A particle starts from point A and travels along the solid curve...

The direction of average velocity between points $A$ and $B$ is along the secant line connecting $A$ and $B$.

The direction of instantaneous velocity at point $B$ is along the tangent to the curve at $B$.

For these two velocities to have the same direction, the secant line passing through $A$ and $B$ must be tangent to the curve at point $B$.

From the given graph, the coordinates of the starting point $A$ are approximately $(1 \text{ m}, 4 \text{ m})$.

We can evaluate the given options by checking if the point lies on the curve and if the secant line is tangent to the curve at that point:

For option (1) $B(3 \text{ m}, 5 \text{ m})$: This point does not lie on the curve shown in the graph.

For option (2) $B(4 \text{ m}, 2 \text{ m})$: The slope of the secant line is $m = \dfrac{2 - 4}{4 - 1} = -\dfrac{2}{3}$. However, at $x = 4 \text{ m}$, the curve is rising, so the tangent has a positive slope. This does not match.

For option (4) $B(2 \text{ m}, 1 \text{ m})$: The slope of the secant line is $m = \dfrac{1 - 4}{2 - 1} = -3$. The curve is concave up in this region, meaning any tangent line will lie below the curve and cannot pass through point $A$.

For option (3) $B(5 \text{ m}, 3 \text{ m})$: The slope of the secant line is $m = \dfrac{3 - 4}{5 - 1} = -\dfrac{1}{4} = -0.25$. Looking at the graph, the point $x = 5 \text{ m}$ is slightly past the local maximum, where the curve is sloping downwards. A tangent line with a gentle negative slope of $-0.25$ perfectly matches the visual representation and passes through $A(1, 4)$.

Thus, the correct position is $x = 5 \text{ m}, y = 3 \text{ m}$.

Answer: $x = 5 \text{ m}, y = 3 \text{ m}$