Welcome to Georgia Highlands College Math 97 and Math 99 instructional video.
In this video segment we’ll be answering the question, how do you solve rational
equation applications, “hint, hint” word problems, involving time in motion.
So we’re talking about that formula that you are probably very familiar with,
distance equals rate times time.
So the first thing I have written on the board here is, I've taken this formula
D equals RT and I’ve arranged it solving for each of the variables because
you’re not ever sure which form of this formula that you’re going to be able or
going to need to use to solve your application problem.
So we have distance equals rate times time and then if I solve for the R, I just
divide both sides by T, so I get rate equals distance over time.
And if I come back here and want to solve for T, I just divide R off of both
sides to get time equals distance divided by the rate. So that’s all three forms
of that particular formula.
So the first thing in this process, besides read the word problem is to identify
the meaning of the variable. And you’re going to decide what to assign a
variable to by a looking at the question that the problem is asking. So what are
you trying to answer here? If you're asked, if you're answering a question, like
if they say, what was the distance? Well obviously you don't know the distance
so that’s where you're going to need to assign your variable.
Then you’re going to set up your DRT chart, and this will make a little bit more
sense when we take a look at our example. You want to use that chart and the
information in the problem to see where you can put an equal sign. So what can
you equate from the chart to create the equation that you’re going to solve?
Then after you figure that out, you actually write the equation, solve it, and
of course, check it to make sure that your solution is actually a solution and
not just an answer.
So let's take a look at this application problem.
All right, so you’re canoeing down the Colorado River, sounds nice, in still
water, so this is not on the river. But in still water your average canoeing
rate is 3 mph. It takes you the same amount of time. Actually that should say
“of time” but I’m not going to redo this whole video because I fixed it by just
writing over the ‘i’. The same amount of time to travel 10 miles downstream with
the current as 2 miles upstream against the current. What is the rate of the
waters? current?
So in looking at the question that we’re going to be answering in working
through this particular application problem, we need to identify our variable.
So what is the rate of the waters’ current? Well, we don't know the rate of the
waters’ current, so we’ll give it the variable X and we want to always note what
we are using that variable to stand for, so I'm just going to put rate of
waters’ current.
So now we know what are variable is standing for. So let's see if we can pick
out information in the problem above to help us fill out this DRT chart below.
So we have two situations, we have you traveling with the current and against
the current.
So, in going back up to our problem, we see it takes the same amount of time if
you travel 10 miles downstream with the current as it does 2 miles upstream
against the current. So when we’re talking about 10 miles downstream and 2 miles
upstream we're talking about a distance.
So we have 10 miles downstream with the current and 2 miles upstream with the
current. Now moving on to our rate column, we want to talk about how fast we’re
moving in each of these situations.
The only piece of information that we have about our rate is that when you're in
still water you can canoe an average of 3 mph, but if you're in the river, if
you're going downstream, you’re going to have the current working with you to
push you faster. If you're coming up stream, you’re going to have the current
working against you which is going to slow you down. We all know that it's no
fun trying to row upstream because you’ve got that current pushing against you.
So when we write our rate, we have to take into consideration the rate of the
current as well. Well the rate of the current is X, we don't know what it is. So
with the current we can go from 3 mph, plus whatever rate the current is helping
us move. Now when we’re coming up against the current, it's working to slow us
down, so we know we can row 3 mph, but we have to subtract the rate of the
current because it's working against us to slow us down.
So now we've expressed both of our rates in terms of how fast we can actually
row, added to or subtracted by the rate of the waters’ current, depending on
which situation if you're going with the current or against it.
Now we use our formula, T equals distance divided by rate because we haven’t
been given any information about time except for one small thing and that's that
it takes the same amount of time to travel down as it does to travel up, and
we'll talk about what that means in regards to your problem in just a moment.
But we can express our time using our distance and our rate.
So notice time equals distance divided by rate, so our time here is 10 over 3
plus X and our time here is 2 over 3 minus X.
Now going back to our problem, where it says it takes the same amount of time.
If something is the same that means you can put an equal sign between them.
So you can use this last column to set up your equation and our equation is 10
over 3 plus X is equal to 2 over 3 minus X. So we have a rational equation here
and we know from previous videos to solve a ration equation you multiply every
term in the equation by the LCD.
The LCD here has a factor of 3 plus X and a factor of 3 minus X. So we simply
multiply every term by 3 plus X times 3 minus X.
And I'll just note again, that that's all happening in the numerator and we do
that so we can come back and say anything divided by itself is one.
And now we're left with 3 minus X times 10 equals 2 times 3 plus X.
So distribution bells and whistles should be going off because that's exactly
what we need to do. 3 times 10 is 30, and negative X times 10 is negative 10 X,
2 times 3 is 6, and 2 times X is positive 2X, so we have a nice, neat, linear
equation to solve here.
So I’m going to subtract 2X off of both sides and subtract 30 off of both sides.
So that takes care of that term 2X and this takes care of that term of 30, and
our resulting equation is negative 12X equals negative 24. Divide both sides by
negative 12 and we get X is 2.
Now with any word problem, you always want to look back up and see if you're
answering the question here. So X is standing for the rate of the waters?
current.
We’ll check the solution before we make our final declaration of what the
solution is. So we go back to the original equation 10 over 3 plus 2 and that
should equal 2 over 3 minus 2. Simplify down each side, let me just give a
little barrier here so things don’t get confusing for you, that’s all to check
on the right-hand side. So we have 10 over 5 equals 2 over 1 and 2 equals 2, so
it works.
Our final answer is the rate of the current is 2 mph.
You always want to make sure you answer in a complete sentence. This just shows
your instructor that you understand what the meaning of your solution is.
So I hope that this has been helpful for you. This is a pretty involved problem
here but you'll see quite a few examples like this, so please make sure you
watch back through and understand where each part of this solution or part of
the problem came from.
If you have any other questions about solving these time in motion problems,
please make sure to contact your Highlands instructor.
Thank you.
