Hello
We must take as starting point in time [tex]t_i = 12~s[/tex], to which corresponds a distance of [tex]S_i = 200~m[/tex]. The final point is instead [tex]t_f=20~s[/tex], to which corresponds a distance of [tex]S_f=600~m[/tex].
In this case, the motion is a straight line in the space-time graph: this means the object is moving by uniform motion. Therefore, the speed is constant and it is equal to the slope of the straight line. So, if we take the initial and final point we mentioned above, the velocity is
[tex]v= \frac{\Delta S}{\Delta t}= \frac{S_f-S_i}{t_f-t_i}= \frac{(600-200)~m}{(20-12)~s}=50~m/s [/tex]
