Dalia flies an ultralight plane with a tailwind to a nearby town in 1/3 of an hour. On the return trip, she travels the same distance in 3/5 of an hour. What is the average rate of speed of the wind and the average rate of speed of the plane?

Initial trip:
Return trip:
Let x be the average airspeed of the plane.
Let y be the average wind speed.
Initial trip: 18 = (x + y)
Return trip: 18 = (x – y)
Dalia had an average airspeed of miles per hour.
The average wind speed was miles per hour.

Respuesta :

Answer:

the average airspeed of the plane = 42 miles per hour

the average wind speed = 12 miles per hour

Step-by-step explanation:

Let x be the average airspeed of the plane.

Let y be the average wind speed.

Distance =time * speed

Initial trip: [tex]18 = \frac{1}{3}(x+y)[/tex]

Return trip: [tex]18 = \frac{3}{5}(x+y)[/tex]

We solve for x  and y

[tex]18 = \frac{1}{3}(x+y)[/tex]

Multiply both sides by 3

54= x+ y

y= 54- x ------------> equation 1

[tex]18 = \frac{3}{5}(x+y)[/tex]

Multiply both side by 5

90 = 3(x-y)

90= 3x- 3y ------------------> equation 2

Plug in y=54-x in second equation

90= 3x- 3(54-x)

90 = 3x - 162 + 3x

90 = -162 + 6x

Add 162 on both sides

252= 6x

Divide both sides by 6

So x= 42

y= 54- x

Plug in 42 for x

y= 54 - 42= 12

the average airspeed of the plane = 42 miles per hour

the average wind speed = 12 miles per hour




The average rate of speed of the wind is 12 miles per second and the average rate of speed of the plan is 42 miles per second.

Given

Dalia flies an ultralight plane with a tailwind to a nearby town in 1/3 of an hour.

On the return trip, she travels the same distance in 3/5 of an hour.

What is the average speed?

The average speed is determined by using the following formula;

[tex]\rm Speed =\dfrac{Distance}{Time}\\\\Distance = Speed \times Time[/tex]

Let x be the average airspeed of the plane.

And y be the average wind speed.

Initial trip: 18 = (x + y)

Return trip: 18 = (x – y)

Dalia flies an ultralight plane with a tailwind to a nearby town in 1/3 of an hour.

Then,

[tex]18 = \dfrac{1}{3} \rm (x + y)[/tex]

On the return trip, she travels the same distance in 3/5 of an hour.

[tex]18 = \rm \dfrac{3}{5} (x-y)[/tex]

Subtract equation 1 from equation 2

[tex]18 -18= \dfrac{1}{3} \rm (x + y) - \dfrac{3}{5}(x-y)\\\\0 = \dfrac{x}{3} + \dfrac{y}{3} - \dfrac{3x}{5}+\dfrac{3y}{5}\\\\ \dfrac{x}{3} + \dfrac{y}{3} - \dfrac{3x}{5}+\dfrac{3y}{5} = 0\\\\ \dfrac{5x+5y-9x+9y}{15}=0 \\\\ -4x+14y = 0\times 15\\\\14 y = 4x\\\\y =\dfrac{4}{14}x\\\\y = \dfrac{2}{7}x[/tex]

Substitute the value of y in equation 1

[tex]18 = \dfrac{1}{3} \rm (x + y)\\\\18 = \dfrac{1}{3} \rm (x + \dfrac{2}{7}x)\\\\18 = \dfrac{x}{3} + \dfrac{2x}{7}\\\\18 = \dfrac{3x+6x}{21}\\\\18 \times 21 = 9x\\\\x = \dfrac{18 \times 21}{9}\\\\x = 2 \times 21\\\\x=42[/tex]

Substitute the value of x = 42 in equation 1

[tex]18 = \dfrac{1}{3} \rm (x + y)\\\\18 = \dfrac{1}{3} \rm (42 + y)\\\\18 \times 3 = 42+y\\\\ 54 = 42+y\\\\y = 54-42\\\\y =12[/tex]

Hence the average rate of speed of the wind is 12 miles per second and the average rate of speed of the plan is 42 miles per second.

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