well, we can find the derivative of position to get velocity, then we can take the derivative of velocity to get acceleration do you know how to find the derivative of t*ln(2t)? we'll be using the product rule
almost, ln(2t) + 1
oops i meant ln xD
no prob, now all we need to do is take the derivative of ln(2t). can you do that?
So the next derivative is 1/t
yup! our last step is to figure out what to plug in for t the problem wants to find the acceleration when the velocity is 0, so we go back to our velocity equation and set it equal to 0, then solve for t ln(2t) + 1 = 0 solve for t
i got 1/2e
right, t = 1/(2e), now we plug that back into our acceleration equation 1/t = 1/(1/(2e) = 1/(0.5e^-1) = 2e
Thank you so much for the help!