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

# 2 Answers

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?).

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!)
**