suppose that the position of a particle as a function of time (in seconds) is given by the formula s(t)=6.5+4t^2-t^4,t>0 the time at which the velocity is zero is= ........ seconds the time at which the acceleration is zero is= ....... seconds the time at which the velocity is maximum is=..... seconds
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briefcaseTeacher (K-12)
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The position function is given as `s(t)=6.5+4t^2-t^4` for t>0
The velocity function is the first derivative of the position function, and the acceleration function is the derivative of the velocity function (or the second derivative of the position function.)
`v(t)=s'(t)=8t-4t^3`
`a(t)=v'(t)=s''(t)=8-12t^2`
(a) The velocity is zero when `8t-4t^3=0`
`4t(2-t^2)=0 ==> 4t=0,2-t^2=0` Since t>0 we have `t=sqrt(2)~~1.41`
The velocity is zero at approximately 1.41 seconds.
(b) The acceleration is zero when `8-12t^2=0`
`t^2=2/3` . Since t>0 `t=sqrt(2/3)~~.82`
The acceleration is zero at approximately .82 seconds.
(c) The velocity is maximum when the acceleration is zero; since t>0 this only occurs at t approximately .82 seconds.
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