Graphs of Motion
Discussion
introduction
Why are there so many equations in this book? Why can't physicists be content with the written word like everyone else? Wouldn't it be easier just to speak directly instead of cloaking ideas behind mathematical cryptograms?
Modern mathematical notation is a highly compact way to encode ideas. Equations can easily contain the information equivalent of several sentences. Galileo's description of an object moving with constant speed (perhaps the first application of mathematics to motion) required one definition, four axioms, and six theorems. All of these relationships can now be written in a single equation.
v̅ =  Δs 
Δt 
When it comes to depth, nothing beats an equation.
Well, almost nothing. Think back to the previous section on the equations of motion. You should recall that the three (or four) equations presented in that section were only valid for motion with constant acceleration along a straight line. Since, as I rightly pointed out, "no object has ever traveled in a straight line with constant acceleration anywhere in the universe at any time" these equations are only approximately true, only once in a while.
Equations are great for describing idealized situations, but they don't always cut it. Sometimes you need a picture to show what's going on — a mathematical picture called a graph. Graphs are often the best way to convey descriptions of real world events in a compact form. Graphs of motion come in several types depending on which of the kinematic quantities (time, displacement, velocity, acceleration) are assigned to which axis.
displacementtime
Let's begin by graphing some examples of motion at a constant velocity. Three different curves are included on the graph to the right, each with an initial displacement of zero. Note first that the graphs are all straight. (Any kind of line drawn on a graph is called a curve. Even a straight line is called a curve in mathematics.) This is to be expected given the linear nature of the appropriate equation. (The independent variable of a linear function is raised no higher than the first power.)
Compare the displacementtime equation for constant velocity with the classic slopeintercept equation taught in introductory algebra.
s =  s_{0}  +  vΔt 
y =  a  +  bx 
Thus velocity corresponds to slope and initial displacement to the intercept on the vertical axis (commonly thought of as the "y" axis). Since each of these graphs has its intercept at the origin, each of these objects had the same initial displacement. This graph could represent a race of some sort where the contestants were all lined up at the starting line (although, at these speeds it must have been a race between tortoises). If it were a race, then the contestants were already moving when the race began, since each curve has a nonzero slope at the start. Note that the initial position being zero does not necessarily imply that the initial velocity is also zero. The height of a curve tells you nothing about its slope.
 On a displacementtime graph…
 slope equals velocity.
 the "y" intercept equals the initial displacement.
 when two curves coincide, the two objects have the same displacement at that time.
In contrast to the previous examples, let's graph the displacement of an object with a constant, nonzero acceleration starting from rest at the origin. The primary difference between this curve and those on the previous graph is that this curve actually curves. The relation between displacement and time is quadratic when the acceleration is constant and therefore this curve is a parabola. (The variable of a quadratic function is raised no higher than the second power.)
s = s_{0} + v_{0}Δt +  1  aΔt^{2}  
2  
y = a + bx + cx^{2}  
As an exercise, let's calculate the acceleration of this object from its graph. It intercepts the origin, so its initial displacement is zero, the example states that the initial velocity is zero, and the graph shows that the object has traveled 9 m in 10 s. These numbers can then be entered into the equation.
s = 


a = 


a = 

When a displacementtime graph is curved, it is not possible to calculate the velocity from it's slope. Slope is a property of straight lines only. Such an object doesn't have a velocity because it doesn't have a slope. The words "the" and "a" are underlined here to stress the idea that there is no single velocity under these circumstances. The velocity of such an object must be changing. It's accelerating.
 On a displacementtime graph…
 straight lines imply constant velocity.
 curved lines imply acceleration.
 an object undergoing constant acceleration traces a portion of a parabola.
Although our hypothetical object has no single velocity, it still does have an average velocity and a continuous collection of instantaneous velocities. The average velocity of any object can be found by dividing the total displacement by the total time.
v̅ =  Δs 
Δt 
This is the same as calculating the slope of the straight line connecting the first and last points on the curve as shown in the diagram to the right. In this abstract example, the average velocity of the object was…
v̅ =  Δs  =  9.5 m  = 0.95 m/s 
Δt  10.0 s 
Instantaneous velocity is the limit of average velocity as the time interval shrinks to zero.
v =  lim  Δs  =  ds 
Δt → 0  Δt  dt 
As the endpoints of the line of average velocity get closer together, they become a better indicator of the actual velocity. When the two points coincide, the line is tangent to the curve. This limit process is represented in the animation to the right.
 On a displacementtime graph…
 average velocity is the slope of the straight line connecting the endpoints of a curve.
 instantaneous velocity is the slope of the line tangent to a curve at any point.
Seven tangents were added to our generic displacementtime graph in the animation shown above. Note that the slope is zero twice — once at the top of the bump at 3.0 s and again in the bottom of the dent at 6.5 s. (The bump is a local maximum, while the dent is a local minimum. Collectively such points are known as local extrema.) The slope of a horizontal line is zero, meaning that the object was motionless at those times. Since the graph is not flat, the object was only at rest for an instant before it began moving again. Although its position was not changing at that time, its velocity was. This is a notion that many people have difficulty with. It is possible to be accelerating and yet not be moving (but only for an instant, of course).
Note also that the slope is negative in the interval between the bump at 3 s and the dent at 6.5 s. Some interpret this as motion in reverse, but is this generally the case? Well, this is an abstract example. It's not accompanied by any text. Graphs contain a lot of information, but without a title or other form of description they have no meaning. What does this graph represent? A person? A car? An elevator? A rhinoceros? An asteroid? A mote of dust? About all we can say is that this object was moving at first, slowed to a stop, reversed direction, stopped again, and then resumed moving in the direction it started with (whatever direction that was). Negative slope does not automatically mean driving backward, or walking left, or falling down. The choice of signs is always arbitrary. About all we can say in general, is that when the slope is negative, the object is traveling in the negative direction.
 On a displacementtime graph…
 positive slope implies motion in the positive direction.
 negative slope implies motion in the negative direction.
 zero slope implies a state of rest.
velocitytime
The most important thing to remember about velocitytime graphs is that they are velocitytime graphs, not displacementtime graphs. There is something about a line graph that makes people think they're looking at the path of an object. A common beginner's mistake is to look at the graph to the right and think that the the v = 9.0 m/s line corresponds to an object that is "higher" than the other objects. Don't think like this. It's wrong.
Don't look at these graphs and think of them as a picture of a moving object. Instead, think of them as the record of an object's velocity. In these graphs, higher means faster not farther. The v = 9.0 m/s line is higher because that object is moving faster than the others.
These particular graphs are all horizontal. The initial velocity of each object is the same as the final velocity is the same as every velocity in between. The velocity of each of these objects is constant during this ten second interval.
In comparison, when the curve on a velocitytime graph is straight but not horizontal, the velocity is changing. The three curves to the right each have a different slope. The graph with the steepest slope experiences the fastest change in velocity. That object has the greatest acceleration. Compare the velocitytime equation for constant acceleration with the classic slopeintercept equation taught in introductory algebra.
v =  v_{0}  +  a̅Δt 
y =  a  +  bx 
You should see that acceleration corresponds to slope and initial velocity to the intercept on the vertical axis. Since each of these graphs has its intercept at the origin, each of these objects was initially at rest. The initial velocity being zero does not mean that the initial position must also be zero, however. This graph tells us nothing about the initial position of these objects. For all we know they could be on different planets.
 On a velocitytime graph…
 slope equals acceleration.
 the "y" intercept equals the initial velocity.
 when two curves coincide, the two objects have the same velocity at that time.
The curves on the previous graph were all straight lines. A straight line is a curve with constant slope. Since slope is acceleration on a velocitytime graph, each of the objects represented on this graph is moving with a constant acceleration. Were the graphs curved, the acceleration would not have been constant.
 On a velocitytime graph…
 straight lines imply uniform acceleration.
 curved lines imply nonuniform acceleration.
 an object undergoing constant acceleration traces a straight line.
Since a curved line has no single slope we must decide what we mean when asked for the acceleration of an object. These descriptions follow directly from the definitions of average and instantaneous acceleration. If the average acceleration is desired, draw a line connecting the endpoints of the curve and calculate its slope. If the instantaneous acceleration is desired, take the limit of this slope as the time interval shrinks to zero, that is, take the slope of a tangent.
 On a velocitytime graph…
 average acceleration is the slope of the straight line connecting the endpoints of a curve.
a̅ =  Δv 
Δt  
 On a velocitytime graph…
 instantaneous acceleration is the slope of the line tangent to a curve at any point.
a =  lim  Δv  =  dv  
Δt → 0  Δt  dt  
Seven tangents were added to our generic velocitytime graph in the animation shown above. Note that the slope is zero twice — once at the top of the bump at 3.0 s and again in the bottom of the dent at 6.5 s. The slope of a horizontal line is zero, meaning that the object stopped accelerating instantaneously at those times. The acceleration might have been zero at those two times, but this does not mean that the object stopped. For that to occur, the curve would have to intercept the horizontal axis. This happened only once — at the start of the graph. At both times when the acceleration was zero, the object was still moving in the positive direction.
You should also notice that the slope was negative from 3.0 s to 6.5 s. During this time the speed was decreasing. This is not true in general, however. Speed decreases whenever the curve returns to the origin. Above the horizontal axis this would be a negative slope, but below it this would be a positive slope. About the only thing one can say about a negative slope on a velocitytime graph is that during such an interval, the velocity is becoming more negative (or less positive, if you prefer).
 On a velocitytime graph…
 positive slope implies an increase in velocity in the positive direction.
 negative slope implies an increase in velocity in the negative direction.
 zero slope implies motion with constant velocity.
In kinematics, there are three quantities: displacement, velocity, and acceleration. Given a graph of any of these quantities, it is always possible in principle to determine the other two. Acceleration is the time rate of change of velocity, so that can be found from the slope of a tangent to the curve on a velocitytime graph. But how could displacement be determined? Let's explore some simple examples and then derive the relationship.
Start with the simple velocitytime graph shown to the right. (For the sake of simplicity, let's assume that the initial displacement is zero.) There are three important intervals on this graph. During each interval, the acceleration is constant as the straight line segments show. When acceleration is constant, the average velocity is just the average of the initial and final values in an interval.
04 s: This segment is triangular. The area of a triangle is onehalf the base times the height. Essentially, we have just calculated the area of the triangular segment on this graph.
Δs =  v̅Δt 
Δs =  ½(v + v_{0})Δt 
Δs =  ½(8 m/s)(4 s) 
Δs =  16 m 
The cumulative distance traveled at the end of this interval is…
16 m
48 s: This segment is trapezoidal. The area of a trapezoid (or trapezium) is the average of the two bases times the altitude. Essentially, we have just calculated the area of the trapezoidal segment on this graph.
Δs =  v̅Δt 
Δs =  ½(v + v_{0})Δt 
Δs =  ½(10 m/s + 8 m/s)(4 s) 
Δs =  36 m 
The cumulative distance traveled at the end of this interval is…
16 m + 36 m = 52 m
810 s: This segment is rectangular. The area of a rectangle is just its height times its width. Essentially, we have just calculated the area of the rectangular segment on this graph.
Δs =  v̅Δt 
Δs =  (10 m/s)(2 s) 
Δs =  20 m 
The cumulative distance traveled at the end of this interval is…
16 m + 36 m + 20 m = 72 m
I hope by now that you see the trend. The area under each segment is the change in displacement of the object during that interval. This is true even when the acceleration is not constant.
Anyone who has taken a calculus course should have known this before they read it here (or at least when they read it they should have said, "Oh yeah, I remember that"). The first derivative of displacement with respect to time is velocity. The derivative of a function is the slope of a line tangent to its curve at a given point. The inverse operation of the derivative is called the integral. The integral of a function is the cumulative area between the curve and the horizontal axis over some interval. This inverse relation between the actions of derivative (slope) and integral (area) is so important that it's called the fundamental theorem of calculus. This means that it's an important relationship. Learn it! It's "fundamental". You haven't seen the last of it.
 On a velocitytime graph…
 the area under the curve equals the change in displacement.
accelerationtime
The accelerationtime graph of any object traveling with a constant velocity is the same. This is true regardless of the velocity of the object. An airplane flying at a constant 600 mph (270 m/s), a sloth walking with a constant speed 1 mph (0.4 m/s), and a couch potato lying motionless in front of the TV for hours will all have the same accelerationtime graphs — a horizontal line collinear with the horizontal axis. That's because the velocity of each of these objects is constant. They're not accelerating. Their accelerations are zero. As with velocitytime graphs, the important thing to remember is that the height above the horizontal axis doesn't correspond to position or velocity, it corresponds to acceleration.
If you trip and fall on your way to school, your acceleration towards the ground is greater than you'd experience in all but a few high performance cars with the "pedal to the metal". Acceleration and velocity are different quantities. Going fast does not imply accelerating quickly. The two quantities are independent of one another. A large acceleration corresponds to a rapid change in velocity, but it tells you nothing about the values of the velocity itself.
When acceleration is constant, the accelerationtime curve is a horizontal line. The rate of change of acceleration with time is a meaningless quantity so the slope of the curve on this graph is also meaningless. Acceleration need not be constant, but the time rate of change of this number has no name. On the surface, the only information one can glean from an accelerationtime graph is the acceleration at any given time.
 On an accelerationtime graph…
 slope is meaningless.
 the "y" intercept equals the initial acceleration.
 when two curves coincide, the two objects have the same acceleration at that time.
 an object undergoing constant acceleration traces a horizontal line.
 zero slope implies motion with constant acceleration.
Acceleration is the rate of change of velocity with time. Transforming a velocitytime graph to an accelerationtime graph means calculating the slope of a line tangent to the curve at any point. (In calculus, this is called finding the derivative.) The reverse process entails calculating the cumulative area under the curve. (In calculus, this is called finding the integral.) This number is then the change of value on a velocitytime graph.
Given an initial velocity of zero (and assuming that down is positive), the final velocity of the person falling in the graph to the right is…
Δv =  aΔt 
Δv =  (9.8 m/s^{2})(1.0 s) 
Δv =  9.8 m/s ≈ 20 mph 
and the final velocity of the accelerating car is…
Δv =  aΔt 
Δv =  (5.0 m/s^{2})(6.0 s) 
Δv =  30 m/s ≈ 60 mph 
 On an accelerationtime graph…
 the area under the curve equals the change in velocity.
There are more things one can say about accelerationtime graphs, but they are trivial for the most part.
phase space
There is a fourth graph of motion that relates velocity to displacement. It is as important as the other three types, but it rarely gets any attention below the advanced undergraduate level. Some day I will write something about these graphs called phase space diagrams, but not today.