Equation Word Problem

Question

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?

Answer 1

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?).

Answer 2

Ronica:

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 S_w= the speed of the wind; Note that this is what we're trying to solve for.

Let S_p= 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 (S_p+S_w) and Distance=1216=(S_p+S_w)_*4

Since the return trip is "against the wind", the total net speed is (S_p-S_w) and Distance=1216=(S_p-S_w)*8

We now have two simultaneous equations that can be solved for S_p and S_w:

From the 1st equation, 1216=4S_p+4S_w  or Sp=(1216-4S_w)/4

Plugging this into the second equation for S_p, 1216=8(1216-4S_w)/4 - 8S_w

or 1216=2432-8S_w-8S_w

or -1216=-16S_w

or S_w=76 miles per hour; This is the answer.

You can now also solve for the speed of the plane to get S_p=228 miles per hour

You can Check your work, plugging S_w and S_p back into the 2 simultaneous equations:

1216=(Sp+Sw)*4

1216=(228+76)*4

1216=1216 (check!)

 

1216=(Sp-Sw)*8

1216=(228-76)*8

1216=1216 (check!)