Answer
a) The body's displacement is $0m$ and the average velocity is $0m/s$.
b) At $t=0$: speed $=6m/s$, acceleration $=-2m/s^2$.
At $t=6$: speed $=6m/s$, acceleration $=-2m/s^2$.
c) The body changes direction at $t=3s$.
Work Step by Step
$$s(t)=6t-t^2$$
a) The body's displacement, $\Delta s$, over the interval $[0,6]$ is $$\Delta s=s(6)-s(0)=(6\times6-6^2)-(6\times0-0^2)=0-0=0m$$
The average velocity, $v_a$, of the body over the interval $[0,6]$ is $$v_a=\frac{\Delta s}{6-0}=\frac{0}{6}=0m/s$$
b) We first look for the velocity of the body at time $t$, $v(t)$, and the acceleration of the body at time $t$, $a(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}(6t-t^2)=6-2t$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}(6-2t)=-2$$
We now examine the speed and acceleration at 2 endpoints of $[0,6]$:
- At $t=0$: $$v(0)=6-2\times0=6m/s$$
So the speed of the body is $|v(0)|=6m/s$
$$a(0)=-2m/s^2$$
- At $t=6$: $$v(6)=6-2\times6=-6m/s$$
So the speed of the body is $|v(6)|=6m/s$
$$a(6)=-2m/s^2$$
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. To find where $v(t)$ changes sign, we first locate the point where $v(t)=0$ then examine the sign before and after that point to determine whether $v(t)$ changes sign there.
$$v(t)=0$$ $$6-2t=0$$ $$t=3s$$
On $[0,3)$ $v(t)\gt0$ and on $(3,6]$, $v(t)\lt0$. Therefore, there is a change of sign in $v(t)$ at $t=3s$, which also means the body has changed direction.