Answer
a. $-9m$ and $-3m/s$.
b. $3m/s$ and $12m/s$. $6m/s^2$ and $-12m/s^2$.
c. The body does not change its direction.
Work Step by Step
a. Given $s(t)=-t^3+3t^2-3t$ and $0\leq t\leq3$, we have $\Delta t=3s$, $s(0)=0m$ and $s(3)=-3^3+3(3)^2-3(3)=-9m$. Thus the body’s displacement is $\Delta s=s(3)-s(0)=-9m$ and the average velocity for the given time interval is $\bar v=\Delta s/\Delta t=-9/3=-3m/s$.
b. Velocity $v(t)=s'(t)=-3t^2+6t-3$; thus, the speeds at the endpoints are $|v(0)|=|-3|=3m/s$ and $|v(3)|=|-3(3)^2+6(3)-3|=12m/s$.
acceleration $a(t)=v'(t)=-6t+6$ thus $a(0)=6m/s^2$ and $a(3)=-6(3)+6=-12m/s^2$.
c. When the body changes its direction, its velocity will pass $0$ in either direction; letting $v(t)=0$, we have $-3t^2+6t-3=0$ or $t^2-2t+1$, which gives $t=1s$.
The graph of $v(t)=-3t^2+6t-3$ is a parabola with a vertex at t=1s and it is concave down. This implies that for all other values of t (other than t=1s) in the interval ,the velocity of the body is negative (except at t=1s where the velocity is 0).
Thus, the body does not change its direction .
