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if s(t) = t*ln(3t), then what is s ' (t) ?

so the acceleration is the second derivative right?

yes of the position function

ok, give me minute

i think it's wrong, but I got 1+ln(3x) for the first derivative

then you find v'

ok, so the secod derivative is 1/x

1/t, yes

okay so now do I plug zero ?

but I have to solve for the acceleration, which is the second derivative.

yes I know

but you need to evaluate the acceleration at the value that makes the velocity 0

so do i solve for x when velocity is zero?

yes solve 0=1+ln(3x)

ok, I got stuck I completely forgot how to solve from this part

|dw:1440985593903:dw|

do I ln both sided to cancel ln ?

1/3e

yes, \[\Large t = \frac{1}{3e}\] now plug that into a(t)

3e

yep, a(t) = 3e is your acceleration when t = 1/(3e)