Rate-Time-Distance Problem 3

>> Alright, here is another rate times time equals distance problem of uniform motion.
So we have a 555 mile trip, so that is why this whole distance in my picture is 555 miles
and you know it is 5 hours, but the car was driven at two speeds.
So, the first part of the trip that was going 105 miles per hour and the second part
of the trip it was going 115 miles per hour.
So, we want to, we have a chart.
So, let's just talk about the first and the second part of the trip.
There is a rate for the first part, a time for the first part and a distance,
and the same thing for the second part of the trip.
Alright, on the first part I am told that it is 105 miles per hour
and on the second part it was 115, alright?
Now, the tricky part, it was 5 hours total, so they didn't each -- it wasn't 5 hours for each,
it might be 3 and 2, 1 and 4, one and a half and three and a half
that is what we don't know, but the time must add up to 5.
So, if you are doing this for the single variable and you can also do this
with two variables if you know how to solve equation to two variables.
But if you are doing it with a single variable if the first part is x and the time has to add
up to 5, so x plus what is going to give you 5,
actually it is 5 minus x. See what happens if you get 5 minus x?
Whatever x is, let's say it was 2 we have to subtract that from 5
to get the 3 hours left here whatever it might be.
Also, you can see if you add, the x's cancel, x plus 5 minus x does add up to 5 hours total.
So, we have x and 5 minus x. It doesn't matter where you put the x
and where you put the 5 minus x, but once you define it, you have to stick with it.
So, now what is the distance for the first part?
We have to get rate times time equals 105x and the second will be 115 times 5 minus x. Alright,
so now I go back up with the picture.
Alright, so in this first part I have got 105x and the second part let's see
if I do the distributive property that is 575 minus 115x.
So, from the picture this piece, right,
that distance plus the second distance here has to add up to 555.
So, that is the formula -- I am sorry, that is the equation we are going to write,
105x plus 575 minus 115x equals 555.
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>> Okay, so on the left hand side I have 105 minus 115 and that is going to be negative 10x.
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>> And then I have to subtract 575 from both sides, so that is going to be negative 20.
And now, I am going to divide by negative 10.
We get x equals 2.
Alright, not done, so now I need to go back and check it by putting it in the chart.
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>> First and second part the rate is still 105 and 115.
Now, the time I now know x, so it is 2.
So, if you went 2 hours for the first half, there is 3 left for the other half
which of course is 5 minus 2 and the distance here we multiply rate times time equals distance
that is 210 and this is 375 -- I am sorry, 345 and does that make sense with the picture?
210 and 345, does that add up to a total of 555 miles.
Let's see if you get the problem, yes.
Okay, so now the question, how long did the car drive at each speed.
Okay, now I have got all information right here.
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>> The car drove 105 miles per hour for 2 hours and at 115 miles per hour for 3 hours.
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>> Isn't this fun?
So, notice it -- give yourself plenty of space to write this out,
to check it, and make sure it all makes sense.
You can see this check chart here, we have got the correct rate, the time really adds
up to 5 hours, the distance really adds up to 555 miles, etcetera.