Question

An airplane flying into a headwind travels the -mile flying distance between Indianapolis, Indiana, and Phoenix, Arizona, in hours. On the return flight, the airplane travels this distance in hours and minutes. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant.

Answer 1

Answer:

the airspeed of the plane = 660 miles/hr

the speed of the wind = 60 miles/hr

Step-by-step explanation:

The values are missing in the given information in the question.

To able to solve this question, we will make some assumptions.

Let's assume that:

The airplane heading into the wind travels 1800 miles in 3 hours.

While returning via the same distance, it travels 2 hours 30 minutes.

We are to determine the plane's airspeed as well as the wind speed.

First, let us say that the speed of the plane is p and the wind speed is q

during the time the plane is traveling for 1800 miles in 3 hrs.

Then, we can say

3p -3q = 1800 ----- (1)

However, in  2hours 30 minutes, we have 2.5 hours

So, when the plane is returning, we have the following equation:

2.5p + 2.5q = 1800 ---- (2)

So, to find p and q, we can use the elimination method.

Simply by multiplying equation (1) with 5 and equation (2) with 6

 (5)3p    -  (5)3q     = (5) 1800

 (6)2.5p  + (6)2.5q = (6) 1800            

  15p -  15q  = 9000

+

 15p + 15q   = 10800      

 30p -  0     = 19800      

30p = 19800

p = 19800/30

p = 660

Since p = 660, then from equation (1) we can obtain the value of q:

So; Using equation (1):

3p -3q = 1800

3(660) - 3q = 1800

1980 - 3q = 1800

-3q = 1800 - 1980

-3q = -180

q = 180/3

q = 60

Therefore, the airspeed of the plane = 660 miles/hr

the speed of the wind = 60 miles/hr