Stationary Objects
The simplest motion that we can come across is that of a stationary object. A stationary object does not move and so its position does not change. Consider an example, Vivian is waiting for a taxi. She is standing two metres from a stop street at t = 0 s. After one ...
Description of Motion
The purpose of this lesson and the next set of lessons is to describe motion, and now that we understand the definitions of displacement, distance, velocity, speed and acceleration, we are ready to start using these ideas to describe how an object or person is moving. We will look at three ways of describing motion:
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words
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diagrams
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graphs
These methods will be described in this section.
We will consider three types of motion: when the object is not moving (stationary object), when the object is moving at a constant velocity (uniform motion) and when the object is moving at a constant acceleration (motion at constant acceleration).
Stationary Objects
The simplest motion that we can come across is that of a stationary object. A stationary object does not move and so its position does not change.
Consider an example, Vivian is waiting for a taxi. She is standing two metres from a stop street at \(t = \text{0}\text{ s}\). After one minute, \(t = \text{60}\text{ s}\), she is still \(\text{2}\) metres from the stop street and after two minutes, at \(t = \text{120}\text{ s}\), she is also \(\text{2}\) metres from the stop street. Her position has not changed. Her displacement is zero (because his position is the same), her velocity is zero (because his displacement is zero) and her acceleration is also zero (because her velocity is not changing).
We can now draw graphs of position vs. time (\(\vec{x}\) vs. \(t\)), velocity vs. time (\(\vec{v}\) vs. \(t\)) and acceleration vs. time (\(\vec{a}\) vs. \(t\)) for a stationary object. The graphs are shown below.
Vivian stands at a stop sign.
Graphs for a stationary object (a) position vs. time (b) velocity vs. time (c) acceleration vs. time.
Vivian's position is \(\text{2}\) metres in the positive direction from the stop street. If the stop street is taken as the reference point, her position remains at \(\text{2}\) metres for \(\text{120}\) seconds. The graph is a horizontal line at \(\text{2}\) \(\text{m}\). The velocity and acceleration graphs are also shown. They are both horizontal lines on the \(x\)-axis. Since her position is not changing, her velocity is \(\text{0}\) \(\text{m·s$^{-1}$}\) and since velocity is not changing, acceleration is \(\text{0}\) \(\text{m·s$^{-2}$}\).
Definition: Gradient
(Recall from Mathematics) The gradient, \(m\), of a line cna be calculated by dividing the change in the \(y\)-value (dependent variable) by the change in the \(x\)-value (independent variable). \(m=\frac{\Delta y}{\Delta x}\)
Since we know that velocity is the rate of change of position, we can confirm the value for the velocity vs. time graph, by calculating the gradient of the \(\vec{x}\) vs. \(t\) graph.
Tip:
The gradient of a position vs. time graph gives the average velocity, while the tangent of a position vs. time graph gives the instantaneous velocity.
If we calculate the gradient of the \(\vec{x}\) vs. \(t\) graph for a stationary object we get:
\begin{align*} v & = \frac{\Delta \vec{x}}{\Delta t} \\ & = \frac{\vec{x}_{f} - \vec{x}_{i}}{t_{f} - t_{i}} \\ & = \frac{\text{2}\text{ m} - \text{2}\text{ m}}{\text{120}\text{ s} - \text{60}\text{ s}} \text{ (initial position } = \text{ final position)} \\ & = \text{0}\text{ m·s$^{-1}$} \text{ (for the time that Vivian is stationary)} \end{align*}
Similarly, we can confirm the value of the acceleration by calculating the gradient of the velocity vs. time graph.
Tip:
The gradient of a velocity vs. time graph gives the average acceleration, while the tangent of a velocity vs. time graph gives the instantaneous acceleration.
If we calculate the gradient of the \(\vec{v}\) vs. \(t\) graph for a stationary object we get:
\begin{align*} a & = \frac{\Delta v}{\Delta t} \\ & = \frac{\vec{v}_{f} - \vec{v}_{i}}{t_{f} - t_{i}} \\ & = \frac{\text{0}\text{ m·s$^{-1}$} - \text{0}\text{ m·s$^{-1}$}}{\text{120}\text{ s} - \text{60}\text{ s}} \\ & = \text{0}\text{ m·s$^{-2}$} \end{align*}
Additionally, because the velocity vs. time graph is related to the position vs. time graph, we can use the area under the velocity vs. time graph to calculate the displacement of an object.
Tip:
The area under the velocity vs. time graph gives the displacement.
The displacement of the object is given by the area under the graph, which is \(\text{0}\) \(\text{m}\). This is obvious, because the object is not moving.
This lesson is part of:
One-Dimensional Motion