Trent flies a plane against a headwind for 5589 miles. The return trip with the wind took 12 hours less time. If the wind speed is 6 mph, how fast does Trent fly the plane when there is no wind?

Question

Trent flies a plane against a headwind for 5589 miles. The return trip with the wind took 12 hours less time. If the wind speed is 6 mph, how fast does Trent fly the plane when there is no wind?

Answer 1

Ralph, let's do this one!

The basic formula is: s = vt where s = distance travelled, v = speed, and t = time

Let Trent's speed without wind be v. Let the time against the wind be t.

Flying against the wind, we have: 5589 = (v - 6) t . . . . . . . . . . Eqn (1)

Flying with the wind: 5589 = (v + 6) (t - 12) . . . . . Eqn (2)

From Eqn (1): 5589 / (v - 6) = t . . . . . . . . . . . . . . . .Eqn(3)

From Eqn (2): 5589 / (v + 6) = t - 12 . . . . . . . . . . . . Eqn (4)

Eqn (3) - Eqn (4): 5589 / (v - 6) - 5589 / (v + 6) = 12

5589 { 1 / (v - 6) - 1 / (v + 6) } = 12

5589 { (v + 6) - (v - 6) } / (v - 6)(v + 6) = 12

5589 ( 12 / (v² - 6²) ) = 12 . . . . . . . . (v - 6)(v + 6) = v² - 6 - diff of 2 squares

5589 / (v² - 6²) = 1

5589 = v² - 36 . . . . . . cross multiplying

5625 = v²

∴ v = √ 5625 = 75 mph

Easy problem, messy equations to solve!