A man flies a small airplane from Fargo to Bismarck, North Dakota ---

Question

A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hour and 12 minutes. What is the plane's speed in still air, and how fast is the wind blowing?

his plane speed equals mph.
the wind speed equals mph.

Answer 1

Plane speed in still air: 120mph

Wind speed: 30pmh

Speed of plane to Bismarck: 90mph

Speed of plane to Fargo: 150mph

Answer 2

Treena,

This is another problem related to distance, so we will utilize the distance formula (d=rt). The issue with this problem is that there are two variables in relation to the rate. The first variable is the plane speed (r) and the second variable is the wind speed (w). We will first set up our equations for the two flights, and then we will address how to solve for both variables.

Flight into the wind: In this flight, the plane speed is working against the wind, so our rate is set up as r-w, since the wind is decreasing our plane speed. This flight took 2 hours (t) and traveled 180 miles (d), so the equation is 180=(r-w)2

Second flight: In this flight, the plane speed is working with the wind, so our rate is set up as r+w, since the wind is increasing our plane speed. This flight took 1.2 hours (t) and traveled 180 miles (d), so the equation is 180=(r+w)1.2

Now, you have two choices, you can use elimination or substitution, in order to address the multiple variables. I will use substitution. Since both flights traveled the same distance, we can set the rt portions of our equations equal to each other and solve for one variable:

rt=rt

(r-w)2=(r+w)1.2

2r-2w=1.2r+1.2w

0.8r=3.2w

r=4w

Now, we can plug in 4w for r, in either equation:

180=(r+w)1.2

180=(4w+w)1.2

180=5w(1.2)

150=5w

30=w

Since we know that r=4w, we can plug in w, to solve for r:

r=4w

r=4(30)

r=120

Now we know our plane speed and wind speed. If we want, we can plug both into our second equation, in order to check the values:

180=(r-w)2

180=(120-30)2

180=90(2)

180=180 This is a true statement, so our values are valid!

Answer 3

The relevant relationship we need to consider is the one between distance, time and speed:

speed = ( distance / time )

The flight west takes 2 hours, and is 180 miles in length, so west speed is easily determined:

Wspeed = 180 miles / 2 hours = 90 mph

Going eastbound, the flight took 1 hour and 12 minutes, which can be re-written as 1.2 hours.  The speed when traveling east:

Espeed = 180 miles / 1.2 hours = 150 mph

When the plane encounters a head wind while traveling west, its forward progress is impeded; the plane travels slower.  Specifically, we can write an equation describing this effect:

Wspeed = Aspeed - wind

Similarly, when the plane travels east, the tail wind increases the forward speed of the plane:

Espeed = Aspeed + wind

Using the above equations, we can solve for Aspeed, the airplane's speed, and wind, the wind speed.

90 = Aspeed - wind

150 = Aspeed + wind, which we can re-write as wind = 150 - Aspeed.

Substituting the second equation into the first:

90 = Aspeed - ( 150 - Aspeed )

90 = Aspeed - 150 + Aspeed.  Adding 150 to both sides:

90 + 150 = Aspeed + Aspeed

240 = 2*Aspeed.  Divide both sides by 2:

240 / 2 = 2*Aspeed / 2

120 = Aspeed

Since we know wind = 150 - Aspeed (from our re-written second equation), solving for wind now that we know Aspeed is simple:

wind = 150 - Aspeed = 150 - 120 = 30

So, to answer the original question, the airline's speed is 120 mph, and the wind speed is 30 mph.