Linear particular motion Calc AB Question!

Test Preparation AP Mathematics & Computer Science
1

<p>I'm having trouble with this question (especially part c). Could anybody help me?</p>

<p>A particle moves along a straight line so that at any time t its position is given by x(t)=2(pi)(t)+cos2(pi)(t).</p>

<p>a) Find the velocity at time t.</p>

<p>b) Find the acceleration at time t.</p>

<p>c) What are all values of t, 0<=t<=3, for which the particle is at rest?</p>

<p>Answers:
a) I just found the derivative which was x'(t)=2pi-2(pi)sin2(pi)(t)</p>

<p>b) I just found the second derivative which was x''(t)=-4(pi)(t^2)cos2(pi)(t)</p>

<p>c) Here is where I am really confused. Is a particle at rest when the velocity or the acceleration is 0? I set acceleration equal to 0 and did this:
1) 0=-4(pi)(t^2)cos2(pi)(t)
2) So now we know that one of the solutions is t=0
3) Then I set cos2(pi)(t) equal to 0 and got t=1/4 and t=3/4. However, since it said all numbers from 0 to 3, I went up in increments of 1/2 and got t=0,1/4,3/4,5/4,7/4,9/4, and 11/4. </p>

<p>Is this right or am I totally off base? Help is much appreciated!</p>

2

<p>For part c: </p>

<p>The particle is at rest when the velocity is 0. Zero velocity means constant position. The particle is not at rest when the acceleration is 0. Zero acceleration means constant velocity (it is moving). </p>

<p>Right?</p>

3

<p>I am genuinely not sure. I thought it would be something easy to google but it’s not. I can’t find an answer anywhere. I’ll solve velocity for 0 now and see what happens.</p>

4

<p>Set V (or the first derivative) equal to 0 to find when the particle is at rest…</p>

5

<p>Okay so I tried that and I got:
1)2pi-2(pi)sin2(pi)(t)=0
2)2pi=2(pi)sin2(pi)(t)
3)1=sin2(pi)(t)</p>

<p>t=1/4, 5/4, 9/4</p>

<p>Is this better?</p>

6

<p>That looks fine to me.</p>

7

<p>Thanks so much!</p>