Question

A homing device left its home to deliver a message. On​ takeoff, the device encountered a tailwind of 10 mph and made the delivery in 15 minutes. The return trip into the wind took 30 minutes. How fast could the device fly in still​ air?

Answer 1

Answer:

  30 miles per hour

Step-by-step explanation:

The relationship between time, speed, and distance is ...

  time = distance/speed

For some distance d and some speed s (in miles per hour), the travel times (in hours) are ...

  1/4 = d/(s+10)

  1/2 = d/(s -10)

Dividing the second equation by the first gives ...

  (1/2)/(1/4) = (d/(s-10))/(d/(s +10))

  2 = (s +10)/(s -10) . . . . simplify

  2s -20 = s +10 . . . . . . multiply by (s -10)

  s = 30 . . . . . . . . . . . add 20-s

  The device flies 30 miles per hour in still air.

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Additional comment

The distance to the delivery point is ...

  d = 1/4(s +10) = 1/4(30 +10) = 10 . . . . miles to the delivery point