Graphs of Motion

The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged

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Discussion

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.

v =  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 awhile.

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.

displacement-time

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 displacement-time equation for constant velocity with the classic slope-intercept equation taught in introductory algebra.

x =  s0  +  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 non-zero 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.

In contrast to the previous examples, let's graph the displacement of an object with a constant, non-zero 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 = s0 + v0Δt +  1 aΔt2
2
y = a + bx + cx2

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 =  s0 + v0Δt +  1  aΔt2
2
a =  2s      
Δt2
a =  2(9 m)  = 0.18 m/s2
(10 s)2

When a displacement-time 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.

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.

v =  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 …

v =  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.

Seven tangents were added to our generic displacement-time graph in the animation 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.

velocity-time

The most important thing to remember about velocity-time graphs is that they are velocity-time graphs, not displacement-time 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 velocity-time 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 velocity-time equation for constant acceleration with the classic slope-intercept equation taught in introductory algebra.

v =  v0  +  aΔt
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.

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 velocity-time 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.

Since a curved line has no single slope we must decide what we mean when asked for the acceleration of an object. 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. These descriptions follow directly from the definitions of average and instantaneous acceleration.

The definition of average acceleration.
a =  Δv
Δt
The definition of instantaneous acceleration.
a =  lim Δv  =  dv
Δt→0 Δt dt
   
   
  • On a velocity-time graph …
    • average acceleration is the slope of the straight line connecting the endpoints of a curve.
  • On a velocity-time graph …
    • instantaneous acceleration is the slope of the line tangent to a curve at any point.

Seven tangents were added to our generic velocity-time graph in the animation below. 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 velocity-time graph is that during such an interval, the velocity is becoming more negative (or less positive, if you prefer).

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 velocity-time graph. But how could displacement be determined? Let's explore some simple examples and then derive the relationship.

Start with the simple velocity-time 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.

 
   Δs =  vΔt    0-4 s: This segment is triangular. The area of a triangle is one-half the base times the height. Essentially, we have just calculated the area of the triangular segment on this graph. The cumulative distance traveled at the end of this interval is …  
 Δs =  ½ (v + v0)Δt 
 Δs =  ½ (8 m/s)(4 s) 
 Δs =  16 m 
  16 m  
 
   Δs =  vΔt    4-8 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. The cumulative distance traveled at the end of this interval is …  
 Δs =  ½ (v + v0)Δt
 Δs =  ½ (10 m/s + 8 m/s)(4 s) 
 Δs =  36 m 
  16 m + 36 m = 52 m  
 
   Δs =  vΔt    8-10 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. The cumulative distance traveled at the end of this interval is …  
 Δs =  (10 m/s)(2 s) 
 Δs =  20 m 
   
  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.

acceleration-time

The acceleration-time 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 acceleration-time 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 velocity-time 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 acceleration-time 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 acceleration-time graph is the acceleration at any given time.

Acceleration is the rate of change of velocity with time. Transforming a velocity-time graph to an acceleration-time 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 velocity-time 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/s2)(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/s2)(6.0 s)
Δv =  30 m/s ≈ 60 mph

There are more things one can say about acceleration-time 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.

Summary

Problems

practice

  1. Complete the worksheet on the first page of worksheet-compare.pdf. Fill each grid space with an appropriately concise answer.

    The answers are on the second page of worksheet-compare.pdf.

         
     
    page 1: the questions   page 2: the answers
         
  2. Work along with this example using worksheet-transform.pdf. The graph below shows velocity as a function of time for some unknown object.
     
    [magnify]
     
    1. What can we say about the motion of this object?
    2. Plot the corresponding graphs of displacement and acceleration as functions of time.
    Solutions …
    1. The height of this graph above the horizontal axis does not show position. This is the most common mistake that students make when interpreting velocity-time graphs.
       
      [magnify]
       
    2. Acceleration is the rate of change of displacement with time. To find acceleration, calculate the slope in each interval.
       
      [magnify]
       
      Plot these values as a function of time. Since the acceleration is constant within each interval, the new graph should be made entirely of linked horizontal segments.
       
      [magnify]
       
    3. Displacement is the product of velocity and time. To find displacement, calculate the area under each interval.
       
      [magnify]
       
      Find the cumulative areas starting from the origin (given an initial displacement of zero) 

       0 s →  0                                =   0 m
 4 s →  0 + 8                            = + 8 m
 8 s →  0 + 8 - 8                        =   0 m
12 s →  0 + 8 - 8 - 16                   = -16 m
16 s →  0 + 8 - 8 - 16 - 8               = -24 m
20 s →  0 + 8 - 8 - 16 - 8 + 0           = -24 m
24 s →  0 + 8 - 8 - 16 - 8 + 0 + 8       = -16 m
30 s →  0 + 8 - 8 - 16 - 8 + 0 + 8 + 24  = + 8 m

      Plot these values as a function of time. Pay attention to the shape of each segment. When the object is accelerating, the line should be curved.
       
      [magnify]
       
  3. Sketch the displacement-time, velocity-time, and acceleration-time graphs for …
    1. an object moving with constant velocity. (Let the initial displacement be zero.)
    2. an object moving with constant acceleration. (Let the initial displacement and velocity be zero.)
    Solutions …
    1. Since the velocity is constant, the displacement time graph will always be straight, the velocity-time graph will always be horizontal, and the acceleration-time graph will always lie along the horizontal axis. When the velocity is positive, the displacement should have a positive slope. When the velocity is negative, the displacement should have a negative slope. When the velocity is zero, all the curves should be horizontal [magnify].
       
      [magnify]
       
    2. Since the acceleration is constant, the displacement time graph will always be a parabola, the velocity-time graph will always be straight, and the acceleration-time graph will always be horizontal. When the acceleration is positive, the velocity should have a positive slope, and the displacement should bend upward. When the acceleration is negative, the velocity should have a negative slope, and the displacement should bend downward. When the acceleration is zero, all the curves should be horizontal.
       
      [magnify]
       
  4. The graph below shows the acceleration of a hydraulic elevator in a four story school building as a function of time.
     
    [magnify]
     
    The graph begins at t = 0 s when the elevator door closed on the second floor and ends at t = 20 s when the door opened on a different floor. Assume that the positive directions for displacement, velocity, and acceleration are upward. Determine …
    1. the maximum speed of the elevator
    2. the duration of the brief jerk experienced by the elevator centered on 17.5 s
    Sketch the corresponding graphs of …
    1. velocity-time
    2. displacement-time
    Determine …
    1. the most likely floor on which the elevator stopped
    Solutions …
    1. A quick glance at the graph shows the elevator accelerating downward over 3 s, then coasting, then accelerating upward for 2 s, coasting again, and then accelerating for a brief burst before stopping. The speed increases, remains constant, decreases, remains constant, and decreases a bit more -- all in the down direction. The greatest speed would happen at the end of the first triangular region on the acceleration-time graph. The area under this bit is the change in velocity from its initial value of zero. Therefore …
       
      Δv = area under a-t graph = area of a triangle
      Δv = ½ bh = ½(3.0 s)(−6.0 m/s2)
      Δv = −9.0 m/s
       
    2. If an elevator is going to work properly it has to stop on a floor to let people off. Therefore, the velocity of the elevator at the end of the graph should be zero. Since change in velocity equal the area under the curve on an acceleration-time graph, we need the total area of the three triangular segments to add up to zero. We've already determined this change for the first triangular segment. Let's repeat it for the second.
       
      Δv = area under a-t graph = area of a triangle
      Δv = ½ bh = ½(2.0 s)(+8.0 m/s2)
      Δv = +8.0 m/s
       
      Add these two areas up and you don't get zero, you get …
       
      Δv = −9.0 m/s + 8.0 m/s
      Δv = −1.0 m/s
       
      Therefore, the area under the remaining segment must be +1.0 m/s to compensate. Use the same concepts as before, but this time solve for the base (change in time) of the triangle instead of the area (change in speed).
       
      Δv =  area under a-t graph = area of a triangle
      +1.0 m/s =  ½ bh = ½ Δt (+4.0 m/s2)
      Δt =  +0.50 s
       
    The best way to construct the graphs for the next two questions is systematically -- beginning from first principles. The rate of change of displacement is called velocity, the rate of change of velocity is called acceleration, and the rate of change of acceleration is called jerk. Yes, you heard me right -- jerk. The straight line segments of the graph we started with correspond to intervals of constant jerk. (If they were curved we'd have non-uniform jerk.) Work backward, integrating the constant value of j to get a, then integrating that to get v, then integrating that to get s (or y in this case since it's a vertical problem).
     
    j = j  =              j
    a = ∫  j dt  =          a0  +  j Δt
    v = ∫  a dt  =      v0  +  a0 Δt   + ½  j Δt2
    y = ∫  v dt  =  y0  +  v0 Δt  + ½  a0 Δt  + ⅙  j Δt3
     
    Apply these equations over and over again. This much computation is best left to a computer. The results are summarized in the table below.
    Hydraulic Elevator
    time (s) jerk (m/s3) acceleration (m/s2) velocity (m/s) position (m)
    0.00 0.0 0.0 0.00 0.000
    1.00 -0.4 0.0 0.00 0.000
    2.50 0.4 -0.6 -0.45 -0.225
    4.00 0.0 0.0 -0.90 -1.350
    13.00 0.8 0.0 -0.90 -9.450
    14.00 -0.8 0.8 -0.50 -10.217
    15.00 0.0 0.0 -0.10 -10.450
    17.25 1.6 0.0 -0.10 -10.675
    17.50 -1.6 0.4 -0.05 -10.696
    17.75 0.0 0.0 0.00 -10.700
    20.00 0.0 0.0 0.00 -10.700

    1. Here's the velocity-time graph for the elevator. The horizontal segments correspond to intervals with no acceleration. The curved segments correspond to intervals with changing acceleration.
       
      [magnify]
       
    2. Here's the displacement-time graph for the elevator. The intervals with acceleration are curved. The intervals without acceleration are straight. The beginning and end of the graph are horizontal since the elevator is stopped.
       
      [magnify]
       
    3. The overall displacement of the elevator was 10.7 m below the second floor. Ceiling heights in a typical residential building are about 3 m. Public buildings like schools tend to have taller floors than homes. My guess is that the ceiling heights in this school are on the order of 5 m. That would mean the elevator stopped in the basement.

worksheets

  1. worksheet-graph-displace.pdf
    The worksheet for this exercise consists of three small and one large displacement-time graph.
    1. Complete the three small displacement-time graphs from the information provided below each graph.
    2. The larger displacement-time graph shows the motion of some hypothetical object over time. Break the graph up into segments and describe qualitatively the motion of the object in each segment. Whenever possible, calculate the velocity of the object as well.
  2. worksheet-graph-velocity.pdf
    The worksheet for this exercise consists of three small and one large velocity-time graph.
    1. Complete the three small velocity-time graphs from the information provided below each graph.
    2. The larger velocity-time graph shows the motion of some hypothetical object over time. Break the graph up into segments and describe qualitatively the motion of the object in each segment. Whenever possible, calculate the acceleration of the object as well.
  3. worksheet-choose-displace.pdf
    The graphs on the accompanying pdf show the displacement of a hypothetical object moving along a straight line. Choose the lettered graph that best represents each of the numbered descriptions. A graph may be used for more than one description or it may not be used at all. Some descriptions may correspond to more than one graph and some may not correspond to any graph at all.
  4. worksheet-choose-velocity.pdf
    The graphs on the accompanying pdf show the velocity of a hypothetical object moving along a straight line. Choose the lettered graph that best represents each of the numbered descriptions. A graph may be used for more than one description or it may not be used at all. Some descriptions may correspond to more than one graph and some may not correspond to any graph at all.

conceptual

  1. conceptual.pdf
    Sketch the displacement-time, velocity-time, and acceleration-time graphs for each of the following scenarios. (Be prepared to explain your sketches.)
    1. An elevator that ascends from the lobby to the 36th floor, stops, descends to the 27th floor, stops, and returns to the lobby.
    2. A basketball is dropped on the court and allowed to bounce up and down several times undisturbed.
    3. A car on a test track performing a zero-to-sixty acceleration test. (This acceleration will not be uniform.)
    4. A race between a tortoise and a hare that unfolds just like the fable of the same name. (An acceleration-time graph is not necessary for this particular problem.)
    5. Two cars are adjacent to each other on a four-lane highway. The first car accelerates uniformly from rest the moment the light changes to green. The second car approaches the intersection already moving and is beside the first car at the instant the light changes. It then continues driving with a constant velocity.
    6. Traffic lights on some streets are timed to facilitate traffic flow at a certain speed. Goofus and Gallant are stopped at a red light on this kind of street. When the light changes Goofus hammers the accelerator until he exceeds the speed limit. He arrives at the first light which is still red and stops. Gallant accelerates at a reasonable rate and never exceeds the speed limit. The second light turns green at just the right instant so that he never needs to brake at an intersection. Goofus and Gallant continue driving this way for three lights.

numerical

  1. The graph to the right

    [magnify]
    shows the altitude of a skydiver initially at rest as a function of time. After 7 s of free fall the skydiver's chute deployed completely, which changed the motion abruptly.
    1. Determine the velocity at the instant …
      1. just before the parachute opened.
      2. just after the parachute opened.
    2. What was the skydiver's acceleration …
      1. from the beginning of the jump to the time just before the parachute opened?
      2. from the time just after the parachute opened to the time when the skydiver landed?
    3. Sketch the corresponding graphs of …
      1. velocity-time.
      2. acceleration-time.
  2. The graph to the right

    [magnify]
    shows the velocity of a skydiver as a function of time. At time t = 0 s the skydiver is located at position y = 0 m at the door of the plane, at t = 8 s the parachute opened, and at t = 12 s the skydiver touched down. Assume that the positive directions for displacement, velocity, and acceleration are downward. Using this information sketch the corresponding graphs of …
    1. displacement-time.
    2. acceleration-time

statistical

  1. special-splits.html
    A split is a time at which the runner reaches a milestone distance in a race. In the 100 m dash, for example, split times are taken every 10 m. Splits for some of the world's fastest sprinters are given on the accompanying webpage. Fit a high order polynomial (fourth, fifth, sixth or higher) to the data for one of these athletes using a data analysis application. Determine the speed of your sprinter as a function of time by taking the derivative of this polynomial. Graph this new function and then analyze it.
    1. What were the runner's initial and final speeds?
    2. What was the runner's maximum speed and when did it occur?
    3. What was the runner's average speed?
    4. Did the runner's speed increase, decrease, or remain roughly the same near the end of the race?
    5. How well do you think this graph describes the actual performance of the runner? Are there any problem regions on the graph? How could the function be modified to improve the fit?
  2. jet-takeoff.txt, jet-landing.txt
    One fine day, a Boeing 717 departed from Mitchell International Airport (MKE) in Milwaukee. Approximately two hours later, it arrived at Laguardia Airport (LGA) in New York. During takeoff and landing, runway positions (in meters) were recorded as a function of time (in seconds) and the data were saved as tab-delimited text files. Using the data in these files and your favorite graphing software …
    1. construct a graph of distance vs. time for…
      1. takeoff and
      2. landing
    2. then add an appropriate curve fit so that you can determine …
      1. the acceleration at takeoff and
      2. the deceleration on landing
    3. and also determine …
      1. the final speed when the airplane left the runway in Milwaukee and
      2. the initital speed when the airplane hit the runway in New York

investigative

  1. The numbered streets in Manhattan above 14th Street are spaced apart such that twenty blocks equal one mile. Ride one of the local trains that runs beneath an avenue for at least five consecutive stations. Using a timer or a wristwatch record the starting and stopping times of the train and the street number of the station until you have reached the fifth station. Translate your data into a displacement-time and velocity-time graph. Include the necessary data tables. Use whatever units you wish. (This investigation can also be performed in other places in a car or a bus if the streets are gridded and you know the grid interval.)

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