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.
$$F_x=ma_x=0.5a_x$$
From the given figure, it is obvious that $a_x$ is not constant all the time.
So, the force versus time graph is given by
- From $t=0$ s to $t=1$ s, $a_x=1$ m/s$^2$, and hence
$$F_x=0.5\cdot1=0.5\;\rm N$$
$\rm \color{red}{(0\;s,0.5\;N)}$, $\rm \color{red}{(1\;s,0.5\;N)}$
- From $t=1$ s to $t=2$ s, the acceleration is not constant, so we need to find it at the end of the line. At $t=2$ s
$a_x=-0.5$ m/s$^2$, and hence
$$F_x=0.5\cdot-0.5=-0.25\;\rm N$$
$\rm \color{red}{(2\;s,-0.25\;N)}$, $\rm \color{red}{(3\;s,-0.25\;N)}$
From $t=3$ s to $t=4$ s, the acceleration is not constant, so we need to find it at the end of the line. At $t=4$ s
$a_x=0$ m/s$^2$, and hence
$$F_x=0.5\cdot0 =0\;\rm N$$
$\rm \color{red}{(4\;s,0\;N)}$
Plugging all these dots into a force versus time graph, as you see below.