Answer
a. forward $0\leq t\lt1$ and $5\lt t\lt7$.
backward $1\lt t\lt5$.
speed up $1\lt t\lt2$ and $5\lt t\lt6$.
slow down $0\leq t\lt1$, $3\lt t\lt5$, and $6\lt t\lt7$.
b. positive $3\lt t\lt6$, negative $0\leq t\lt2$, $6\lt t\lt7$. zero $2\lt t\lt3$, $7\lt t\lt9$.
c. $t=0$ and $2\leq t\leq 3$
d. $7\leq t\leq 9$
Work Step by Step
a. The particle moved forward when its velocity was positive, which is given in the interval of $0\leq t\lt1$ and $5\lt t\lt7$.
The particle moved backward when its velocity was negative, which was given in the interval of $1\lt t\lt5$.
The particle was speeding up when its speed (absolute value of velocity) was increasing, which was given in the interval of $1\lt t\lt2$ and $5\lt t\lt6$.
The particle was slowing down when its speed (absolute value of velocity) was decreasing, which was given in the interval of $0\leq t\lt1$, $3\lt t\lt5$, and $6\lt t\lt7$.
b. As acceleration $a$ is the derivative of velocity; a positive acceleration will be indicated by a positive slope on the velocity curve. Thus for positive $a$, $3\lt t\lt6$; for negative $a$, $0\leq t\lt2$, $6\lt t\lt7$, and for $a=0$, $2\lt t\lt3$, $7\lt t\lt9$.
c. As speed is the absolute value of the velocity, the maximum speed happened at $t=0$ and $2\leq t\leq 3$
d. The particle would stand still for more than an instant when its speed was zero for an interval; this can be identified as $7\leq t\leq 9$
