My Math Forum problem in motion Along coordinate line > help me

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 October 30th, 2010, 12:45 PM #1 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 problem in motion Along coordinate line > help me Hi all
 October 30th, 2010, 01:05 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: problem in motion Along coordinate line > help me You are given the displacement , where $0\le t\le 3$ thus $v(t)=\dfrac{ds}{dt}=t^3-3t^2+2t$ $v(0)=0$ $v(3)=3^3-3\cdot3^2+2\cdot 3=6$ To find where the velocity changes, we first find all points where v = 0 within the given interval. $t^3-3t^2+2t=0$ $t(t-1)(t-2)=0$ Thus t = 1 and t = 2 are critical points. So we check point in the three intervals the two critical points divide the given interval into. v(.5) is positive v(1.5) is negative v(2.5) is positive Thus, the particle changes direction at t = 1 and t = 2.
 October 30th, 2010, 01:38 PM #3 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 Re: problem in motion Along coordinate line > help me thanks very much now i understand >>>
 October 30th, 2010, 01:42 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: problem in motion Along coordinate line > help me We put those numbers into the velocity function $v(t)=t^3-3t^2+2t$. Even easier is to put them into its factored form: $t(t-1)(t-2)$ and look at the signs of each factor. for .5 we have (+)(-)(-) = positive for 1.5 we have (+)(+)(-) = negative for 2.5 we have (+)(+)(+) = positive And actually, we don't really need to check since all of the roots are of multiplicity 1, but I thought checking would be helpful to show the change in direction.
 October 30th, 2010, 01:46 PM #5 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 Re: problem in motion Along coordinate line > help me thanks very much now i understand >>>

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### motion along a coordinate line

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