Fall+Week+09

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Thursday 10/13



 * Particle Motion Solutions:**

1995 #2. (excuse the poor handwriting). [Anybody mind telling me how to make this a link?] Thanks for posting this, Philip. I believe the domain mentioned in the problem is [0, 5] so you need to check the sign of v(t) for the interval (3pi/2, 5) to complete the answer to part a) The rest of it looks good.  Ms. Gentry  - I think I got it. -Phillip.  You sure did!

1994 #4 I think this is right, but I'm not suite sure about part c), I know that it would be a minimum as close to zero as possible because of the domain of natural log and that specific velocity function, but I'm not sure I showed it properly. Good Job on a) and b.See below the graphic for comments on c) Ms. G

Your answer would be OK if it was impossible to have a negative velocity. Unfortunately that is not the case. So....If you want to show the velocity is a minimum, the graph of velocity will have a horizontal tangent line at the minimum, and the derivative of velocity will change from negative to positive. To figure out when this happens, first set v '(t) = 0. v ' (t) = a(t) = ln(t) = 0 e ^0 = t t = 1. For values of t < 1, ln(t) < 0. For values of t > 0, ln(t) > o so the derivative of the velocity function changes form negative to positive at t =0. Therefore, the minimum velocity occurs when t = 1. The question asked for the minimum velocity so now evaluate v(1). v(t) = t (-1 - ln(t)) so v(1) = 1( -1 - ln(1) = -1. Answer: Minimum velocity is -1