Over a year ago I wrote an article “Algebra Help: How to Solve Boat-in-the-River Word Problems” that has become very popular. A variation on this kind of problem is the airplane-with-a-tail-wind problem. (Or problems that describe how fast and far an airplane can go with the wind versus against the wind. Sometimes instead of talking about flying “with a tail wind” the problem will talk about flying “against a head wind.”)

Both of these kinds of problems are just a variation of the dreaded uniform motion word problems which I’ve discussed in two previous articles. With the airplane with the tail wind problems, we assume that the airplane has a uniform speed in still air (or when there is no wind) and that the speed of the wind (or the velocity of the wind) is constant and is going in one direction, like the current of a river. The Algebra student will be presented with a math word problem in which he must solve for the speed of the airplane (in still air or with no wind, implied), or the velocity of the wind, or the time spent going against the wind or the time going with the wind or the distance the plane has traveled, or some combination of these variables.

With these problems, I recommend using the following variables:

S = speed of the airplane in still air (no wind)

W = speed of the wind (or the velocity of the wind)

Tw = time spent going with the wind

Ta = time spent going against the wind

Dw = distance gone with the wind

Da = distance gone against the wind

As with other uniform motion word problems, an important algebraic equation to remember is:

Distance = Rate * Time

With the airplane with the tail wind math problem, we modify this slightly when we are going with the wind to get:

Dw= (S + W) * Tw

That is, the speed (or rate) of the airplane going with the wind is the speed of the airplane in still air plus the speed of the wind (or velocity of the wind). That is because when one is going with the wind, the wind works with you.

When an airplane is going against the wind (into a head wind), the wind works against you so we have to change our distance equation to reflect this by subtracting the speed of the wind from the speed of the airplane (in still air) to get our rate:

Da = (S – W) * Ta

The best way to understand how to solve these is to work through a couple examples.

**Sample Airplane-with-a-Tail-Wind Word Problem #1:** *An airplane went 340 miles in 2 hours with the wind, but it took 4.25 hours to make the same course against the wind. Find the speed of the plane and velocity of the wind.*

With these kinds of word problems I recommend to the Algebra student that you always start out by writing down the two essential distance equations, before anything else:

Dw = (S + W) * Tw

Da = (S – W) * Ta

Secondly we need to write down the information we know from dissecting the text of the word problem.

Dw = Da = 340

Tw = 2

Ta = 4.25

Now, we just need to plug in the information we have into our two main distance equations:

340 = (S + W ) * 2

340 = (S – W) * 4.25

340 = 2*S + 2*W

340 = 4.25*S – 4.25*W

Solve for S in terms of W:

170 = S + W

S = 170 – W

Plug this equation into the second distance equation and then solve for W:

340 = 4.25*(170-W) – 4.25*W

340 = 722.5 – 4.25*W – 4.25*W

-382.5 = -8.5*W

W = 45

S = 170 – 45 = 125

The speed of the plane is 125 mph while the velocity of the wind is 45 mph.

**Sample Airplane-with-a-Tail-Wind Word Problem #2:** *A plane has a 4-hour supply of gasoline. How far can it fly from an airport and return if the speed out is 120 mph and the speed in is 80 mph?*

Write down the two essential distance equations:

Dw = (S + W) * Tw

Da = (S – W) * Ta

Next, write down the information we can dissect from the word problem. First, let’s look at that gasoline, which is really just a tricky way of describing time:

Tw + Ta = 4

Since we are flying away from the airport and then returning, the two distances are the same.

Dw = Da = D

To simplify matters, we will just use D for our distance variable.

Next, let’s look at the “speed out”. Since it is the faster of the two speeds, we know we are going with the wind.

S + W = 120

Lastly let’s look at the “speed in.” This must be against the wind.

S – W = 80

Before we plug in our information into our two distance equations, we need to write one of the time variables in terms of the other.

Tw = 4 – Ta

Now we can plug:

D = 120*(4-Ta)

D = 80*Ta

Simplify:

D = 480 – 120*Ta

Make the two equations equal to each other and solve for Ta:

80*Ta = 480 – 120*Ta

200*Ta = 480

Ta = 2.4

Solve for D:

D = 80*2.4 = 192 miles

Blessings!

**Source**

Virgil S. Mallory. A First Course in Algebra (1943)

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