Answer
a. $\frac{9}{4}m$ and$\frac{3}{4}m/s$.
b. $0m/s$ and $6m/s$.
$2m/s^2$ and $11m/s^2$.
c. $t=1s$ and $t=2s$
Work Step by Step
a. Given $s(t)=t^4/4-t^3+t^2$ and $0\leq t\leq3$, we have $\Delta t=3s$, $s(0)=0m$ and $s(3)=3^4/4-3^3+3^2=9/4m$. Thus the body’s displacement is $\Delta s=s(3)-s(0)=\frac{9}{4}m$ and the average velocity for the given time interval is $\bar v=\Delta s/\Delta t=9/12=\frac{3}{4}m/s$.
b. Velocity $v(t)=s'(t)=t^3-3t^2+2t$, thus speeds at the endpoints are $|v(0)|=0m/s$ and $|v(3)|=|3^3-3(3)^2+2(3)|=6m/s$.
Acceleration $a(t)=v'(t)=3t^2-6t+2$, thus $a(0)=2m/s^2$ and $a(3)=3(3)^2-6(3)+2=11m/s^2$.
c. When the body changes its direction, its velocity will pass $0$ in either direction; letting $v(t)=0$, we have $t^3-3t^2+2t=0$; discard $t=0$ as a solution to get $t^2-3t+2=0$, which gives $t=1, 2$. Thus, the body changed its direction at $t=1s$ and $t=2s$
