2-1 Position, Displacement, and Dis

2-1 Position, Displacement, and Distance
In describing an object’s motion, we should first talk about position – where is the object? A position is a vector because it has both a magnitude and a direction: it is some distance from a zero point (the point we call the origin) in a particular direction. With one-dimensional motion, we can define a straight line along which the object moves. Let’s call this the x-axis, and represent different locations on the x-axis using variables such as and , as in Figure 2.1.
Figure 2.1: Positions = +3 m and = –2 m, where the + and – signs indicate the direction.
If an object moves from one position to another we say it experiences a displacement.
In Figure 2.1, we defined the positions = +3 m and = –2 m. What is the displacement in moving from position to position ? Applying Equation 2.1 gives
. This method of adding vectors to obtain the displacement is shown in Figure 2.2. Note that the negative sign comes from the fact that the displacement is
directed left, and we have defined the positive x-direction as pointing to the right.
Figure 2.2: The displacement is –5 m when moving from position to position . Equation 2.1,
Displacement: a vector representing a change in position. A displacement is measured in length units, so the MKS unit for displacement is the meter (m).
We generally use the Greek letter capital delta (!) to represent a change. If the initial position is and the final position is we can express the displacement as:
. (Equation 2.1: Displacement in one dimension)
the displacement equation, tells us that the displacement is
arrow on the axis is the displacement, the vector sum of the vector
, as in the figure. The bold and the vector .
To determine the displacement of an object, you only have to consider the change in position between the starting point and the ending point. The path followed from one point to the other does not matter. For instance, let’s say you start at and you then have a displacement of
8 meters to the left followed by a seco
nd displacement of 3 meters right. You again end up at ,
as shown in Figure 2.4. The total distance traveled is the sum of the magnitudes of the individual displacements, 8 m + 3 m = 11 m. The net displacement (the vector sum of the individual
displacements), however, is still 5 meters to the left: . Chapter 2 – Motion in One Dimension Page 2 - 2