Velocity, acceleration, derivatives without function I'm having trouble with an exercise about position, velocity and acceleration. 
Apologies for the formatting but the prompt has a graph. 

So I was able to fill out the table up to a certain point, but I"m having issues understanding how I can find the velocity if I don't have the function. I know that I should use the derivatives, but without having the expression of the function, how can I find it?
Thanks a lot for your help,
 A: If the velocity is constant, which is indicated by the fact that the position-time graph is a straight line, you can just take any two points off the graph and use $v=\frac {\Delta s}{\Delta t}$.  Any two points on the segment will do.  For example, from $t=2$ to $t=6$ it moves from $+6$ to $-10$, so the velocity is $\frac {-10-(+6)}{6-2}=-4$
A: You just look at the slope of the graph. An upward slope indicates motion to the right of the origin/starting point, or positive velocity, and vice-versa. As an example, in the interval $0 < t < 2$, the object travels $6$ meters in $2$ seconds, hence a velocity of $3 \frac{m}{s}$. You continue finding the slope of each “section” of the graph. Just remember that velocity depends on whether the slope is upward or downward, while speed is a scalar, not a vector, so it’s always positive.
Finally, acceleration is the derivative of velocity with respect to time. A linear graph for position means constant velocity, hence acceleration is $0$.
For instance:
$$s(t) = nt \implies v(t) = n \implies a(t) = 0$$
Notice how you need one higher power for position (something involving $t^2$) so you get a non-zero acceleration, such as
$$s(t) = \frac{1}{2}at^2 \implies v(t) = at \implies a(t) = a$$
for some constant $a$. 
Hence, for a linear position-time graph, you just find the slope, which becomes the velocity.
