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How to find the at rest position of a particle when trig functions

  1. Mar 11, 2014 #1
    so my given: s(t)=cos(pie8*t/4)

    took the derivative= velocity function

    then, v(t)= -pie/4 *sin(pie*t/4)

    When is the particle at rest? v(t)=0

    now, 0= -pie/4 *sin(pie*t/4)


    im lost here. I know it's very simple im just over thinking. What do I do from here?

    thanks
     
  2. jcsd
  3. Mar 12, 2014 #2
    figured it out. had to remind myself that the period of sin and cosin is -2pie->2pie.

    then use the formula forget what its called. but 2pie/b

    2pie/pie/4 = 8. then when divide the interval into 4 sub intervals and we have 0, 2, 4, 6, 8.

    remembering that sin is negative at 0, pie, 2pie. so velocity is zero at t=0,4,8
     
  4. Mar 12, 2014 #3

    jbunniii

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    $$-\frac{\pi}{4} \sin \frac{\pi t}{4} = 0$$
    is true if and only if ##\sin(\pi t/4) = 0##. For what values of ##x## is ##\sin(x) = 0##?
     
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