Graphical Analysis of One-Dimensional Motion

Graphs of Motion Where Acceleration is Not Constant

Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 2.51. Time again starts at zero, and the initial velocity is 165 m/s. (This was the final velocity of the car in the motion graphed in Figure 2.48). Acceleration gradually decreases from $5.0 \, m/s^{2}$ to zero when the car hits 250 m/s. The velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.

Two line graphs of jet car velocity and acceleration, respectively. First line graph is of velocity over time. Line graph has a positive slope that decreases over time and flattens out at the end. Second line graph is of acceleration over time. Line has a negative slope that increases over time until it flattens out at the end. The line is not smooth, but has several kinks.

Figure 2.51 Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure 2.48 ends. (a) The velocity gradually approaches its top value. The slope of this graph is acceleration. It is plotted in the final graph. (b) Acceleration gradually declines to zero when velocity becomes constant. Notice in each of the three graphs that the acceleration drops down to zero and the velocity levels out. This results in a position-time graph that is almost linear. A close-up of the position time graph would show a slight curvature, as indicated in the velocity graph.

Example 2.19 Calculating Acceleration from a Graph of Velocity versus Time

Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the $v$ vs. $t$ graph in Figure 2.51(a).

Strategy

The slope of the curve at $t=25 \, s$ is equal to the slope of the line tangent at that point, as illustrated in Figure 2.51(a).

Solution

Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, $a$ .

$\begin{gathered} \text { slope }=\frac{\Delta v}{\Delta t}=\frac{(260 \mathrm{~m} / \mathrm{s}-210 \mathrm{~m} / \mathrm{s})}{(51 \mathrm{~s}-1.0 \mathrm{~s})} \\ a=\frac{50 \mathrm{~m} / \mathrm{s}}{50 \mathrm{~s}}=1.0 \mathrm{~m} / \mathrm{s}^{2} \end{gathered}$

Discussion

Note that this value for $a$ is consistent with the value plotted in Figure 2.51(b) at $t=25 \mathrm{~s}$ .

A graph of position versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.