Graphical Analysis of One-Dimensional Motion
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 displacement, 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
A straight-line graph. The equation for a straight line is
Graph of Displacement vs. Time (a = 0, so v is constant)
Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have
Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats.</figcaption> </figure>
Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity
Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.
The slope of the graph of displacement
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 displacement of 25 m at 0.50 s and 2000 m at 6.40 s. Its displacement at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.
Find the average velocity of the car whose position is graphed in Figure.
The slope of a graph of
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.)
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
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 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 displacement and velocity are initially 200 m and 15 m/s, respectively.<figure class="ui-has-child-figcaption" id="import-auto-id3596921" style="width: 840px;">
Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an
A U.S. Air Force jet car speeds down a track. (credit: Matt Trostle, Flickr)</figcaption> </figure>
The graph of displacement versus time in Figure(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 displacement-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(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(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure(c).
Calculate the velocity of the jet car at a time of 25 s by finding the slope of the
The slope of an
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, where Q is the point at
1. Find the tangent line to the curve at
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,
This is the value given in this figure’s table for
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
The slope of a graph of velocity
Additional general information can be obtained from Figure and the expression for a straight line,
In this case, the vertical axis
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. Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure.) Acceleration gradually decreases from
Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure ends. (a) The slope of this graph is velocity; it is plotted in the next graph. (b) The velocity gradually approaches its top value. The slope of this graph is acceleration; it is plotted in the final graph. (c) Acceleration gradually declines to zero when velocity becomes constant.</figcaption> </figure>
Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the
The slope of the curve at
Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope,
Note that this value for
A graph of displacement 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.
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?<figure id="import-auto-id3504346" style="width: 782px;">
(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.<figure id="import-auto-id1666671" style="width: 782px;">
- Graphs of motion can be used to analyze motion.
- Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
- The slope of a graph of displacement
xvs. time tis velocity v.
- The slope of a graph of velocity
vvs. time tgraph is acceleration a.
- Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.
(a) Explain how you can use the graph of position versus time in Figure to describe the change in velocity over time. Identify (b) the time (
(a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure. (b) Identify the time or times (
(a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure. (b) Based on the graph, how does acceleration change over time?<figure id="import-auto-id1778975" style="width: 782px;">
(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure. (b) Identify the time or times (
<figure id="import-auto-id1447833" style="width: 782px;">
Consider the velocity vs. time graph of a person in an elevator shown in Figure. Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.<figure id="import-auto-id2006890" style="width: 782px;">
A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
Problems & Exercises
Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.
(a) By taking the slope of the curve in Figure, verify that the velocity of the jet car is 115 m/s at
</figure> <figure id="import-auto-id4101417" style="width: 782px;">
Using approximate values, calculate the slope of the curve in Figure to verify that the velocity at
Using approximate values, calculate the slope of the curve in Figure to verify that the velocity at
By taking the slope of the curve in Figure, verify that the acceleration is
Construct the displacement graph for the subway shuttle train as shown in [link](a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.
</figure> <figure id="import-auto-id4128350" style="width: 782px;"></figure> <figure id="import-auto-id4151339" style="width: 782px;"></figure>
A graph of
(a) 6 m/s
(b) 12 m/s
(d) 10 s
Figure shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.<figure id="import-auto-id4035681" style="width: 782px;"></figure>
- independent variable
- the variable that the dependent variable is measured with respect to; usually plotted along the
- dependent variable
- the variable that is being measured; usually plotted along the
- the difference in
y-value (the rise) divided by the difference in x-value (the run) of two points on a straight line
y-value when x= 0, or when the graph crosses the y-axis