Flight Path. In this problem you will use a cosine function to model the flight path of a plane that begins its descent from 20000 ft when it is 45 miles from the airport. Find the fuction p(x)= ACos(wx)+B that that gives the altitude in ft when the plane is x miles from the runway. your function must satisfy P(0)=0, P'(0)=0 P(45)=20000 P'(45)=0. if the plane is maintaining a constant horizontal velocity of 250mph, what is the maximum acceleration experienced by the passengers?

To find the function p(x), we need to determine the values of A, B, C, and w. Let's break down the given conditions step by step:

1. P(0) = 0: This condition means that when the plane is 0 miles from the runway, the altitude is 0 ft. Using the function p(x) = ACos(wx) + B, substitute x = 0 and set the equation equal to 0:
ACos(0) + B = 0

Since Cos(0) = 1, the equation simplifies to:
A + B = 0

We have our first equation: A + B = 0

2. P'(0) = 0: This condition represents the derivative of the altitude function at x = 0, which represents the initial velocity of the plane. Taking the derivative of p(x) with respect to x:
P'(x) = -AwSin(wx)

Substituting x = 0 and equating it to 0:
-AwSin(0) = 0

Since Sin(0) = 0, the equation simplifies to:
0 = 0

This condition provides no new information and can be disregarded.

3. P(45) = 20000: This condition represents the altitude of the plane when it is 45 miles from the runway. Substituting x = 45 into the altitude function:
ACos(45w) + B = 20000

We have our third equation: ACos(45w) + B = 20000

4. P'(45) = 0: This condition represents the derivative of the altitude function at x = 45, which gives us information about the acceleration at that point. Taking the derivative of p(x) and substituting x = 45:
P'(x) = -AwSin(wx)

-AwSin(45w) = 0

Since Sin(45w) = 0, the equation simplifies to:
0 = 0

This condition also provides no new information and can be disregarded.

Now, we have three equations to solve for A, B, and C:

1. A + B = 0
2. ACos(45w) + B = 20000

From the first equation, we can solve for A in terms of B as A = -B. Substituting this into the second equation:
-BCos(45w) + B = 20000

Rearranging the equation:
BCos(45w) = B - 20000

At this point, it is not possible to determine the exact values of A, B, and C since we have two unknowns (w and B) and only one equation. Additional information or assumptions are needed to find precise values.

Moving on to the final part of the question: finding the maximum acceleration experienced by the passengers.

Given that the horizontal velocity of the plane is 250 mph, we can find the horizontal distance (x) traveled by the plane using the formula:
x = v * t

where v is the velocity (250 mph) and t is the time taken to travel that distance.

The time taken to travel the distance of 45 miles is:
t = x / v = 45 miles / 250 mph ≈ 0.18 hours

Since the passengers are experiencing a constant horizontal velocity, the time t is the same for all altitudes.

To find the maximum acceleration, we need to determine the second derivative of the altitude function (P''(x)) and evaluate it at x = 45.

Taking the second derivative of p(x) = ACos(wx) + B:
P''(x) = -A^2w^2Cos(wx)

Substituting x = 45:
P''(45) = -A^2w^2Cos(45w)

Note that without any specific values or constraints on w or A, we cannot determine the exact value of the maximum acceleration experienced by the passengers.