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.

Let the speed of the wind be [tex]x \rm \; km \cdot h^{-1}[/tex], and the speed of the plane in still air be [tex]y\rm \; km \cdot h^{-1}[/tex]. 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 [tex]\rm 960\; km[/tex] from Ottawa to with a tail wind (that is: at a ground speed of [tex]x + y[/tex].)
It takes the plane four hours to travel [tex]\rm 960\; km[/tex] from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of [tex]-x + y[/tex].)

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

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

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

[tex]\left\{ \begin{aligned}&4 \times 3 (x + y) = 4\times960 && 4 \times (1) \\ &3\times 4(-x + y) = 3\times 960 && 3 \times (2) \end{aligned}\right.[/tex].

[tex]\left\{ \begin{aligned}&12 x + 12y = 4\times960 && 4 \times (1) \\ &- 12x + 12y = 3\times 960 && 3 \times (2) \end{aligned}\right.[/tex].

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

[tex](12 - (-12) x = 1 \times 960[/tex].

[tex]24 x = 960[/tex].

[tex]x = 40[/tex].

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

In other words,

Wind speed: [tex]\rm 40\; km \cdot h^{-1}[/tex].
Speed of the plane in still air: [tex]\rm 320\; km \cdot h^{-1}[/tex].

adefunkeadewole adefunkeadewole · Answer 2 · 2020-03-28T02:34:25+01:00

Answer:

Wind speed = 40 km/hr

Speed of the plane in still air = 280km/hr

Step-by-step explanation:

Formula to use

Distance = Rate multiplied by Time

D = R × T

Speed of the plane in still air = a

Speed of the wind = b

Plane with the wind (tailwind) is a + b where the time = 3hours

Plane against the wind is a - b, where the time = 4hours

This would give us 2 equations

3 × (a+ b) = 960

3a + 3b = 960 ...... Equation 1

4 × (a- b) = 960

4a - 4b = 960 ........ Equation 2

Solve this like you would with system of equations:

We solve the equation using elimination method.

Mulitply equation 1 by 4 and equation 2 by -3

4 × (3a + 3b = 960)

12a + 12b = 3840 ..... Equation 4

-3 × (4a- 4b = 960)

-12a + 12b= -2880 ........ Equation 5

Therefore we have

12a + 12b = 3840 ..... Equation 4

-12a + 12b= -2880 ........ Equation 5

24b = 960

b = 960÷ 24

b = 40

Wind Speed = 40km/hr

To find the speed of air, we use equation 2

4a - 4b = 960 ........ Equation 2

Substituting 40 which is wind of speed for b in equation 2 we would have

4a - 4(40) = 960

4a - 160 = 960

4a = 960 + 160

4a = 1120

a = 1120 ÷ 4

a = 280

The speed of the small plane in still air is 280 km/hr