Answer
a) The velocity is $0$ at $t=1$ and $t=3$, and $a(1)=-2m/s^2$ and $a(3)=2m/s^2$.
b) The body moves forward on $(-\infty,1)\cup(3,\infty)$ and moves backward on $(1,3)$.
c) The velocity is increasing on $(-\infty,2)$ and decreasing on $(2,\infty)$.
Work Step by Step
$$v(t)=t^2-4t+3=(t-1)(t-3)$$
First, we need to find $a(t)$.
$$a(t)=\frac{dv}{dt}=\frac{d}{dt}(t^2-4t+3)=2t-4$$
a) The velocity is $0$ when $v(t)=0$, which are when either $t=1$ or $t=3$.
- For $t=1$: $a(1)=2\times1-4=-2m/s^2$
- For $t=3$: $a(3)=2\times3-4=2m/s^2$
b) The moving direction of the body corresponds to the sign of $v(t)$. In detail,
- The body moves forward when $v(t)\gt0$.
- The body moves backward when $v(t)\lt0$.
From part a), we learn that $v(t)=0$ either at $t=1$ or $t=3$. Taking these points into account, we examine the following intervals for the sign of $v(t)$:
- On $(-\infty,1)$: both $(t-1)$ and $(t-3)$ are negative, so $v(t)\gt0$. The body moves forward.
- On $(1,3)$: $(t-1)$ is positive while $(t-3)$ is negative, so $v(t)\lt0$. The body moves backward.
- On $(3,\infty)$: both $(t-1)$ and $(t-3)$ are positive, so $v(t)\gt0$. The body moves forward.
c) Whether the velocity increases or decreases corresponds to the sign of the acceleration $a(t)$. In detail,
- The velocity is increasing when $a(t)\gt0$.
- The velocity is decreasing when $a(t)\lt0$.
$a(t)=0$ when $2t-4=0$, or $t=2$. Taking this point into account, we examine the following intervals for the sign of $a(t)$:
- On $(-\infty,2)$: $(2t-4)\lt0$ so $a(t)\lt0$. The velocity is decreasing.
- On $(2,\infty)$: $(2t-4)\gt0$ so $a(t)\gt0$. The velocity is increasing.
