A twin-engined aircraft can fly 1216 miles from city A to city B in 4 hours with the wind and make the return trip in 8 hours against the wind. What is the speed of the wind?
Equation Word Problem
1) Calculate the speed on the way from A to B;
2) Calculate the speed on the way back, from B to A;
3) Subtract the latter from the former, this will be twice the speed of the wind (why?).
Lets convert this word problem into algebraic terms:
Let D=the distance between cities, (note that the distance traveled is the same in both directions)
Let Sw= the speed of the wind; Note that this is what we're trying to solve for.
Let Sp= the speed of the plane
The basic formula to be applied is Distance=Speed*Time
Since the first trip is "with the wind", the total net speed is (Sp+Sw) and Distance=1216=(Sp+Sw)*4
Since the return trip is "against the wind", the total net speed is (Sp-Sw) and Distance=1216=(Sp-Sw)*8
We now have two simultaneous equations that can be solved for Sp and Sw:
From the 1st equation, 1216=4Sp+4Sw or Sp=(1216-4Sw)/4
Plugging this into the second equation for Sp, 1216=8(1216-4Sw)/4 - 8Sw
or Sw=76 miles per hour; This is the answer.
You can now also solve for the speed of the plane to get Sp=228 miles per hour
You can Check your work, plugging Sw and Sp back into the 2 simultaneous equations: