Question

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

Answer 1

Answer:airspeed of the plane is 520 mph

speed of the wind is 20 mph

Step-by-step explanation:

Let x represent the airspeed of the airplane.

Let y represent the speed of the wind.

Since the airplane was flying into the headwind, the total speed would be x - y miles per hour.

The flying distance between New York City and Albuquerque, New Mexico is 1800 miles.

Time taken to travel this distance is 3 hours 36 minutes = 3 + 36/60 = 3.6 hours

Distance = speed × time

Therefore

1800 = 3.6(x - y)

1800 = 3.6x - 3.6y - - - - - - - - - - 1

On the return flight, it is assumed that the airplane flew in the direction of the wind. The total speed would be (x + y) miles per hour. The same distance is traveled in 3 hours and 20 minutes = 3 + 20/60 = 3.33 hours. Therefore

1800 = 3.33(x + y)

1800 = 3.33x + 3.33y - - - - - - - - - - 2

Multiplying equation 1 by 3.33 and and equation 2 by 3.6, it becomes

6000 = 12x - 12y

6480 = 12x + 12y

Adding both equations, it becomes

12480 = 24x

x = 12480/24 = 520

Substituting x = 520 into equation 1, it becomes

1800 = 3.6 × 520 - 3.6y

1800 = 1872 - 3.6y

3.6y = 1872 - 1800 = 72

y = 72/3.6 = 20

Answer 2

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