The position $s$ at time $t$ of a moving object is given by $$s(t)=\frac{1}{3}t^3-\frac{3}{2}t^2$$ Assume the object moves in a straight line and that $t$ is greater or equal to zero. Measure $s$ in inches and $t$ in seconds.
When is the object speeding up? When is is the object slowing down?
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EDIT(to be clearer)
You have to look for intervals where both the velocity and the acceleration are pointing towards the same direction. To do so:
Find the first derivative of $s(t)=\dfrac{t^3}{3}-\dfrac{3t^2}{2}$ to find critical points. Separate the intervals in positive and negative ones given these critical points. Do the same with the second derivative. Then: if the velocity and acceleration are both positive or negative, then select that interval as the one in which the objetct speeds up.
Warning!: Please don't confuse positive velocity with speeding up. Only the acceleration will tell you if the object is slowing down or speeding up.
Figure: $|v(t)|$ from which it can be seen that there is an interval where the object actually slows down.
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The velocity at time $t$ is $t^2-3t$. This is positive if $t\gt 3$ and negative if $0\lt t\lt 3$.
The speed is $3t-t^2$ for $0\lt t\lt 3$ and $t^2-3t$ for $t\gt 3$.
We interpret "speeding up" as meaning that the speed is increasing. In the interval from $0$ to $3$, the derivative of the speed is $3-2t$. This is positive if $0\lt t\lt \frac{3}{2}$.
For $t\gt 3$ the derivative of the speed is $2t-3$, which is positive for all $t\ge 3$.
Thus the particle is speeding up in the interval $0\lt t\lt \frac{3}{2}$ and also in the interval $3\le t\lt \infty$.
The instantaneous velocity is the derivative
$$v(t) = \frac{ds}{dt} = t^2 - 3t$$
Can you take it from here?