Answer
See the graph below.
Work Step by Step
We know, from Newton's second law, that
$$\sum F_x=ma_x$$
And in this case, there is only one force $F_x$ exerted on a 500-g (0.5 kg) object.
$$F_x=ma_x=0.5a_x$$
Thus,
$$a_x=2F_x$$
From the given figure, it is obvious that $F_x$ is not constant all the time.
So, the acceleration versus time graph is given by
- At $t=0$ s to $t=1$ s, $F_x=0$ N, and hence
$$a_x=2\cdot 0 =0\;\rm m/s^2$$
$\rm \color{red}{(0\;s,0\;m/s^2)}$, $\rm \color{red}{(1\;s,0\;m/s^2)}$
- At $t=2$ s, $F_x=-0.5$ N, and hence
$$a_x=2\cdot -0.5 =-1\;\rm m/s^2$$
$\rm \color{red}{(2\;s,-1\;m/s^2)}$
- At $t=3$ s, $F_x= 0.5$ N, and hence
$$a_x=2\cdot 0.5 = 1\;\rm m/s^2$$
$\rm \color{red}{(3\;s, 1\;m/s^2)}$
- At $t=4$ s, $F_x= 1.5$ N, and hence
$$a_x=2\cdot 1.5 = 3\;\rm m/s^2$$
$\rm \color{red}{(4\;s, 3\;m/s^2)}$
Plugging all these dots into the acceleration versus time graph, as you see below.