Answer
a. $t=0.71, 3.3$ sec.
b. $v\lt0$ during $0\leq t\lt0.71$ and $3.3\lt t\leq4$ sec, $v\gt0$ during $0.71\lt t\lt 3.3$ sec.
c. $t=0.71, 3.3$ sec.
d. $|v|$ increasing during $0.71\lt t\lt 2$ and $3.3\lt t\leq 4$ sec, $|v|$ decreasing during $0\leq t\lt 0.71$ and $2\lt t\lt 3.3$ sec,
e. $|v|=7$) at $t=0, 4$ sec, $|v|=0$ at $t=0.71, 3.3$ sec.
f. $|s|=10.3$ at $t=3.3$ sec.
Work Step by Step
Given $s(t)=-t^3+6t^2-7t+4$, $0\leq t\leq 4$ , we have $v(t)=-3t^2+12t-7$, $a(t)=-6t+12$ , and we can graph the functions as shown in the figure.
a. The object will be momentarily at rest when its velocity is zero. With $v(t)=0$, we have $t=0.71, 3.3$ sec.
b. It moves left ($v\lt0$) during $0\leq t\lt0.71$ and $3.3\lt t\leq4$ sec, and it moves right ($v\gt0$) during $0.71\lt t\lt 3.3$ sec.
c. It changes direction when its velocity crosses with the x-axis, so we have $t=0.71, 3.3$ sec.
d. It speeds up ($|v|$ increasing) during $0.71\lt t\lt 2$ and $3.3\lt t\leq 4$ sec, and it slows down ($|v|$ decreasing) during $0\leq t\lt 0.71$ and $2\lt t\lt 3.3$ sec,
e. It is moving fastest (highest speed $|v|=7$) at the beginning and the end $t=0, 4$ sec, and it is moving slowest ($|v|=0$) at $t=0.71, 3.3$ sec.
f. It is farthest from the axis origin $|s|=10.3$ at $t=3.3$ sec.
