For the velocity time graph shown in figure in time interval from t = 0 to t = 5 s. The ratio of distance to magnitude of displacement is

The total area under the v-t graph will give us the displacement of particle.

Distance is given by the addition of individual areas under the graph.

$\frac{d}{\left|\overrightarrow{s}\right|}=\frac{(\frac{1}{2}\times 4\times 10+\frac{1}{2}\times 1\times 5)}{(\frac{1}{2}\times 4\times 10-\frac{1}{2}\times 1\times 5)}=\frac{22.5}{17.5}=\frac{9}{7}$