Algebra Help: How to Solve Airplane-With-A-Tail-Wind Word Problems

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)
T_w = time spent going with the wind
T_a = time spent going against the wind
D_w = distance gone with the wind
D_a = 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:

D_w= (S + W) * T_w

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:

D_a = (S - W) * T_a

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:

D_w = (S + W) * T_w
D_a = (S - W) * T_a

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

D_w = D_a = 340
T_w = 2
T_a = 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:

D_w = (S + W) * T_w
D_a = (S - W) * T_a

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:

T_w + T_a = 4

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

D_w = D_a = 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.

T_w = 4 - T_a

Now we can plug:

D = 120*(4-T_a)
D = 80*T_a

Simplify:

D = 480 - 120*T_a

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

80*T_a = 480 - 120*T_a
200*T_a = 480
T_a = 2.4

Solve for D:

D = 80*2.4 = 192 miles

Blessings!

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