Christie pilots her plane for 360 mi against a headwind in 33 hrs. The flight would take 22 hrs and 15 mins with a tailwind of the same speed. Find the headwind and the speed of the plane in still air.

no

Let's assume the speed of the plane in still air is represented by "p" (in miles per hour), and the speed of the headwind is represented by "h" (in miles per hour).

To find the headwind and the speed of the plane in still air, we can set up two equations based on the given information.

1. Against the headwind:
The time taken to travel the distance of 360 miles is 33 hours. The effective speed of the plane against the headwind is (p - h) mph.
Using the formula: Speed = Distance / Time, we can write the equation as:
(p - h) = 360 / 33

2. With the tailwind:
The time taken to travel the same distance of 360 miles is 22 hours and 15 minutes, which is equivalent to (22 + 15/60) hours, or (22.25) hours. The effective speed of the plane with the tailwind is (p + h) mph.
Using the formula: Speed = Distance / Time, we can write the equation as:
(p + h) = 360 / 22.25

Now, we have a system of two equations:
(p - h) = 360 / 33
(p + h) = 360 / 22.25

To solve this system, we can use the method of elimination.

Multiplying both sides of the first equation by 33, and multiplying both sides of the second equation by 22.25, we get:
33(p - h) = 360
22.25(p + h) = 360

Expanding these equations, we have:
33p - 33h = 360
22.25p + 22.25h = 360

Now, let's solve this system of equations:

Multiply the first equation by -22.25, and the second equation by 33:
-22.25(33p - 33h) = -22.25(360)
33(22.25p + 22.25h) = 33(360)

Expanding these equations, we get:
-735.75p + 735.75h = -8010
735.75p + 735.75h = 11880

Adding the two equations together, we have:
0p + 1471.5h = 3870

Dividing both sides of the equation by 1471.5, we find:
h = 3870 / 1471.5
h ≈ 2.63 mph

So, the headwind speed is approximately 2.63 mph.

To find the speed of the plane in still air (p), we can substitute the value of the headwind (h) into either of the two original equations. Let's use the first equation:

(p - h) = 360 / 33
p - 2.63 = 10.91

Adding 2.63 to both sides, we find:
p = 13.54

Therefore, the speed of the plane in still air is approximately 13.54 mph.

To solve this problem, we need to set up a system of equations using the given information.

Let's assume that the speed of the plane in still air is represented by "p" (in miles per hour) and the speed of the headwind (or tailwind) is represented by "w" (in miles per hour).

Now, let's set up the first equation using the information about flying against the headwind:
Distance = Speed × Time
360 mi = (p - w) × 33 hrs

Next, let's set up the second equation using the information about flying with the tailwind:
Distance = Speed × Time
360 mi = (p + w) × (22 hrs + 15 mins)

Since 15 mins is equal to 15/60 = 0.25 hrs, we can convert 22 hrs and 15 mins to 22.25 hrs:
360 mi = (p + w) × 22.25 hrs

Now, we have a system of two equations:

1) 360 = (p - w) × 33
2) 360 = (p + w) × 22.25

To solve this system, we can use the method of substitution or elimination.

Let's rearrange Equation 1 to solve for p - w:
p - w = 360/33
p - w = 10.909 (approximately)

Next, we can substitute this value into Equation 2 and solve for p + w:
360 = (10.909 + 2w) × 22.25
360 = 242.447 + 44.625w
360 - 242.447 = 44.625w
117.553 = 44.625w
w = 117.553/44.625
w ≈ 2.63 (approximately)

Now, we can substitute the value of w back into Equation 1 to find p:
p - 2.63 = 10.909
p = 10.909 + 2.63
p ≈ 13.54 (approximately)

Therefore, the speed of the headwind is approximately 2.63 miles per hour and the speed of the plane in still air is approximately 13.54 miles per hour.

d = rt

360 = (x -t)33
360= (x+t)22.25

360 = 33x-33t
360 = 22.5x - 22.5t

Can you solve from here?