s = (t^2-1)^3
s' = velocity = 3(t^2-1)^2(2t)
= 6t(t^2-1)^2
acceleration is the second derivative of s, so differentiate again using the product rule, sub in t=2 and you are done
this was on the math test that i got wrong...
The position equation for the movement of a particle is given by s=(t^2-1)^3 when s is measured in feet and t is measured in seconds. Find the acceleration at two seconds...
i found the first derivative but i think it's wrong: (2t)^3
help!!!
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