Net displacement of the particle in 10 s,

S = BC – OA = 40 – 60 = – 20 m

Total time taken t = 10 s

$\therefore \text{ Average velocity of the particle }{\mathrm{v}}_{\mathrm{avg}}=\frac{\mathrm{S}}{\mathrm{t}}=\frac{-20}{10}=-2 \mathrm{m}/\mathrm{s}$

Thus magnitude of average velocity of the particle is 2 m/s.