Answer
a. $-2m/s^2$ and $2m/s^2$
b. $0\leq t\lt1$ or $ t\gt3$ sec. $1\lt t\lt 3$ sec.
c. $t\gt2$ sec. $0\leq t\lt2$ sec.
Work Step by Step
a. Given $v(t)=t^2-4t+3$ and $ t\geq0$, we have acceleration $a(t)=v'(t)=2t-4$.
When $v(t)=0$, we have $t^2-4t+3=0$, which leads to $t=1, 3$.
At these values, $a(1)=2(1)-4=-2m/s^2$ and $a(3)=2(3)-4=2m/s^2$
b. Assume when $v\gt0$, the body is moving forward; we have $t^2-4t+3\gt0$ or $(x-1)(x-3)\gt0$ which leads to the solution of $0\leq t\lt1$ or $ t\gt3$ sec.
When it is moving backward $v\lt0$, we can get $1\lt t\lt 3$ sec.
c. When the velocity is increasing, we have $a\gt0$; thus $2t-4\gt0$ which leads to $t\gt2$ sec.
When the velocity is decreasing, we have $a\lt0$; thus $2t-4\lt0$ which leads to $0\leq t\lt2$ sec.
