The figure shows the position-time (x-t) graph of one-dimensional...

From the graph, it is a straight line so, it represents uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph. Initial velocity, (v_1=\frac{2}{2}=1 \mathrm{~ms}^{-1}) Final velocity, (v_2=\frac{-2}{2}=-1 \mathrm{~ms}^{-1}) $$\begin{aligned} & & \mathrm{p}_i & =m \mathrm{v}_1=0.4 \mathrm{~N}-\mathrm{s} \\ & & \mathrm{p}_f & =m \mathrm{v}_2=-0.4 \mathrm{~N}-\mathrm{s} \\ & & \mathrm{J} & =\mathrm{p}_f-\mathrm{p}_i=-0.4-0.4=-0.8 \mathrm{~N}-\mathrm{s} \\ \therefore & & |\mathrm{J}| & =0.8 \mathrm{~N}-\mathrm{s} \end{aligned}$$)$