Answer
(a) See the graph.
(b) $v_{av}=4~m/s$
(c) $average~speed=4~m/s$
Work Step by Step
(a) See the graph.
(b) $x=(6~m/s)t+(-2~m/s^{2})t^{2}$
For $t=0$:
$x=(6~m/s)0+(-2~m/s^{2})0^{2}=0$
For $t=1~s$:
$x=(6~m/s)(1~s)+(-2~m/s^{2})(1~s)^{2}=6~m-2~m=4~m$
$v_{av}=\frac{x_{1}-x_{0}}{t_{1}-t_{0}}=\frac{4~m-0}{1~s-0}=4~m/s$
(c) Let's find the vertex of the parabola:
$h=-\frac{b}{2a}=-\frac{6~m/s}{2\times(-2~m/s^{2})}=1.5~s$. The vertex is not between $t=0$ and $t=1~s$
The particle moves 4 m in the positive direction (from $t=0$ to $t=1~s$). So, the total distance traveled is equal to 4 m. Now, let's find the average speed:
$average~speed=\frac{distance}{elapsed~time}=\frac{4~m}{1~s}=4~m/s$
