where should a pilot start descent

[jessicaxutie]

mn

 

[Deleted member 4993]

An approach path for an aircraft landing is shown in the figure and satisfies the following conditions:
i. The cruising altitude is h when descent starts at a horizontal distance l from touchdown at the origin.
ii. The pilot must maintain a constant horizontal speed v throughout descent.
iii. The absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity).
1. Find a cubic polynomial p(x) = ax[sup:15ioqpup]3[/sup:15ioqpup] + bx[sup:15ioqpup]2[/sup:15ioqpup] + cx + d that satisfies condition (i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and touchdown.
2. Use conditions (ii) and (iii) to show that . 6hv[sup:15ioqpup]2[/sup:15ioqpup]/l[sup:15ioqpup]2[/sup:15ioqpup] ?k
3. Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k = 860 mi/h2. If the cruising altitude of a plane is 35,000 ft and the speed is 300 mi/h, how far away from the airport should the pilot start descent?

Duplicate post:

http://answers.yahoo.com/question/index ... 057AAEvCWc

http://www.artofproblemsolving.com/Foru ... p?t=135604

Where is the figure?

Please show your work - indicating exactly where you are stuck - so that we can begin to help you.

 

[jessicaxutie]

here is a more helpful site:http://www.webster.ac.at/faculty/wazir/homepages/PlaneDescent.PDF
actually i don't know where to start. i know that you have to find the derivative but after that i don't know what else to do.

 

[Deleted member 4993]

This is a project problem - and you need to show your understnding of the problem.

Start with the fact that

p(0) = 0

so what are the conditions on p'(x)?

 

[jessicaxutie]

would that be 3ax[sup:1wadstzg]2[/sup:1wadstzg]+2bx+c ?

 

[Deleted member 4993]

would that be 3ax[sup:1re9ybid]2[/sup:1re9ybid]+2bx+c ?<<< That is the expression for p'(x)

However, you want to impose condition on p'(x) like we put condition on p(x) at x = 0 (from the figure).

 

[jessicaxutie]

so it is 3a[sup:2ssrh15o]2[/sup:2ssrh15o]+2b

 

[Deleted member 4993]

so it is 3a[sup:28shkegp]2[/sup:28shkegp]+2b<<<< How is that ? What is that?

 

[jessicaxutie]

i not sure .. i thought x=0.

 

[Deleted member 4993]

You need a face-to-face tutor.

 

[jessicaxutie]

plane descent

qgd

 

[Deleted member 4993]

duplicate post:

http://answerboard.cramster.com/calculu ... 397-0.aspx

 

[stapel]

mn

I guess this poster figures she's gotten everything she's going to get from us, and is now posting compilations of tutors' work on yet-another tutoring site, as though it were her own. Nice.

Eliz.

 

[jessicaxutie]

Re:

mn

I guess this poster figures she's gotten everything she's going to get from us, and is now posting compilations of tutors' work on yet-another tutoring site, as though it were her own. Nice.

Eliz.

of course, i have everything that i need. this site or who ever "us" is, did not even help. and i didn't even say that this is my own work. so you need to recheck before posting something up.

 

[stapel]

i didn't even say that this is my own work.

You asked the new bunch of tutors to "check my work".

this site or who ever "us" is, did not even help.

So I guess, by "help", you mean "do it for you"...? Because you certainly received here many replies and many attempts to assist you, and I'm sure you don't mean to insult those tutors' efforts.

My best wishes to you in your studies.

Eliz.