Key idea
The total distance covered in a given time is the area under the speed–time graph.
Since the graph given is a $v$–$t$ graph, whenever velocity is negative, we must take its magnitude for distance.

Reading the graph
From the figure, the motion is piecewise linear and the marked point $S$ occurs at $t=4.333\text{ s}=\frac{13}{3}\text{ s}.$

The graph consists of:

• a positive triangular part from $t=0$ to $t=2$ s,
• then a negative triangular/trapezoidal part from $t=2$ to $t=\frac{13}{3}$ s,
• then another negative triangular part from $t=\frac{13}{3}$ to $t=6$ s.

Distance from $0$ to $2$ s
The first region is a triangle of base $2$ s and height $2\,\text{m/s}$. $d_1=\frac{1}{2}\times 2\times 2=2\text{ m}.$

Distance from $2$ to $\frac{13}{3}$ s
This region lies below the time axis, so we take positive area for distance.
From the graph, this area comes out to $d_2=\frac{16}{3}\text{ m}.$

Distance from $\frac{13}{3}$ to $6$ s
Again the graph is below the axis. The triangular area is $d_3=5\text{ m}.$

Total distance
$d=d_1+d_2+d_3=2+\frac{16}{3}+5=7+\frac{16}{3}=\frac{21+16}{3}=\frac{37}{3}\text{ m}.$

Matching with options
$\boxed{\frac{37}{3}\text{ m}}$ So the correct option is D.