Answer
(a) moving away $(0,1.5),(3.5,5.5),(8,11),(12,13.5)sec$,
moving toward the origin $(1.5,3.5),(5.5,8),(11,12),(13.5,16)sec$.
(b) $t=0,3.5,12,16sec$.
(c) $t=1.5,5.5,8,11,13.5sec$.
(d) positive $(0,1.5),(5.5,8),(11,13.5)sec$, negative $(1.5,5.5),(8,11),(13.5,16)sec$.
(These are approximate values read from the graph.)
Work Step by Step
With on the given curve, we can estimate the velocity $v(t)=s'(t)$ and acceleration $a(t)=s''(t)$ as shown in the figure.
(a) The object is moving away from the origin when $s\gt0, v\gt0$ or $s\lt0, v\lt0$ (same signs) which gives time intervals of $(0,1.5),(3.5,5.5),(8,11),(12,13.5)sec$, and is moving toward the origin when $s\gt0,v\lt0$ or $s\lt0, v\gt0$ (opposite signs) which gives time intervals of $(1.5,3.5),(5.5,8),(11,12),(13.5,16)sec$.
(b) The velocity equals to zero when the curve reaches extrema, which happens at times of $t=0,3.5,12,16sec$.
(c) The acceleration equals to zero at times when there is an inflection point, which gives $t=1.5,5.5,8,11,13.5sec$.
(d) The acceleration is positive when the curve is concave up, which gives regions of $(0,1.5),(5.5,8),(11,13.5)sec$, The acceleration is negative when the curve is concave down, which gives regions of $(1.5,5.5),(8,11),(13.5,16)sec$.
Please note the numbers can be a little different if you read the graph more accurately.
