x(t) = -t + t^3/6
v(t) = dx/dt = -1 + t^2/2
I think you can now answer the questions if you think a bit.
Suppose an object moves along the x-axis so that its position at time $t$ is $x=-t+\frac{t^3}{6}$.
(a) Find the velocity, $v(t)=x(t)$, of the object.
(b) What is $v(0)$? What does this say about the direction of motion of the object at time $t=0$?
(c) When is the object at the origin? What is the velocity of the object when it is at the origin?
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