Section 2.9 Chapter Summary
Position of an object is described with respect to a reference point in space, which is usually taken to be the origin of a coordinate system. In one-dimensional motion along \(x\)-axis of a Cartesian coordinate system, the position of a particle is its \(x\)-coordinate. The position of a particle is a function of time: \(x(t)\text{.}\) The change in position over a time interval is called displacement over that interval. For instance, if the positions of a particle are \(x_1\) and \(x_2\) at times \(t_1\) and \(t_2\) respectively, then the displacement during time \(t_1\) to \(t_2\) is given by \(x_2-x_1\text{.}\) The displacement of motion along an axis can be positive or negative depending upon whether \(x_2\) is to the right of \(x_1\) or to the left of \(x_1\text{.}\)
The average \(x\)-velocity is the ratio of \(x\)-displacement and the time interval.
\begin{equation*}
v_x^{ave} = \frac{x_2-x_1}{t_2-t_1}.
\end{equation*}
The average velocity during an interval will be positive or negative depending upon the displacement. The average speed is the ratio of total distance traveled during an interval and the time interval. Since average speed depends on total distance, it is always positive. If an object turns around during its motion, then the average speed will differ from the magnitude of the average velocity.
The instantaneous velocity is the rate of change of position at a particular instant in time. For motion along \(x\)-axis, it is equal to the slope of \(x\) vs \(t\) plot. The slope of \(x\) vs \(t\) is also equal to the derivative of \(x(t)\) with respect to \(t\text{.}\) The instantaneous speed is equal to the magnitude of instantaneous velocity.
\begin{align*}
\end{align*}
The average acceleration is defined as the change in velocity divided by the time interval. Thus, if the instantaneous velocities of an object are \(v_1\) and \(v_2\) at times \(t_1\) and \(t_2\) respectively. Then average acceleration during \(t_1\) to \(t_2\) is
\begin{equation*}
a_x^{ave} = \frac{v_2-v_1}{t_2-t_1}.
\end{equation*}
The instantaneous acceleration is the rate of change of instantaneous velocity at a particular instant in time.
\begin{equation*}
a_x = \frac{dv_x}{dt} = \text{slope of v vs t plot}.
\end{equation*}
Constant acceleration motion provides a particularly useful case. The changes in velocity and position over a time interval \(t = 0\) to \(t = t\) are given by the following equations where subscript \(x\) are dropped for the sake of brevity.
| 1. \(v=v_0 + a t \) |
| 2. \(x=x_0+v_0t + \frac{1}{2}a t^2 \) |
| 3. \(v^2=v_0^2+2a\left(x-x_0\right)\) |
Free fall or free rise is an example of constant acceleration motion, where the acceleration is equal to \(g\) or \(9.81 m/s^2\) and pointed towards the center of earth. The vertical motion (up or down) is usually described by selecting y-axis pointed up. This turns the constant acceleration of free fall or rise equal to minus g, i.e. in the constant acceleration equations.
| 1. \(v=v_0 - g t \) |
| 2. \(x=x_0+v_0t - \frac{1}{2}g t^2 \) |
| 3. \(v^2=v_0^2-2g\left(x-x_0\right)\) |
Variable accelerations are described by approximating the variable acceleration by a series of constant accelerations in infinitesimally time intervals. We find that the change in velocity \(v_2-v_1\) over a time interval \(t_1\) to \(t_2\) equals the area under the \(a\) vs \(t\) curve, which is exactly equal to the definite integral of \(a(t)\) with respect .
\begin{equation*}
v_2-v_1 = \sum_{i=1}^Na_i\Delta t = \text{Area under a(t) vs t curve} = \int_{t_a}^{t_2} a(t) dt.
\end{equation*}
