The velocity-time graph of a linear motion is shown in figure. The distance from the starting point after 8 sec is

Distance from the starting point is equal to displacement.

Displacement = area of v-t graph

                            = $\frac{1}{2}[4+2]4-\frac{1}{2}[4+3]\times 2 = 5 \mathrm{m}$

Note: Here distance travelled is not asked, but the distance from the starting point.