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

### 3 Answers by Expert Tutors

Priti S. | Math Tutor with Patience & KnowledgeMath Tutor with Patience & Knowledge
5.0 5.0 (448 lesson ratings) (448)
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We have 2 situation one with wind and one against wind

Speed = x , W= Wind

*case 1       ( With Wind)                     *case 2      ( Against Wind)

t=3 hr  d= 2160 m.  Rate = x + W            t=4 hr  d= 2160 m.  Rate = x - W

Distance = Rate * Time

2160 =  (x+W) * 3                                                      2160 = (x-W) * 4

2160 / 3 = (x+W)                                                        2160/4 = x-W

720 = x+W                                                                  540 = x-W

NOW 2 equations solve by substitution or elimination.  ( i did Elimination)

720 = x+W

540 = x-W

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1260 = 2x    (W was eliminated) solve for x

x = 630  Speed of the Airplane

now solve for W. W = 90  Speed of the Wind

4.9 4.9 (233 lesson ratings) (233)
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Hello again, Allison -- mph with wind is 2160/3 or 720 mph, facing wind is 2160/4 or 540 mph.

"No wind" MUST be in the middle: 630 mph, with wind boost of 90 mph & wind drag of same 90 mph :)

Ralph L. | Algebra I, II, Visual Basic, Beginning C++ tutorAlgebra I, II, Visual Basic, Beginning C...
4.0 4.0 (1 lesson ratings) (1)
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let A be the speed of airplane, W be the speed of the wind

first case (with the wind):

3 * (A + W) = 2160               eq. 1

second case (against the wind):

4 * (A - W) = 2160                 eq. 2

let's use eq. 1 to solve for A:

3A = 2160 - 3W  --> A = 720 - W   eq. 3

let's substitute A into eq. 2:

4 * (720 - W) - 4W = 2160

2880 - 4W - 4W = 2160

-8W = 2160 - 2880

-8W = -720

W = 90

so the speed of wind is 90 miles/hour

let's use eq. 3 to solve for A:

A = 720 - 90 = 630

therefore A = 630 miles/hour and W = 90 miles/hour