Using the tail-to-head method and a scale of 1 cm represents 10 N, draw an accurate vector diagram to determine the magnitude of the resultant force acting on the box.

Jack starts at point A.
Jack's movement ends at point C.

Let point A be the origin, so its position is $x_A = 0 \text{ m}$.
Point B is $70 \text{ m}$ East of A. We define East as the positive direction.
So, the position of B is $x_B = +70 \text{ m}$.
Jack runs from B to C due West. The problem states that the distance between point B and point C is $70 \text{ m}$. Since West is the negative direction:
The position of C is $x_C = x_B - 70 \text{ m}$.

Displacement ($\Delta x$) is the change in position from the initial point to the final point.

Note: The information about Jack's velocity ($5 \text{ m}\cdot\text{s}^{-1}$) and time ($10 \text{ s}$) for the B to C segment would imply a distance of $5 \text{ m/s} \times 10 \text{ s} = 50 \text{ m}$. However, the problem explicitly states "The distance between point B and point C is 70 m". We use the explicitly stated distance to define the positions of the points.

Jack's displacement is $\text{0 m}$.