Systems of linear equations uniform motion

# Systems of linear equations uniform motion

>>All right, so these are
uniform motion problems with the current,
against the current, and what’s happening is you have
a boat, and it’s going upstream, which means the current
will be going against it. So you can kind of
think of the boat, and the current’s pushing
the opposite direction. Downstream the current’s
going with the boat. So if you’re in this situation,
the boat is slowing down. In this situation
the boat is speeding up because the current
is literally pushing it. All right, so if x is the
speed of the boat in calm water and y is the speed of
the stream in calm water, then the actual effective rate
of the boat would be x minus y, because the current is
subtracting speed from the boat. Going downstream the effect
of speed would be x plus y, because the current is
actually pushing the boat, and there’s your
setup for the begin. Now all we have to do is
go through the word problem and find out the distance
traveled in each situation and the time traveled
in each situation. So when Dwayne drives his
boat to work upstream, the trip takes 45 minutes. Now, with rates, we want miles
per hour, not miles per minute. So that 45 minutes would
have to be divided by a 60 to give us 0.75, and if we
plug it in a calculator, and there you go, .75. When he comes home
downstream, it takes 30 minutes. So again out of 60 and that’s
a clean division of 0.5. So now we have its hours. The distance to work
is 90 miles. So whether he comes
home or goes to work, the distance is 90 miles, and what we’ve just done
now is created our formula, and it’s distance
equals rate times time. So for the first
situation, which is upstream, we have an equation of this, and for the second
equation we have… All right, and then now we
have our linear equations, and before, we would
actually distribute this, but since these are nice, what we’re going to
do is divide them. So what’s 90 divided by .75? 120. So we get 120 equals x minus y, and we’ll do the same thing
here, 90 divided by .5. That’s 180, x plus y,
and now if you notice, the y’s are opposites. So we can just add them up,
and we get two x’s equals 300, which would allow us to divide. So we get 150 equals x, and
what x represents is the speed of the boat. It’s a very, very fast boat. So the boat, the boat’s speed, is 150 miles per hour. Then how do we find
the current speed? Just take any of the
equations and plug it in. So we have 120, or
we’ll do the 180. So 180 equals x plus
y. Plug in our x. Oops, that’s the y. So
that means the speed of the current is
30 miles per hour. That’s it. So that gives us an effective
rate of 180 miles per hour for the boat going with
the current upstream — I mean, downstream. Then going against the current
would be 120 miles per hour. All right, let’s try one more, and then I’ll give
you one to try. So a jet plane flying through
the wind went 2100 miles in four hours. Against the wind the plane could
fly only 1760 miles per hour. Find the rate of the
plane in calm air and the rate of the wind. So that’s where we have the x
and the y, and we always want to make x the rate
of the object. So in this case it’s
a plane in calm. I don’t even care about the air. All right, and the other one
is just the rate of the wind, and it’s the same situation. So if it goes against the wind,
you would be subtracting them, and if you go with the wind,
you would be adding them. Now we can fill in the rest. So a jet plane flying with
the wind, that’s this one, went 2100 miles in four hours. Against the wind the
plane could only fly 1760 in the same amount of time. So that gives us
our two equations. 1760 equals x minus y times 4, and 2100 equals x
plus y times 4. Then again you want
to divide them. So we take 1760 divided
by 4 and we get 440. Then the same thing. 2100 divided by 4 is 525, and you notice the y’s
are nice and clean. So we get 2x equals,
that’s 965 divided by 2. So x equals 482.5. And what does x stand
for, again? It’s the rate of the
plane in calm air. So rate of plane in calm
air is 482.5 miles per hour. Make sure there are miles here. Yup, miles per hour. To find the rest,
just plug it back in, so 525 equals 482.5 plus y. So
y would equal, just subtract it, 525 minus 482.5, 42.5. So rate of wind is 42.5 miles
per hour, and there you go. That’s our answer. So hopefully you kind of get
the idea of these systems of linear equations using
currents and wind speed. Let’s have you try
one right here. This is for a rowing team. Hit pause and give it a shot. All right, and so this is
what you should have so far. All right, let’s figure… So here’s what you should