SOLUTION: A sightseeing plane travels into a 60 mph headwind and returns, going 150 miles out and coming 150 miles back in 2 hours. a. How fast would the plane have gone in still air?

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Question 1196968: A sightseeing plane travels into a 60 mph headwind and returns, going 150 miles out and coming 150 miles back in 2 hours.
a. How fast would the plane have gone in still air?
b. How far could the plane have gone in 2 hours had there been no wind?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
60 mph AGAINST headwind?
150 mph WITH the headwind

What's the "in 2 hours"? Whole round trip or just the time for returning?


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Reading more carefully, "... travels INTO a 60 mph headwind,..."
The round trip of 2 hours
                 SPEED       TIME              DISTANCE
INTO WIND        r-60        150/(r-60)         150
WITH WIND        r+60        150/(r+60)         150
Total                         2

150%2F%28r-60%29%2B150%2F%28r%2B60%29=2-------Solve for r.

Steps will lead to r%5E2-150r-3600=0.

Solution for quadratic equation:

r=%28150%2B-+sqrt%28150%5E2%2B4%2A3600%29%29%2F2
r=%28150%2B-+sqrt%2836900%29%29%2F2
r=%28150%2B-+sqrt%289%2A41%2A100%29%29%2F2
r=%28150%2B-+3%2A10%2Asqrt%2841%29%29%2F2
r=75%2B15%2Asqrt%2841%29

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


I see no ambiguity in the given information, as the other tutor seems to.

The plane goes 150 miles out and 150 miles back in 2 hours, when the wind is 60mph.

Let x be the speed of the plane in still air; then write and solve the "time" equation that says time out plus time back is 2 hours, where time = distance/rate.

150%2F%28x-60%29%2B150%2F%28x%2B60%29=2

I won't bother going through a formal algebraic solution, since the answer is not a "nice" number. Use a graphing calculator to graph the expression for the total time and find where the total time is 2 hours.