Question

Rod is a pilot for Crossland Airways. He computes his flight time against a headwind for a trip of 2900 miles at 5 hr. The flight would take 4 hr and 50 min if the headwind were half as great. Find the headwind and the plane’s air speed.

Please help me solve. I'm unable to solve. I have no clue!

Answer 1

speed first trip = s - h
speed second trip = s - 0.5 h

distance = rate * time
2900 = (s-h)(5)
2900 = (s-0.5h)(4 50/60) = (s - 0.5h)(4.833)

Answer 2

P = plane's speed.
W = wind speed.
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The plane traveled 2900 miles in 300 minutes (5 hr x 60 min/hr = 300 min).
distance = rate x time
rate = d/t = 2900 miles/300 min = 9.667 mi/min.

One equation is P-W = 9.667 miles/min.

If the wind speed is 1/2 that, it takes 4 hrs and 50 min (290 minutes).
rate = d/t = 2900/290 = 10 miles/min.

Second eqution is P-0.5W = 10.

Two equations in P and W. Solve for P and W. If I didn't make an error (check me out on the equations and arithmetic), P = 10.337 = 10.0 miles/min.

Check.
First flight
10.333 - 0.667 = 9.667 mi/min.
and 9.667 mi/min x 300 min = 2900 miles.

If wind is 1/2 that, it will be 0.337
10.337 - 0.337 = 10 mi/min x 290 min = 2900 miles.

You many change miles/min to miles/hr if you wish but I thought 4 hours and 50 minutes would be better in minutes.
W = 0.667 miles/min.

Answer 3

Two equations in P and W. Solve for P and W. If I didn't make an error (check me out on the equations and arithmetic), P = 10.337 = 10.0 miles/min
<B>I made a typo here. P = 10.337 miles/min.</b>