1. Use area under the $v$–$t$ graph

The position (displacement from origin) at time $t=5\text{ s}$ is the area under the velocity–time graph from $t=0$ to $t=5$:

$x(5)=\int_0^5 v\,dt$

2. Read the graph in parts

From the given graph, the motion is piecewise:

• From $t=0$ to $t=2\text{ s}$: velocity increases linearly from $0$ to $4\text{ m/s}$.
• From $t=2$ to $t=3\text{ s}$: velocity is constant at $4\text{ m/s}$.
• From $t=3$ to $t=5\text{ s}$: velocity decreases linearly from $4\text{ m/s}$ to $0$.

So the displacement is the sum of areas:

• left triangle,
• rectangle,
• right triangle.

3. Calculate each area

(i) Triangle from $0$ to $2$ s

$A_1=\frac12 \times 2 \times 4=4\text{ m}$

(ii) Rectangle from $2$ to $3$ s

$A_2=1 \times 4=4\text{ m}$

(iii) Triangle from $3$ to $5$ s

$A_3=\frac12 \times 2 \times 4=4\text{ m}$

4. Total displacement

$x(5)=A_1+A_2+A_3=4+4+4=12\text{ m}$

5. Compare with options

The computed position is

$\boxed{12\text{ m}}$

This does not match any listed option. Hence the stored answer $\text{B: }9\text{ m}$ appears inconsistent with the graph interpretation above.