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 12-16-2005, 11:04 PM #1 Member   Join Date: Sep 2005 Posts: 765 how to find maximum speed??? a particle moves on the x-axis in such a way that its position at time t is given by x=(2t-1)((t-1)^2) i know that speed is absolute value of velocity and when velocity is 0 t= 2/3 and 1 so during the interval (2/3 , 1) when is the speed at a maximum?? Reply
 12-16-2005, 11:51 PM #2 Junior Member   Join Date: Sep 2005 Posts: 155 well you have the equation x=(2t-1)((t-1)^2)... so take the derivative of that to find eq for speed... first d second plus second d first... v(t) = (2t - 1)(2t - 2) + ((t - 1)^2)(2) v(t) = (4t^2 - 6t + 2) + (2t^2 - 4t +2) v(t) = 6t^2 -10t +4 so then to find where speed is greatest, find the critical numbers (where deriv is 0 or undef). and i know i'm working with the velocity eq... so we'll count min's as well as max's to find "max speed" v'(t) = 12t - 10 0 = 12t - 10 t = 5/6 so plug in 5/6 to velocity v(5/6) = 6*(5/6)^2 - 10(5/6) - 4 ..........= 25/6 - 50/6 + 24/6 ..........= -1/6 take absolute value, and the max speed is 1/6 units/sec i guess you didn't need to know what the max speed was, but i did it anyway. but the t value for the max is 5/6. -will Reply
 12-17-2005, 12:40 AM #3 Member   Join Date: Sep 2005 Posts: 765 also just to make sure, on what interval is the particle moving to the left? i got (2/3,1) is that right? Reply
 12-17-2005, 03:32 AM #4 Junior Member   Join Date: Nov 2005 Posts: 136 Is that an open or closed interval? If it's a close interval you must check the endpoints for speed. abs(v(2/3)) = abs(v(1)) < abs(v(5/6)) K, answer for max speed still occurs at time = 5/6 The particle moves to the left when velocity is negative, so the entire interval, yes. Reply
 12-17-2005, 03:38 AM #5 Member   Join Date: Sep 2005 Posts: 765 entire interval meaning [2/3,1] or (2/3,1) Reply
 12-17-2005, 04:16 AM #6 Junior Member   Join Date: Nov 2005 Posts: 136 Actually, we use [2/3,1] to mean 2/3<=t<=1 and (2/3,1) to mean 2/3
 12-17-2005, 10:11 AM #7 Junior Member   Join Date: Sep 2005 Posts: 155 you don't need to check the endpoints in this one because he said straight out that the speed at the endpoints IS zero. Reply
 12-17-2005, 10:15 AM #8 Junior Member   Join Date: Nov 2005 Posts: 136 Ouch, totally missed that. =( But c'mon, who is never been guilty of not reading a question completely? =P Reply
 12-17-2005, 05:49 PM #9 Member   Join Date: Sep 2005 Posts: 765 so is it [2/3,1] or (2/3,1)? which one is it? the question was asking, during what interval of time is the particle moving to the left? Reply
 12-17-2005, 06:48 PM #10 Junior Member   Join Date: Nov 2005 Posts: 136 At t = 2/3 and t = 1, since velocity is zero, you are in the process of changing directions (you are not moving to the left). So you move to the left at 2/3

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