<p>Can anybody give me a tip on solving this problem?</p>
<p>A particle moves along the y-axis so that its velocity at any time t>or equal to 0 is given by v(t)=tcost^2. At time t=0, the position of the particle is y=3. For what values of t, when t is between 0 and 5 and can equal both, is the particle moving upward?</p>
<p>-Okay so i know that the particle moves upward when velocity is positive, but I am confused about the y-axis part. When I graph it, it crosses the y-axis in only one spot, at 0.</p>
<p>Thanks for any help!</p>
<p>take derivative of velocity to find acceleration
Solve for zeros of acceleration
plug'n chug, find the range of numbers that give positive acceleration.</p>
<p>That is when the particle moves up.</p>
<p>*note
1. I dont think original position matters.
2. Try not to rely on your graphing calculator. Most of the time it only dampens your ability to learn.</p>
<p>phoenix - Don't you mean when the velocity is positive, not acceleration? When the acceleration is positive that just means that the velocity is increasing. When the Velocity is positive that means displacement is increasing, a.k.a. the particle is moving up.</p>
<p>Yeah, I don't think acceleration matters in this problem. I think you just use techniques of graphing and find where the equation is increasing and decreasing. So just set v(t)=0, find the roots of the equation, and then do the little test thing to find where it is increasing.</p>
<p>ok i now see that I did it right in the first place. Thanks for your help!</p>