Answer
a. $t=6.25$ sec.
b. down $6.25\lt t\leq 12.5$ sec, up $0\leq t\lt 6.25$ sec,
c. $t=6.25$ sec.
d. speed up $6.25\lt t\leq 12.5$ sec, slows down $0\leq t\lt 6.25$ sec,
e. 200ft/s, $t=0, 12.5$ sec, v=0 $t=6.25$ sec.
f. 625ft at $t=6.25$ sec.
Work Step by Step
Given $s=200t-16t^2$ ft, we have $v=200-32t$ ft/s, $a=-32ft/s^2$ (negative means downward), where $0\leq t \leq 12.5$, 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 $t=6.25$ sec.
b. It moves down during $6.25\lt t\leq 12.5$ sec, and it moves up during $0\leq t\lt 6.25$ sec,
c. It changes direction when its velocity crosses with the x-axis, so we have $t=6.25$ sec.
d. It speeds up during fall down $6.25\lt t\leq 12.5$ sec, and it slows down during moving up $0\leq t\lt 6.25$ sec,
e.It is moving fastest (highest speed 200ft/s) at the beginning and at the end $t=0, 12.5$ sec, and it is moving slowest (v=0) at the highest point $t=6.25$ sec.
f. It is farthest from the axis origin at the highest point of 625ft at $t=6.25$ sec.
