Answer
a. $0m$, $0m/s$.
b. $6m/s$, $6m/s$, $-2m/s^2$, $-2m/s^2$.
c. $t=3s$
Work Step by Step
a. Given $s(t)=6t-t^2$ and $0\leq t\leq6$, we have $s(0)=0$ and $s(6)=6(6)-6^2=0$. Thus the body’s displacement is $\Delta s=s(6)-s(0)=0m$ and average velocity for the given time interval is $\bar v=\Delta s/\Delta t=0m/s$.
b. Velocity $v(t)=s'(t)=6-2t$, thus the speeds at endpoints are $|v(0)|=|6-0|=6m/s$ and $|v(6)|=|6-2(6)|=6m/s$.
acceleration $a(t)=v'(t)=-2$, thus $a(0)=a(6)=-2m/s^2$.
c. When the body changes its direction, its velocity will pass $0$ in either direction; letting $v(t)=0$, we have $6-2t=0$ and $t=3s$. Thus, the body changed its direction at $t=3s$
