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Calculus (Derivatives)

Two particles are moving in straight lines. The displacement (in meters) of particle 1 is given by the function e^(4cos(t)) , where t is in seconds. The displacement (in meters) of particle 2 is given by the function -(t^3)/(3) - (t^2)/(2) + 2 , where t is in seconds. Find the first positive time at which the particles have(approximately) the same velocity.

A.) t = 1.569 seconds
B.) t = 0 seconds
C.) t = 2.366 seconds
D.) t = 0.588 seconds
E.) t = 1.011 seconds

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Mishaka

  1. v1 = -4 sint e^4cost
    v2 = -t^2 - t
    so when does 4 sin t e^4 cos t = t^2+t ??
    try approximately those answers to see when v1=v2
    A 3.96 and 4.16
    B 4 and 0 no
    C the rest look pretty far off
    try A accurately
    4.0288 and 4.0307
    It is A

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    Damon

  2. Yes the answer is t=1.569
    the first positive time which the two particles have the same velocity is at 1.569 seconds. checck this with nderv on your cal. easier to find the difference to find where its derivative is 0.

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    mathssssss

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