Posted by T.Gracken on April 30, 2001 at 10:37:06:
In Reply to: Re: find acceleration posted by dan on April 29, 2001 at 19:19:36:
: The problem reads: A particle moves along the curve given by y=sqrt(t^3+1). Find the acceleration when t=2.
: Kevin wrote:second derivative again. first derivate would be 1/2(t^3+1)^(-1/2)(3t^2).
: Would this workout to be 1/(3t)^(-1/2)(6t)?
: I have tried this a few different ways but have not gotten the correct answer yet (multi choice)
: Thank you very much, I have been working on this all day
Recall that if s(t) is a function that represents the POSITION of an object at a given time "t", then
v(t) = s '(t) is the VELOCITY of the object at a given time "t", AND...
a(t) = v '(t) is the ACCELERATION of the object at a given time "t".
In your case, y is the position function, so determine the second derivative of y with respect to t and evaluate when t=2.