The velocity (v) versus time (t) plot of a particle is shown in the figure, for a time interval of 40 s. The total distance travelled by the particle and the and the average velocity during this period are, respectively___________, 

From the velocity-time graph:

• Positive triangular region from 0 to 20 s
• Negative triangular region from 20 to 40 s

Distance travelled is total area under the speed-time graph.

First triangle:

$$\text{Area}_1=\frac{1}{2}\times20\times5=50\ \text{m}$$

Second triangle (take magnitude):

$$\text{Area}_2=\frac{1}{2}\times20\times5=50\ \text{m}$$

Total distance:

$$50+50=100\ \text{m}$$

Net displacement:

$$50-50=0$$

Average velocity:

$$v_{\text{avg}}=\frac{\text{displacement}}{\text{total time}}$$

$$v_{\text{avg}}=\frac{0}{40}=0\ \text{m/s}$$