A plane flies 500 miles one way into?

Question

a wind of 16mph then returns with a wind of 16mph. If it takes 5hrs to complete the trip, how fast can the plane travel in no wind?

Answer 1

Let's call the velocity with no wind v. You are solving the sum of the timings for 500 miles with the wind and 500 miles against the wind is 5 hours:

500 / (v - 16) + 500 / (v + 16) = 5 hours

Let's divide by 5 up-front, the multiply by (v - 16)(v + 16) to get rid of all the denominators. That will leave us with a quadratic to solve:

100 / (v - 16) + 100 / (v + 16) = 1
100 (v + 16) + 100 (v - 16) = (v + 16)(v - 16)
200v = v^2 -256
v^2 - 200v - 256 = 0

Using the quadratic formula:

v = (200 +/- sqrt((-200)^2 - 4*1*(-256))) / 2
v = 100 +/- sqrt (40,000 + 1,024)/2
v = 100 +/- sqrt(41,024)/2

v = 100 + 101.27, v = 100 - 101.27
v = 201.27, v = -1.27

Ignoring the negative answer, it's 201.27.

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Checking the answer:

500 / (201.27 + 16) + 500 / (201.27 - 16) =
2.301 + 2.699 =
5

Answer 2

With problems like this, start out by making variables for the things that might be useful in solving the problem.

Make a variable "v" to stand for the plane's velocity when there's no wind. That's the number we're trying to find.

The speed against the wind is "v–16"; so the time for Leg 1 of the trip is:

T1 = Distance/Speed = (500 mi)/(v–16)

The speed WITH the wind is "v+16"; so the time for Leg 2 is:

T2 = (500 mi)/(v+16)

We know that the total time (T1+T2) is 5 hours. So:

(500 mi)/(v–16) + (500 mi)/(v+16) = 5 hours

Now the problem is set up. Just use algebra to solve for "v".