<p>A particle moves along the x-axis so that at any time t is greater than or equal to 0 its velocity is given by v(t) = ln(t+1) - 2t^2 + 4t -1.
What is the net displacement of the particle between t=0 and t=2? Answer -.704. Explanations? Thanks!</p>
<p>Ummm...you integrate the velocity using a calculator. Pretty straightforward.</p>
<p>that gives total distance traveled, not net displacement..</p>
<p>No. That gives the displacement. The distance traveled is obtained by integrating the absolute value of the velocity formula.</p>
<p>so anti differentiate the function ?</p>
<p>With your calculator. Yes.</p>
<p>you wouldnt need a caculator, but thats how you do it. And because the area under the x-axis is negative, the integral shows the displacement, not the distance traveled.</p>
<p>Since the answer is -.704, I assumed that this is the calculator section of the test.</p>
<p>I don't know what I'm doing wrong, but when I integrate that (a=0, b=2) I come up with 1.9625.</p>
<p>?</p>
<p>That's what I get too. So unless I am seriously missing something, I think 1.9625 is the right answer. When you graph the velocity equation you can see that the integral will be positive.</p>
<p>Thank god. I thought I was being an idiot and forgot how to integrate.</p>