<p>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)</p>
<p>i know that speed is absolute value of velocity and when velocity is 0 t= 2/3 and 1</p>
<p>so during the interval (2/3 , 1) when is the speed at a maximum??</p>
<p>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</p>
<p>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"</p>
<p>v'(t) = 12t - 10
0 = 12t - 10
t = 5/6</p>
<p>so plug in 5/6 to velocity</p>
<p>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</p>
<p>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.</p>
<p>-will</p>
<p>also just to make sure, on what interval is the particle moving to the left? i got (2/3,1) is that right?</p>
<p>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</p>
<p>The particle moves to the left when velocity is negative, so the entire interval, yes.</p>
<p>entire interval meaning [2/3,1] or (2/3,1)</p>
<p>Actually, we use [2/3,1] to mean 2/3<=t<=1 and (2/3,1) to mean 2/3<t<1</p>
<p>So, you see, with the first notation you actually have extremas at the endpoints. Hence why you needed to check if speed was greatest there.</p>
<p>you don't need to check the endpoints in this one because he said straight out that the speed at the endpoints IS zero.</p>
<p>Ouch, totally missed that. =(
But c'mon, who is never been guilty of not reading a question completely? =P</p>
<p>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?</p>
<p>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<t<1 or (2/3,1).</p>