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anonymous

  • 5 years ago

Let s=30/(t^2+12) be the position function of a particle moving along a coordinate line, where s is in feet and t is in seconds. (a) Find the maximum speed of the particle for t>=0 . If appropriate, leave your answer in radical form. Speed (ft/sec): ? (b) Find the direction of the particle when it has its maximum speed.

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  1. anonymous
    • 5 years ago
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    If t is >= 0, plug in 0 in the equaion and you'll get : s(0) = 30/12 ft/sec I guess ^_^

  2. anonymous
    • 5 years ago
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    I don't think that is the correct, answer. Give me a few minutes, I will do this problem.

  3. anonymous
    • 5 years ago
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    Ok, to obtain the maximum speed you would have to take the second derivative of s(t) So, s''(t) = 0 will give you the maximum speed, s'(t)=0 will give you where the speed is zero not maximum. (this is why the above answer is incorrect). s''(t) = 0 solving for t will be +2 and -2 choosing t>=0 then t = +2 therefore plug it back into the velocity equation s'(t) which is going to be at time t=2 s'(2)= -15/32 Feet/Second, that would be the velocity. speed would just be the absolute value, so speed at time t=2 is 15/32 b) the direction will be in the negative direction since the velocity is negative, and the initially we assume the object to be at coordinate (0,5/2) given from s(0)

  4. anonymous
    • 5 years ago
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    for your reference, s'[t]=-((60 t)/(12 + t^2)^2) s''[t]=[(240 t^2)/(12 + t^2)^3] - [ 60/(12 + t^2)^2]

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