More plane speed

Question

 An airplane takes 3 hours to travel a distance of  2160 miles with the wind.  The return trip takes 4 hours against the wind.  Find the speed of the plane in still air and the speed of the wind. 

Answer 1

Distance formula says distance = rate*time or d = rt

 

Our distance both directions is the same.  d = 2160 miles

 

Travelling with the wind our rate is plane speed + wind speed.

Let r = rate of speed of the plane, w = wind speed

 

Equation 1:  2160 = (r+w)3

 

Travelling against the wind the overall speed is r-w so

 

Equation 2:  2160 = (r-w)4

 

Let's just rearrange those two equations

1:   3(r+w) = 2160

2:   4(r-w) = 2160

 

multiplying through we have:

 

1:   3r+3w = 2160

2:   4r-4w = 2160

 

multiply equation 1 by 4 and equation 2 by 3 and

add the result to eliminate the w variable and solve for r

 

12r + 12w = 8640       4(equation 1)

12r - 12w  = 6480

----------------------

24r           = 15120     divide both sides by 24

 

r = 15120/24

r = 630

 

The rate of the plane in still air is 630 mile per hour

 

Substitute that back into one of the original equations.

Let's use equation 1:   3(r+w) = 2160

 

3(630 + w) = 2160

630+w = 720

w = 720 - 630

w = 90

 

The wind speed is 90mph