Question

A small plane took 3 hours to fly 960 km from Ottawa to Halifax with a tail wind. On the return trip, flying into the wind, the plane took 4 hours. Find the wind speed and the speed of the plane in still air.

Answer 1

• Wind speed:
  `\rm 40\; km \cdot h^(-1)`.
• Speed of the plane in still air:
  `\rm 320\; km \cdot h^(-1)`.

Explanation:

This problem involves two unknowns:

• wind speed, and
• speed of the plane in still air.

Let the speed of the wind be
`x \rm \; km \cdot h^(-1)`, and the speed of the plane in still air be
`y\rm \; km \cdot h^(-1)`. It takes at least two equations to find the exact solutions to a system of two variables.

Information in this question gives two equations:

• It takes the plane three hours to travel
  `\rm 960\; km` from Ottawa to with a tail wind (that is: at a ground speed of
  `x + y`.)
• It takes the plane four hours to travel
  `\rm 960\; km` from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of
  `-x + y`.)

Create a two-by-two system out of these two equations:

`\left\{ \begin{aligned}&3(x + y) = 960 && (1) \\ &4(-x + y) = 960 && (2) \end{aligned}\right.`.

There can be many ways to solve this system. The approach below avoids multiplying large numbers as much as possible.

Note that this system is equivalent to

`\left\{ \begin{aligned}&4 * 3 (x + y) = 4*960 && 4 * (1) \\ &3* 4(-x + y) = 3* 960 && 3 * (2) \end{aligned}\right.`.

`\left\{ \begin{aligned}&12 x + 12y = 4*960 && 4 * (1) \\ &- 12x + 12y = 3* 960 && 3 * (2) \end{aligned}\right.`.

Either adding or subtracting the two equations will eliminate one of the variables. However, subtracting them gives only
`1 * 960` on the right-hand side. In comparison, adding them will give
`7 * 960`, which is much more complex to evaluate. Subtracting the second equation (
`3 * (2)`) from the first (
`4 * (1)`) will give the equation

`(12 - (-12) x = 1 * 960`.

`24 x = 960`.

`x = 40`.

Substitute
`x` back into either equation
`(1)` or
`(2)` of the original system. Solve for
`y` to obtain
`y = 320`.

In other words,

• Wind speed:
  `\rm 40\; km \cdot h^(-1)`.
• Speed of the plane in still air:
  `\rm 320\; km \cdot h^(-1)`.