Graphical Analysis of One-Dimensional Motion

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Course: PHYS101: Introduction to Mechanics
Book: Graphical Analysis of One-Dimensional Motion
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Date: Monday, July 15, 2024, 10:08 AM

Description

Read this text for an introduction on graphical position, velocity, and acceleration with regards to one another. As you read, pay attention to Figure 2.46 which is an example of a linear graph is the graph of position versus time when acceleration is zero.

See an example of this type of graph in Figure 2.47. In this graph, we can determine the slope by picking two different points on the line, taking the change in y-value, and dividing it by the change in x-value between those two points. In this case, the unit for slope is m/s, which is the unit for velocity. Therefore, the slope for a graph of position versus time with zero acceleration is the average velocity of that object. See how to calculate the average velocity of an object from this type of graph in Example 2.17.

When acceleration is a non-zero constant, the graph of position versus time is no longer linear. You can see an example of this type of graph in Figure 2.48. Note that while the position versus time graph is not linear, the velocity versus time graph is linear. In the position versus time graph, the slope at any given point is the instantaneous velocity of the object. The instantaneous slope can be determined by drawing a tangent line at the desired point along the graph and determining slope. Pay attention to the tangent lines drawn in Figure 2.48 (a).

To determine instantaneous velocity at a given time when acceleration is a non-zero constant, take a look at Example 2.18. We can determine instantaneous velocities at multiple points along a position-time graph with constant non-zero acceleration and make a table relating these instantaneous velocities to the specified time along the x-axis where we found them. Then, we can use that table to plot velocity versus time. This process is demonstrated in Figure 2.48 (a) and (b). The slope of this linear graph has units  m/s^{2} , which are acceleration units. Therefore, the slope of the velocity versus time graph is acceleration.

Introduction

A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of position, velocity, and acceleration versus time to illustrate one-dimensional kinematics.


Slopes and General Relationships

First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we call the horizontal axis the x-axis and the vertical axis the y-axis, as in Figure 2.46, a straight-line graph has the general form

y=m x+b.

Here m is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter b is used for the y-intercept, which is the point at which the line crosses the vertical axis.

Graph of a straight-line sloping up at about 40 degrees.

Figure 2.46 A straight-line graph. The equation for a straight line is y=mx+b.


Graph of Position vs. Time (a = 0, so v is constant)

Time is usually an independent variable that other quantities, such as position, depend upon. A graph of position versus time would, thus, have x on the vertical axis and t on the horizontal axis. Figure 2.47 is just such a straight-line graph. It shows a graph of position versus time for a jet-powered car on a very flat dry lake bed in Nevada.

Line graph of jet car displacement in meters versus time in seconds. The line is straight with a positive slope. The y intercept is four hundred meters. The total change in time is eight point zero seconds. The initial position is four hundred meters. The final position is two thousand meters.

Figure 2.47 Graph of position versus time for a jet-powered car on the Bonneville Salt Flats.

Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity \bar{v} and the intercept is position at time zero-that is, x_{0}. Substituting these symbols into y=m x+b gives

x=\bar{v} t+x_{0}

or

x=x_{0}+\bar{v} t.

Thus a graph of position versus time gives a general relationship among displacement(change in position), velocity, and time, as well as giving detailed numerical information about a specific situation.

The Slope of X VS. T

The slope of the graph of position x vs. time t is velocity v.

\text { slope }=\frac{\Delta x}{\Delta t}=v

Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.

From the figure we can see that the car has a position of 525 m at 0.50 s and 2000 m at 6.40 s. Its position at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.

 


Source: Rice University, https://openstax.org/books/college-physics/pages/2-8-graphical-analysis-of-one-dimensional-motion
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.

Example 2.17 Determining Average Velocity from a Graph of Position versus Time: Jet Car

Example 2.17 Determining Average Velocity from a Graph of Position versus Time: Jet Car

Find the average velocity of the car whose position is graphed in Figure 2.47.

Strategy

The slope of a graph of x vs. t is average velocity, since slope equals rise over run. In this case, rise = change in position and run = change in time, so that

\text { slope }=\frac{\Delta x}{\Delta t}=\bar{v}.

Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger).

Solution

1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points).

2. Substitute the x and t values of the chosen points into the equation. Remember in calculating change (\Delta) we always use final value minus initial value.

\bar{v}=\frac{\Delta x}{\Delta t}=\frac{2000 \mathrm{~m}-525 \mathrm{~m}}{6.4 \mathrm{~s}-0.50 \mathrm{~s}},

yielding

\bar{v}=250 \mathrm{~m} / \mathrm{s}.

Discussion

This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.

Graphs of Motion When a is Constant but a≠0

The graphs in Figure 2.48 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the position and velocity are initially 200 m and 15 m/s, respectively.

Three line graphs. First is a line graph of displacement over time. Line has a positive slope that increases with time. Second line graph is of velocity over time. Line is straight with a positive slope. Third line graph is of acceleration over time. Line is straight and horizontal, indicating constant acceleration.

Figure 2.48 Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an x vs. t graph is velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point. (b) The slope of the v vs. t graph is constant for this part of the motion, indicating constant acceleration. (c) Acceleration has the constant value of 5.0 \, m/s^{2} over the time interval plotted.


A photo shows a U S Air Force jet speeding down a track with a lot of smoke around it.

Figure 2.49 A U.S. Air Force jet car speeds down a track


The graph of position versus time in Figure 2.48(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a position-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line.

Tangent lines are shown for two points in Figure 2.48(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 2.48(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 2.48(c).

Example 2.18 Determining Instantaneous Velocity from the Slope at a Point: Jet Car

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the x vs. t graph in the graph below.

A graph of displacement versus time for a jet car. The x axis for time runs from zero to thirty five seconds. The y axis for displacement runs from zero to three thousand meters. The curve depicting displacement is concave up. The slope of the curve increases over time. Slope equals velocity v. There are two points on the curve, labeled, P and Q. P is located at time equals ten seconds. Q is located and time equals twenty-five seconds. A line tangent to P at ten seconds is drawn and has a slope delta x sub P over delta t sub p. A line tangent to Q at twenty five seconds is drawn and has a slope equal to delta x sub q over delta t sub q. Select coordinates are given in a table and consist of the following: time zero seconds displacement two hundred meters; time five seconds displacement three hundred thirty eight meters; time ten seconds displacement six hundred meters; time fifteen seconds displacement nine hundred eighty eight meters. Time twenty seconds displacement one thousand five hundred meters; time twenty five seconds displacement two thousand one hundred thirty eight meters; time thirty seconds displacement two thousand nine hundred meters.

Figure 2.50 The slope of an x vs. t graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.

Strategy

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 2.50, where Q is the point at t=25 \, s.

Solution

1. Find the tangent line to the curve at t=25 \, s.

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope, v.

\text { slope }=v_{\mathrm{Q}}=\frac{\Delta x_{\mathrm{Q}}}{\Delta t_{\mathrm{Q}}}=\frac{(3120 \mathrm{~m}-1300 \mathrm{~m})}{(32 \mathrm{~s}-19 \mathrm{~s})}

Thus,

v_{\mathrm{Q}}=\frac{1820 \mathrm{~m}}{13 \mathrm{~s}}=140 \mathrm{~m} / \mathrm{s}.

Discussion

This is the value given in this figure's table for v at t=25 \mathrm{~s}. The value of 140 \mathrm{~m} / \mathrm{s} for v_{\mathrm{Q}} is plotted in Figure 2.50. The entire graph of v vs. t can be obtained in this fashion.

Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a v vs. t graph, rise = change in velocity \Delta v and run = change in time \Delta t.

The Slope of V vs. T

The slope of a graph of velocity v vs. time t is acceleration a.

\text { slope }=\frac{\Delta v}{\Delta t}=a

Since the velocity versus time graph in Figure 2.48(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 2.48(c).

Additional general information can be obtained from Figure 2.50 and the expression for a straight line, y=m x+b

In this case, the vertical axis y is V, the intercept b is v_{0}, the slope m is a, and the horizontal axis x is t. Substituting these symbols yields

v=v_{0}+a t

A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.

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.

Check Your Understanding

A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?

Line graph of velocity versus time. The line has three legs. The first leg is flat. The second leg has a negative slope. The third leg also has a negative slope, but the slope is not as negative as the second leg.

Figure 2.52


Solution

(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.

(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

A line graph of acceleration versus time. There are three legs of the graph. All three legs are flat and straight. The first leg shows constant acceleration of 0. The second leg shows a constant negative acceleration. The third leg shows a constant negative acceleration that is not as negative as the second leg.

A line graph of acceleration versus time. There are three legs of the graph. All three legs are flat and straight. The first leg shows constant acceleration of 0. The second leg shows a constant negative acceleration. The third leg shows a constant negative acceleration that is not as negative as the second leg.

Figure 2.53