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
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.)
(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.
The position function in black, the velocity function in red, and the acceleration function in green: