### Video Transcript

Water in a canal, 5.4 meters wide
and 1.8 meters deep, is flowing with the speed of 25 kilometers per hour. How much area can it irrigate in 40
minutes if 10 centimeters of standing water is required for irrigation?

Let’s start out by sketching this
canal. Looking at a cross section of the
water as it flows in this canal, we know the canal is 5.4 meters wide and 1.8 meters
deep. And along with that, the water
flows with a speed we’ll call 𝑆, where 𝑆 is given as 25 kilometers every hour.

Knowing all this, we want to figure
out how much area the water in this canal can irrigate in a time of 40 minutes. We can label this irrigated area
𝐴. And the problem statement tells us
that for this area there is 10 centimeters of standing water required.

That means that our area sits on
top of a depth of water of 10 centimeters. In other words, there’s some total
volume of water 𝑉 which is equal to this area multiplied by a depth of 10
centimeters.

This takes us back to our canal,
we’re given a cross-sectional area of the canal along with a flow rate and a time
during which the water is flowing. We’re also able to calculate a
total volume of water. We can consider it this way: in a
time of 40 minutes with this flow rate of 25 kilometers an hour, some length of
water in the canal — we’ve called it capital 𝐿 — will flow past this point.

We want to solve for 𝐿 because if
we can, then we’ll be able to calculate the total volume of water that flows through
the canal, which will help us solve for the area that’s irrigated. To start solving for 𝐿, let’s
recall that an average speed 𝑆 is equal to a distance travelled divided by the time
it took to travel that distance.

In our case, we can write that 𝑆,
the flow rate of the water in the canal, is equal to 𝐿 divided by 𝑡, the time that
the water is flowing. If we multiply both sides of this
expression by the time 𝑡, we see that on the right-hand side that term cancels out
of numerator and denominator.

This means that the 𝐿 that we want
to calculate to solve for the total volume of water that flows 𝑉 is equal to our
flow rate times the time that the water is flowing. That’s equal to 25 kilometers per
hour multiplied by the time of 40 minutes. We see though that there’s an issue
with combining these two values: the units aren’t consistent. One has time units of hours, but
the other has time units of minutes.

To match them up, we can recall
that one hour is equal to 60 minutes. When we make that substitution, we
see that the units of time cancel out and the fraction 40 divided by 60 reduces to
two-thirds. In other words, the length of water
that flows along the canal in 40 minutes is equal to 25 kilometers times
two-thirds.

There is one last change we’ll want
to make to this number though before inserting it into our expression for the volume
𝑉. We can see that the other
dimensions of this volume are given in units of meters. But our length currently is in
units of kilometers. We can recall that one kilometer is
equal to 1000 meters, which means that 𝐿 is equal to 25 kilometers times two-thirds
or 25000 meters times two-thirds, which is equal to 50000 meters divided by
three.

We can now take this value for 𝐿
and plug it into our equation for the volume 𝑉. We now have an expression for the
total volume of water that flows through the canal in 40 minutes. And this is equal to the area that
we’re able to irrigate multiplied by a depth of 10 centimeters.

Our next task then is to set these
two equations equal to one another and solve for the area 𝐴. Looking at our combined equation,
we may notice that there’s a units mismatch between the left side, which has units
of meters, and the right side, which we see has units of centimeters. To correct this mismatch, we can
recall that one meter is equal to 100 centimeters. This means we can replace 10
centimeters with one over 10 meters.

And now, in order to isolate the
area 𝐴 that we want to solve for on one side of this equation, we’ll multiply both
sides of the equation by 10 inverse meters. On the right side of our equation,
this cancels out everything, except the area 𝐴. And on the left side of our
equation, if we multiply by the factor of 10 and group all the units of meters, we
find a result of 18 times 5.4 times 50000 over three square meters.

Now, all we need to do is calculate
this left side and that will give us our area 𝐴 in units of square meters. To simplify this calculation, we
can multiply 5.4 by 10, removing one of the zeros from the 50000, so that the value
in parentheses is now 18 times 54 times 5000 over three.

One way to proceed in simplifying
this number further is to realize that three divides evenly into 54. It goes into 54 18 times. Now, our value is 18 times 18 times
5000. We can now calculate what 18
squared is by hand. When we carry out this
multiplication, we find a final result of 324. Next, we can expand the numbers in
the parentheses from 324 times 5000 to 324 times five times 1000. Multiplying 324 by five, we find
the result of 1620.

Looking at our expression in
parentheses now, we count one, two, three, four zeros, which means we can we rewrite
it as 162 times 10 to the fourth and that’s meters squared. This is the area that the canal
could irrigate in a time of 40 minutes.