5.3: Displacement
After having discussed how the position of a point mass \(i\) can be described in space using coordinate systems and vectors, we now turn to how its position evolves in time, and introduce the important concepts of displacement, velocity
and acceleration that are essential for analysing dynamics. For this purpose we compare the position \(\overrightarrow{\boldsymbol{r}}_{i}\left(t_{1}\right)\) at a first time \(t_{1}\) to \(\overrightarrow{\boldsymbol{r}}_{i}\left(t_{2}\right)\) at a later time \(t_{2}=t_{1}+\Delta t\), with \(\Delta t>0\). The difference between these vectors, the change in position vector, is the displacement vector as can be seen in Figure 5.1.
The displacement vector \(\Delta \overrightarrow{\boldsymbol{r}}_{i, 12}\) describes the displacement in space of a point mass between times \(t_{1}\) and \(t_{2}\).
\[\Delta \overrightarrow{\boldsymbol{r}}_{i, 12}=\overrightarrow{\boldsymbol{r}}_{i}\left(t_{2}\right)-\overrightarrow{\boldsymbol{r}}_{i}\left(t_{1}\right) \tag{5.15} \label{5.15}\]
The initial and final position are separated by a displaced distance \(\Delta r_{i, 12}=\) \(\left|\Delta \overrightarrow{\boldsymbol{r}}_{i, 12}\right|\). The displaced path distance \(\Delta s_{i, 12}\) of the mass, which is the displaced distance as measured along the path (instead of along a straight line) can be determined by the equation:
\[\Delta s_{i, 12}=s_{i}\left(t_{2}\right)-s_{i}\left(t_{1}\right) \tag{5.16} \label{5.16}\]
when the trajectory of the mass is parametrised by a path curve as discussed in the previous section 5.2.3.
The total travelled distance \(\Delta s_{\mathrm{T}, i, 12}\) is not the distance between the initial and final positions, but depends on the path taken. The travelled distance can be larger than the displaced distance \(\Delta s_{i, 12} \mid\) if the mass reverses direction on its path and travels back and forth along the same path curve in multiple segments, in which case the distances \(\left|\Delta s_{i, 12}\right|\) along each of these segments (with different sign \(\Delta s_{i, n m}\) ) should be summed up to determine \(\Delta s_{\mathrm{T}, i, 12}\).
If the path is a straight line, the displaced path distance \(\left|\Delta s_{i, 12}\right|\) is equal to the displaced distance \(\left|\Delta \overrightarrow{\boldsymbol{r}}_{i, 12}\right|\), but when the path is curved, the displaced path distance is always longer, because a straight line is the shortest distance between two points: \(\left|\Delta s_{i, 12}\right| \geq\left|\Delta \overrightarrow{\boldsymbol{r}}_{i, 12}\right|\).
The infinitesimal displacement vector that corresponds to an infinitesimally small time difference \(\mathrm{d} t\) with \(t_{1}=t\) and \(t_{2}=t+\mathrm{d} t\) is found to be:
\[\mathrm{d} \overrightarrow{\boldsymbol{r}}_{i}=\overrightarrow{\boldsymbol{r}}_{i}(t+\mathrm{d} t)-\overrightarrow{\boldsymbol{r}}_{i}(t) \tag{5.17} \label{5.17}\]
The closer you look at a small segment of a smooth trajectory, the more it resembles a straight line. This can be seen in Figure 5.1 by comparing the dashed line and the straight arrow between \(\overrightarrow{\boldsymbol{r}}_{i}(6 \mathrm{~s})\) and \(\overrightarrow{\boldsymbol{r}}_{i}(7 \mathrm{~s})\). Therefore the infinitesimally travelled distance \(\mathrm{d} s_{i}=s_{i}(t+\mathrm{d} t)-s_{i}(t)\) becomes equal to the length of the infinitesimal displacement vector: \(\left|\mathrm{d} s_{i}\right|=\left|\mathrm{d} \overrightarrow{\boldsymbol{r}}_{i}\right|\).