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Help with Calculus Problem?
A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 27 t^2 + 108 t where s is measured in feet and t in seconds.
Find the velocity (in ft/sec) of the particle at time t=0: I figured out that this is 108.
The particle stops moving (i.e. is in a rest) twice, once when t=A and again when t=B where A < B. I also figured out that A is 3 and B is 6.
What is the position of the particle at time 18? I'm pretty sure that this is 4860.
Finally, what is the TOTAL distance the particle travels between time 0 and time 18? This is the part that I'm stuck on because I'm not sure how to calculate total distance.
-Thanks for the help.
1 Answer
- Christophe GLv 41 decade agoFavorite Answer
the total distance is the integral of the function.
An integral is a summation of continuous numbers.
You need to calculate the integral of s(t) between 0 and 18.
