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Posted 2002-11-11, 04:12 PM in reply to LordZpider's post "Calculus: Related rates"
Quote:
Originally posted by Titusfied
Heh, this is a very easy problem dude.

Therefore, when the pile is at 9.847 feet high, 100 minutes have passed, and (9.847/100) = 0.09847 feet per minute.

That is the answer you were looking for, no?
First, not an easy problem.
Second, 9.874/100 is change in distance divided by change in time. Think.... that's average velocity.

Here is how I would do it:

I decided to track the radius starting with your radius at 0 at the origin.
0-------->+

Okay, here's a couple of equations: (` is initial)
2a(x-x`) = (v)(v) - (v`)(v`)
x = x` + (v`)(t) + (a)(t)(t)/2

Initial velocity is 0
Initial displacement is 0 since that's where I set my axis.
Plug in numbers and you get respectively:
2a(9.847 - 0) = (v)(v)
9.847 = 0 + (a)(5000) (100seconds squared divided by 2)

Solve for a and plug into other equation:
2(.00197)(9.847) = (v)(v)

Solve for v:
v = .19694 feet/minute
---------------------------------
Check my work, that may or may not be the right method to do this problem. Hope I helped.
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