} = 2*3*x\]\[\cancel{2}*3*\frac{1}{\cancel{2}} + 2*\cancel{3}*\frac{1}{\cancel{3}} = 2*3*x\]\[3*1+2*1 = 2*3*x\]\[3+2=6x\]\[5=6x\]\[x=\frac{5}{6}\]

Now if we had done the common denominator route, it might have gone something like this:

\[\frac{1}{2} + \frac{1}{3} = x\]\[\frac{1}{2} *\frac{3}{3}+ \frac{1}{3}*\frac{2}{2} = x\]\[\frac{3*1}{2*3}+\frac{2*1}{2*3} = x\]\[\frac{3}{6}+\frac{2}{6} = x\]\[x=\frac{2+3}{6} = \frac{5}{6}\]

Answer 14

I showed a few more steps on the clearing approach because it might not be familiar, whereas you said you've done common denominators before.

So, we have the option of multiplying our fractions by fractions like \[\frac{s+6}{s+6}\] to get a common denominator, or just multiplying everything by \[(s+6)(s-6)\]and knowing that we'll cancel out many of the terms.

\[\frac{a}{s+6} + \frac{a}{s-6} = 9\]\[(s+6)(s-6)\frac{a}{s+6} + (s+6)(s-6)\frac{a}{s-6} = 9(s+6)(s-6)\]\[a(s-6) + a(s+6) = 9(s+6)(s-6)\]\[as -6a + as + 6a = 9(s^2-36)\](the right hand because it is a difference of squares)
\[2as = 9(s^2-36)\]\[9s^2-9*36 -2as = 0\]\[9s^2-2(22.5)s-9*36 = 0\]We can solve that with the quadratic formula:
\[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Here we have \(a = 9, b = -45, c = -324\) so that gives us \[x =\frac{ -(-45) \pm\sqrt{(-45)^2-4(9)(-324)}}{2*9} = \frac{45\pm\sqrt{13689}}{18} = \frac{45\pm117}{18}\]We don't care about the negative answer, only the positive one (we're talking speed here), so \[s=9\]
is our answer.

Answer 15

Checking our answer:

against current, boat goes 9-6 = 3 mph, taking 22.5/3 = 7 1/2 hours
with current, boat goes 9+6 = 15 mph, taking 22.5/15 = 1 1/2 hours
7 1/2 hours + 1 1/2 hours = 9 hours, which is the time allowed, so our answer checks out.

Answer 16

If we'd done it by making a common denominator, the final part would have been much the same.
\[\frac{a}{s+6}*\frac{s-6}{s-6} + \frac{a}{s-6}*\frac{s+6}{s+6} = 9\]\[\frac{a(s-6)+a(s+6)}{(s+6)(s-6)} = 9\]\[as -6a + as + 6a = 9(s+6)(s-6)\]and you should recognize that from above.

Answer 17

Does that make sense? Do you understand how I came up with the solution?

Answer 18

On the quadratic equation I'm not sure where -45 and -324 came from

Answer 19

Okay, I took a small shortcut there, hoping that you would see what I did.

\[9s^2-2(22.5)s-9*36 = 0\]\[9s^2-45s-324 = 0\]Form for quadratic formula is \[ax^2+bx+c=0\]so \(a=9, b=-45,c=-324\)

Make sense?

Answer 20

Okay. Yes, the makes sense. I just have to show all my steps and I'm just making sure I have all of them.

Answer 21

It would be very beneficial for you to try to work the problem yourself without looking at what we did here. Show all your steps by doing them :-)

Answer 22

Okay. Thanks.