Calculus help, anyone? Position, velocity, acceleration

If the position of the particle is given by s(t) = (1/3)t^3 - t^2 - 3t + 4 , v(t)= t^2 - 2t - 3 and a(t) = 2t - 2 .....

1) The particle changes direction at time(s): ________.

2) The total distance traveled from 0 < t < 6 is: ______.

3) The total displacement from 0 < t < 6 is ______.

4) The average velocity from 0 < t < 6 is ______ .

5) The average speed from 0 < t < 6 is ______.

I don't really care if you give me the answers... I would just like to know how to do these problems because I don't understand them!! Thanks!! :)

They give you all three of position, velocity, and acceleration, but you should be aware that each of those is the derivative of the previous one.

1) Changing direction means the velocity changes from positive to negative, or vice versa. To do this it has to go through zero. So, the time when the direction changes is the time when v(t) = 0.
Solve the quadratic t^2 - 2t - 3 = 0

3) Displacement is the difference in position. The displacement from time t = 0 to t = 6 is s(6) - s(0).

4) Average velocity is (total displacement) / time.
5) Average speed is (total distance) / time.


2) Distance is more complicated than displacement because you count the entire distance that you've gone, even if you backtracked. For example, running all the way around a baseball diamond you travel about 100m but have a displacement of 0.
Use the answer to question (1) to break up the time into intervals in which the particle is moving in each direction. Add up the absolute value of the displacements over each interval.

Another way to look at this: speed is the absolute value of velocity, and distance the integral of speed. You can find the total distance by integrating |v(t)|. But this integral is essentially the same as evaluating s(t) over the two (or more) intervals.

1) The particle changes direction at time(s): ________.