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An airplane flying into a headwind travels the 1800-mile flying distance between New York City and Albuquerque, New Mexico in 3 hours and 36 minutes. On the return flight, the same distance is traveled in 3 hours and 20 minutes. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant.

airspeed of the plane mph
speed of the wind mph

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marcthemathtutor

Answer:

plane speed = 520 mph

wind speed = 20 mph

Step-by-step explanation:

let the speed of the wind be w mph and the speed of the plane be p mph

flying from NY to NM,

time taken = 3 hr 36 min = 3.6 hr

average ground speed = total distance / time taken

= 1800 / 3.6 = 500 mph

since it is a headwind, the plane speed will be slowed down by the head wind, resulting in a lower ground speed.

we can write the following equation:

ground speed = plane speed - wind speed

500 = p - w ------(eq1)

flying from NY to NM (return flight),

time taken = 3 hr 20 min = 3.333 hr

average ground speed = total distance / time taken

= 1800 / 3.333 = 540 mph

on the return trip, the headwind becomes a tailwind, hence the total ground speed would be faster than the plane's air speed , we can write the following equation:

ground speed = plane speed + wind speed

540 = p +  w ------(eq2)

with these 2 systems of equations, we can solve for p & w using either substitution of elimination method.

eventually you will end up with

plane speed = 520 mph

wind speed = 20 mph

   Speed of airplane is 520 miles per hour and speed of the wind is 20 miles per hour.

    Let the speed of an airplane = x miles per minute

And the speed of the wind = y miles per minute

Distance between New York City and Albuquerque = 1800 miles

Airplane covers the distance between New York City and Albuquerque in 3 hours 36 minutes Or 3.6 hours.

In return journey, airplane covers this journey in 3 hours 20 minutes Or 3.33 hours.

Since, airplane takes more time from New York City to Albuquerque,

Therefore, airplane covered this distance against the wind,

And the speed of the airplane against the wind = (x - y) miles per hour

Expression for the speed,

Speed = [tex]\frac{\text{Distance}}{\text{Time}}[/tex]

(x - y) = [tex]\frac{1800}{3.6}[/tex]

x - y = 500 ------- (1)

Speed of airplane in return journey = (x + y) miles per hour

And the equation will be,

x + y = [tex]\frac{1800}{3.33}[/tex]

x + y = 540 ------- (2)

By adding equation (1) and (2),

(x - y) + (x + y) = 500 + 540

2x = 1040

x = 520 miles per hour

From equation (1),

520 - y = 500

y = 20 miles per hour

       Therefore, speed of airplane is 520 miles per hour and the speed of wind is 20 miles per hour.

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