Answer
(a) See the graph.
(b) $v_{av}=-2~m/s$
(c) $average~speed=2.16~m/s$
Work Step by Step
(a) See the graph.
(b) $x=(-5~m/s)t+(3~m/s^{2})t^{2}$
For $t=0$:
$x=(-5~m/s)0+(3~m/s^{2})0^{2}=0$
For $t=1~s$:
$x=(-5~m/s)(1~s)+(3~m/s^{2})(1~s)^{2}=-5~m+3~m=-2~m$
$v_{av}=\frac{x_{1}-x_{0}}{t_{1}-t_{0}}=\frac{-2~m-0}{1~s-0}=-2~m/s$
(c) From mathematics:
The equation of a parabola: $y=ax^2+bx+c$
The vertex: (h, k), where $h=-\frac{b}{2a}$ and $k=\frac{4ac-b^{2}}{4a}$
(Let $t_v$ be the time in the vertex of the parabola and $x_v$ the position in the vertex of the parabola).
To physics: ($y~-> x,~x~->t,~h~->t_v,~k~->x_v$)
The equation of a parabola : $x=at^2+bt+c=c+bt+at^2$
The vertex: ($t_v,~x_v$), where $t_v=-\frac{b}{2a}$ and $x_v=\frac{4ac-b^{2}}{4a}$
Notice that the $a$ in the parabola equation does not means acceleration.
Let's find the vertex of the parabola, that is, let's find the moment ($t_v$) when the particle reaches the lowest position ($x_v$).
$t_v=-\frac{b}{2a}=-\frac{-5~m/s}{2\times3~m/s^{2}}=0.83~s$. $t_v$ is between $t=0$ and $t=1~s$
$4ac-b^{2}=4(-2~m/s^{2})0-(-5~m/s)^{2}=-25~m^{2}/s^{2}$
$x_v=\frac{4ac-b^{2}}{4a}=\frac{-25~m^{2}/s^{2}}{4\times3~m/s^{2}}=-2.08~m$
The particle moves 2.08 m in the negative direction (from $t=0$ to $t=0.83~s$). Then it moves 0.08 m in the positive direction (from $t=0.83~s$ to $t=1~s$). So, the total distance traveled is equal to 2.16 m. Now, let's find the average speed:
$average~speed=\frac{distance}{elapsed~time}=\frac{2.16~m}{1~s}=2.16~m/s$
