Displacement Vector and resultant vector

Question

I am struggling with the concept of displacement. From my understanding displacement can be found for 1D motion along the x-axis as $\Delta x= x_{f}-x_{i}$. For example someone walks $1\,\mathrm{m}$ to the left then $1\,\mathrm{m}$ to the right then $2\,\mathrm{m}$ to the left and finally $3\,\mathrm{m}$ to the right. I know that I can find the net displacement by adding them all up as $-1+1-2+3$ and get $1\;\mathrm{m}$ to the right as my resultant displacement. My question is why can't I use $\Delta x$ and use the $3$ as my final position and $0$ as my initial position? I know that it clearly doesn't give me the correct answer but why does that not work?

Answer 1

Responding to your question and follow up comment, displacement is indeed final position minus initial position. But in your example, the numbers you are listing are in fact displacements, not positions.

Let us number your positions as follows:

\begin{array} {r|r} \hline x_0 & 0 \\ x_1 & -1 \\ x_2 & 0 \\ x_3 & -2 \\ x_4 & 1 \\ \end{array}

Those are the positions after each step.

The displacements after each step are given by $\Delta x = x_{current} - x_{previous}$, or in other words:

\begin{array} {r|r} \hline \Delta x_1 & -1 \\ \Delta x_2 & 1 \\ \Delta x_3 & -2 \\ \Delta x_4 & 3 \\ \end{array}

Which are the numbers you presented. You can easily verify that the current position is the cumulative sum of all previous displacements, and that any given displacement is the difference between the current and previous positions. The total displacement is $x_4-x_0 = 1$, but subtracting the individual displacements like $\Delta x_4 - \Delta x_1$ does not give anything meaningful.