Derivatives help!?

Question

A particle moves in such a way that
x(t)= 1/3t^3-3t^2+8t is the position of the particle when t is greater than 0.

when t=3, what is the total distance the particle has traveled? include turning point and correct total distance!
ahh this is a 1973 Free Response AP CAlculus ab question and i have no clue how to solve it!!!!!!!! please help soon! :)

Answer 1

Integrate x(t) from t=0 to t=3.

Int x(t) = 1/3(3t^2) -3(2t) +8 = t^2 -6t +8

X(0) = 8 and x(3) = 3^2 -6(3) +8 = 9-18+8 =-1

So the particle travelled along the curve 8-1 = 7 units of distance in 3 units of time.
)

Answer 2

f(t) = 1/(3t^3-3t^2+8t)
The easy way to get the derivative for this is to re-write the function as
f(t) = (3t^3-3t^2+8t)^(-1) and then use the rule for a function raised to a power [ g(x) = (h(x))^n => g'(x) = (h(x))^(n-1) * h'(x) ] I'll let you work out the details ☺

The total distance will be the integral of f(t)∙dt from 0 to 3

Have fun ☺

Doug