$\begin{array}{l}
\,For\,the\,body\,starting\,from\,rest\\
{x_1} = 0 + \frac{1}{2}a{t^2} \Rightarrow {x_1} = \frac{1}{2}a{t^2}\\
For\,the\,body\,moving\,with\,{\rm{constant}}\,speed\\
{x_2} = vt\\
\therefore \,{x_1} – {x_2} = \frac{1}{2}a{t^2} – \,vt \Rightarrow \frac{{d\left( {{x_1} – {x_2}} \right)}}{{dt}} = at – v\\
at\,\,t = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_1} – {x_2} = 0\\
For\,t < \frac{v}{a};\,the\,slope\,is\,negative\,\\
For\,t = \frac{v}{a};\,the\,slope\,is\,zero\,\\
For\,t > \frac{v}{a};\,the\,slope\,is\,positive
\end{array}$