Answer
a. $-6m/s^2$ and $6m/s^2$
b. $3m/s$
c. $6m$
Work Step by Step
a. Given $s(t)=t^3-6t^2+9t$ and $0\leq t\leq3$, we have
velocity $v(t)=s'(t)=3t^2-12t+9$ and acceleration $a(t)=v'(t)=6t-12$.
When $v(t)=0$, we have $3t^2-12t+9=0$ and $t^2-4t+3=0$ which leads to $t=1, 3$.
At these values, $a(1)=6(1)-12=-6m/s^2$ and $a(3)=6(3)-12=6m/s^2$
b. Letting $a(t)=0$, we have $6t-12=0$; thus $t=2$ and the speed $|v(2)|=|3(2)^2-12(2)+9|=3m/s$
c. At $t=0$, $s(0)=0, v(0)=9m/s$, the object was moving to the right. At $t=1$, $s(1)=4m, v(1)=0m/s$ and the object was not moving, but will change direction. At $t=2$, $s(2)=2^3-6(2)^2+9(2)=2m, v(2)=-3m/s$ the object was moving to the left. The total distance traveled by the body in $0\leq t\leq 2$ is given by $|s(1)-s(0)|+|s(2)-s(1)|=4+2=6m$
