Answer
a) The body's displacement is $-9m$ and the body's average velocity is $-3m/s$.
b) At $t=0$: speed $=3m/s$, acceleration $=6m/s^2$
At $t=3$: speed $=12m/s$, acceleration $=-12m/s^2$
c) The body does not change direction during the given interval.
Work Step by Step
$$s(t)=-t^3+3t^2-3t$$
a) The body's displacement, $\Delta s$, over the interval $[0,3]$ is $$\Delta s=s(3)-s(0)=(-3^3+3\times3^2-3\times3)-(-0^3+3\times0^2-3\times0)$$ $$\Delta s=-9-0=-9m$$
The average velocity, $v_a$, of the body over the interval $[0,3]$ is $$v_a=\frac{\Delta s}{3-0}=\frac{-9}{3}=-3m/s$$
b) We first look for the velocity of the body at time $t$, $v(t)$, and the acceleration of the body at time $t$, $a(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}(-t^3+3t^2-3t)=-3t^2+6t-3$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}(-3t^2+6t-3)=-6t+6$$
We now examine the speed and acceleration at 2 endpoints of $[0,3]$:
- At $t=0$: $$v(0)=-3\times0^2+6\times0-3=-3m/s$$
So the speed of the body is $|v(0)|=3m/s$
$$a(0)=-6\times0+6=6m/s^2$$
- At $t=3$: $$v(3)=-3\times3^2+6\times3-3=-27+18-3=-12m/s$$
So the speed of the body is $|v(3)|=12m/s$
$$a(3)=-6\times3+6=-12m/s^2$$
c) We know that if
- $v(t)\gt0$: the body moves forward.
- $v(t)\lt0$: the body moves backward.
So the body will change direction when $v(t)$ changes sign.
However, here we have $$v(t)=-3t^2+6t-3=-3(t^2-2t+1)=-3(t-1)^2$$
Since $-3(t-1)^2\le0$ for all $t\in[0,3]$, $v(t)$ cannot change sign in the interval, so as a result, the body will not change direction in the given interval.
