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 October 30th, 2010, 01:46 PM #1 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 velocity of a body moving along the s-axis v = t^2-4t +3 >> hello ! porticle motion at time t>=0 , the velocity of a body moving along the s-axis v = t^2-4t +3 a ) Find the body acceleration each time the volicty is zero . b ) When the body moving forward ? backward ? c) when is the body's velocity increasing ? deceasing ? ----------- the answer>>>>> [color=#0000FF]a ) v = 0 2t-4 = 00 t = 2 acceleration = dv/dt = 2 b) I'm not sure for this answer : v = (2)^2 -4(2) + 3 = -1 then the body moving backward c ) body's velocity is deceasing [/color]
 October 30th, 2010, 03:50 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: velocity of a body moving along the s-axis v = t^2-4t +3 We are given $v(t)=t^2-4t+3$ a) Find the body's acceleration each time the velocity is zero. $a(t)=\dfrac{dv}{dt}=2t-4$ $v(t)=(t-1)(t-3)=0$ Thus, the velocity is zero at t = 1 and t = 3. $a(1)=-2$ $a(3)=2$ b) When is the body moving forward? backward? v(0) = (-)(-) = positive, moving forward on [0,1) v(2) = (+)(-) = negative, moving backward on (1,3) v(4) = (+)(+) = positive, moving forward on (3,?) c) When is the body's velocity increasing? decreasing? We equate a(t) to zero, the check the interval(s). a(t) = 2(t - 2) = 0 a(t) is zero at t = 2. a(0) = negative, velocity decreasing on [0,2) a(3) = positive, velocity increasing on (2,?)
 October 30th, 2010, 07:15 PM #3 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 Re: velocity of a body moving along the s-axis v = t^2-4t +3 Hi all the velocity is zero at t = 1 and t = 3. why you talke 3 why didny take 1 to 2 ?
 October 30th, 2010, 07:34 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: velocity of a body moving along the s-axis v = t^2-4t +3 Well, part (a) asks what the acceleration is when the velocity is zero. Factoring the given v(t) we find $v(t)=t^2-4t+3=(t-1)(t-3)=0$ Thus, the velocity is zero when t = 1 and t = 3. So we use those values for t in a(t) which we find by differentiating v(t) with respect to t, which is $a(t)=2t-4$ 2 is not a zero for v(t), but it is a zero for a(t), which tells us when the velocity changes the sign of its rate of change. Graph of velocity: Graph of acceleration:
 October 30th, 2010, 11:19 PM #5 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 Re: velocity of a body moving along the s-axis v = t^2-4t +3 I forget how we can get (t-1 ) (t-3 ) = 0 helo ne in this step .
 October 31st, 2010, 01:30 AM #6 Member   Joined: Oct 2010 Posts: 32 Thanks: 0 Re: velocity of a body moving along the s-axis v = t^2-4t +3 thanks now I understand all

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