Mathematics, 17.11.2019 06:31 soonerlady19

Asmall 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.

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Answer from: 1slimeyshyy

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

Step-by-step explanation:

This problem involves two unknowns:

wind speed, andspeed 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

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

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

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

$(12 - (-12) x = 1 \times 960$ .

24 x = 960 .

x = 40 .

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

In other words,

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

Answer from: krab38

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

Answer from: Quest

The answer would be choice b

Answer from: Quest

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x