The velocity $$(v)$$ and time $$(t)$$ graph of a body in a straight line motion is shown in the figure. The point $$S$$ is at $$4.333$$ seconds. The total distance covered by the body in $$6$$ s is:

Area above the time axis (trapezoid from $$t = 0$$ to $$t = \frac{13}{3}$$):

$$\text{Area}_1 = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times \left[\frac{13}{3} + (3 - 2)\right] \times 4 = 2 \times \left(\frac{13}{3} + 1\right) = \frac{32}{3}\text{ m}$$

Area below the time axis (triangle from $$t = \frac{13}{3}$$ to $$t = 6$$):

$$\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{5}{3} \times 2 = \frac{5}{3}\text{ m}$$

Total distance covered: $$d = \text{Area}_1 + \text{Area}_2 = \frac{32}{3} + \frac{5}{3} = \frac{37}{3}\text{ m}$$