Illustration of Distance and Displacement
If we use the same situation as earlier we can explore the concepts in more detail. Consider our description of the location of the houses, school and the shop. Komal walks to meet Kevin at his house before walking to school. What is Komal's displacement ...
Illustration of Distance and Displacement
If we use the same situation as earlier in the previous lesson, we can explore the concepts in more detail. Consider our description of the location of the houses, school and the shop.
Komal walks to meet Kevin at his house before walking to school. What is Komal's displacement and what distance did he cover if he walks to school via Kevin's house?
Komal covers a distance of \(\text{400}\) \(\text{m}\) to Kevin's house and another \(\text{500}\) \(\text{m}\) from Kevin's house to the school. He covers a total distance of \(\text{900}\) \(\text{m}\). His displacement, however, is only \(\text{100}\) \(\text{m}\) towards the school. This is because displacement only looks at the starting position (his house) and the end position (the school). It does not depend on the path he travelled.
To calculate his distance and displacement, we need to choose a reference point and a direction. Let's choose Komal's house as the reference point, and towards Kevin's house as the positive direction (which means that towards the school is negative). We would do the calculations as follows:
\begin{align*} \text{Distance } (D) & = \text{path travelled } \\ & = \text{400}\text{ m} + \text{500}\text{ m} \\ & = \text{900}\text{ m} \end{align*}\begin{align*} \text{Displacement } \left(\Delta \vec{x}\right) & = \vec{x}_{f} - \vec{x}_{i} \\ & = -\text{100}\text{ m} + \text{0}\text{ m} \\ & = -\text{100}\text{ m} \\ & = \text{100}\text{ m} \text{ (in the negative } x \text{ direction)} \end{align*}Very often in calculations you will get a negative answer. For example, Komal's displacement in the example above, is calculated as \(-\text{100}\) \(\text{m}\). The minus sign in front of the answer means that his displacement is \(\text{100}\) \(\text{m}\) in the opposite direction (opposite to the direction chosen as positive in the beginning of the question). When we start a calculation we choose a frame of reference and a positive direction. In the first example above, the reference point is Komal's house and the positive direction is towards Kevin's house. Therefore Komal's displacement is \(\text{100}\) \(\text{m}\) towards the school. Notice that distance has no direction, but displacement has a direction.
Kevin walks to school with Komal and after school walks back home. What is Kevin's displacement and what distance did he cover? For this calculation we use Kevin's house as the reference point. Let's take towards the school as the positive direction.
\begin{align*} \text{Distance } (D) & = \text{ path travelled} \\ & = \text{500}\text{ m} + \text{500}\text{ m} \\ & = \text{1 000}\text{ m} \end{align*}\begin{align*} \text{Displacement } \left(\Delta \vec{x}\right) & = \vec{x}_{f} - \vec{x}_{i} \\ & = \text{0}\text{ m} + \text{0}\text{ m} \\ & = \text{0}\text{ m} \end{align*}It is possible to have a displacement of \(\text{0}\) \(\text{m}\) and a distance that is not \(\text{0}\) \(\text{m}\). This happens whenever you end at the same point you started.
Differences Between Distance and Displacement
The differences between distance and displacement can be summarised as:
Distance |
Displacement |
|
1. depends on the path |
1. independent of path taken |
|
2. always positive |
2. can be positive or negative |
|
3. is a scalar |
3. is a vector |
This lesson is part of:
One-Dimensional Motion