Answer
a) See the graph below.
b) $v_x=\rm -2\; m/s$
c) $v_x=\rm -2\; m/s$
d) At $x=-2\;\rm m$
e) At $x=2\;\rm m$
f) See the graph below.
Work Step by Step
a)
We can draw this graph by using any software calculator but the author here asks us to plug points by calculating the distance each 0.5 s.
See the graph below.
b)
We draw the slope line at $t=1$ s, as you see in the second figure below. And it seems that the slope is about 2 m/s.
Thus, the velocity at $t=1$ s is
$$v_{x}=\dfrac{x_2-x_1}{t_2-t_1}=\dfrac{-3-(-1)}{2-1}=\color{red}{\bf -2}\;\rm m/s$$
c)
Find the velocity at $t=1$ s by using derivatives.
$$x=t^2-4t+2$$
take the derivative relative to $dt$;
$$\overbrace{ \dfrac{d}{dt}\left(x\right) }^{{\color{blue}{\;v_x}}}=\dfrac{d}{dt}\left(t^2-4t+2\right)$$
$$v_x=2t-4\tag 1$$
At $t=1$ s;
$$v_x=2\cdot 1-4=\color{red}{\bf -2}\;\rm m/s$$
which is the same result we found above by the slope.
d)
There is only one turning point which is at $t=2$ s, as we see in the figures below. At $t=2$ s, the particle will be at
$$x=\color{red}{\bf -2}\;\rm m$$
e)
To find the position of the particle when its velocity is 4 m/s, we need to use equation (1) above and solve for $t$.
And after finding at what time its velocity is 4 m/s, we can use the given formula to find its position.
$$t=\dfrac{v_x+4}{2}=\dfrac{4+4}{2}=\color{green}{4}\;\rm s$$
Hence,
$$x=4^2-(4\cdot 4)+2=\color{red}{\bf 2}\;\rm m$$
f) The motion graph of the particle is shown below.
