Mark H. answered • 09/07/18
Tutor
New to Wyzant
Taught at computer camps (ages 8-15) to major corporations
I'll get you started:
Let d be the distance traveled by the airplane per trip. We know d. d = 600 miles
Let a (sorry, no good choices for a variable name for time) be the time it took going with the wind. We know a. a = 4 hours.
Let b be the time it took going against the wind to make the same trip of distance d. We know b. b = 6 hours.
So we can calculate the speed the plane is traveling during a single trip by the simple formula: d / T = s. Where d is the distance traveled, T is the time of a particular trip (one time you substitute a for T, and b the other time). So you have two speeds. One with the wind and on against the wind.
The key here is that the plane with no wind travels at one speed. The plane travels without any help or hindrance of the wind (lets call this n, the normal speed the plane travels). So if we let w be the speed of the wind, then during the first trip with the wind (taking time a), the total speed the plane travels is the normal speed plus the wind speed, lets call that w. So using d / Ta, where Ta=a. We get an sa (speed for the first trip). Since the plane is traveling with the wind, it travels at it's normal speed n, plus the speed of the wind. So sa = n + w. On the return trip you have a different s (plugging T = b into sb = d / Tb). The sb is against the wind so this sb = n - w.
You now have two equations where s values are known (and is a different value either based off of a or b), and two unknowns, n (the speed of the plane in still air) and w (the speed of the wind). These are exactly the two numbers the word problem is asking you for. Hope this helps.
