PAGE EDITOR(S): Adrian Diaz-Granados, Ned De Leo, Nick Brayer, Annika Wreder, Alex Hernandez

  • Displacement:shortest distance between starting point and ending point of a motion, including difection.
    • Unit = meters
  • Average Velocity:displacement covered in a time interval
    • The sign of the velocity (indicating the direction) comes from the sign of the displacement
    • Unit = meters/seconds
  • Average Speed:total distance traveled in a period of time (never negative!!!!)
    • Unit = meters/seconds
  • Average Acceleration:the rate of change of velocity; how fast the velocity changes
    • Unit = meters/seconds²




Which of the follow sentences contains an example of instantaneous velocity?

“The car covered 500 kilometers in the first 10 hours of its northward journey.”

“Five seconds into the launch, the rocket was shooting upward at 5000 meters per second.”

“The cheetah can run at 70 miles per hour.”

“Moving at five kilometers per hour, it will take us eight hours to get to the base camp.”

“Roger Bannister was the first person to run one mile in less than four minutes.”

Instantaneous velocity has a magnitude and a direction, and deals with the velocity at a particular instant in time. All three of these requirements are met only in B. A is an example of average velocity, C is an example of instantaneous speed, and both D andE are examples of average speed.



Position vs. Time GraphsPosition vs. time graphs give you an easy and obvious way of determining an object’s displacement at any given time, and a subtler way of determining that object’s velocity at any given time.
external image YvsTgraph.gif
Any point on this graph gives us the position of the object at a particular moment in time. For instance, the point at (2,–2) tells us that, two seconds after it started moving, the object was two centimeters to the left of its starting position, and the point at (3,1) tells us that, three seconds after it started moving, the object is one centimeter to the right of its starting position.Let’s read what the graph can tell us about the object’s movements. For the first two seconds, the object is moving to the left. Then, in the next second, it reverses its direction and moves quickly to y = 1. The object then stays still at y = 1 for three seconds before it turns left again and moves back to where it started.Calculating VelocityWe know the object’s displacement, and we know how long it takes to move from place to place. Armed with this information, we should also be able to determine the object’s velocity, since velocity measures the rate of change of displacement over time. If displacement is given here by the vector y, then the velocity of the ant is
external image phy.total77.gif
If you recall, the slope of a graph is a measure of rise over run; that is, the amount of change in the y direction divided by the amount of change in the x direction. In our graph, external image phy.total78.gif is the change in the y direction and external image phy.total79.gif is the change in the x direction, so v is a measure of the slope of the graph. For any position vs. time graph, the velocity at time t is equal to the slope of the line at t. In a graph made up of straight lines, like the one above, we can easily calculate the slope at each point on the graph, and hence know the instantaneous velocity at any given time.We can tell that the ant has a velocity of zero from t = 3 to t = 6, because the slope of the line at these points is zero. We can also tell that the object is cruising along at the fastest speed between t = 2 and t = 3, because the position vs. time graph is steepest between these points. Calculating the object’s average velocity during this time interval is a simple matter of dividing rise by run, as we’ve learned in math class.
external image NEWNEWvelocity.gif
Average VelocityHow about the average velocity between t = 0 and t = 3? It’s actually easier to sort this out with a graph in front of us, because it’s easy to see the displacement at t = 0 andt = 3, and so that we don’t confuse displacement and distance.
external image NEWavg.vel=ch.gif
Average SpeedAlthough the total displacement in the first three seconds is one centimeter to the right, the total distance traveled is two centimeters to the left, and then three centimeters to the right, for a grand total of five centimeters. Thus, the average speed is not the same as the average velocity of the object. Once we’ve calculated the total distance traveled by the object, though, calculating its average speed is not difficult:
external image 5cm-3sec.gif
Curved Position vs. Time GraphsThis is all well and good, but how do you calculate the velocity of a curved position vs. time graph? A few points on the graph will probably be labeled, and you will have to identify which point has the greatest or least velocity. Remember, the point with the greatest slope has the greatest velocity, and the point with the least slope has the least velocity. The turning points of the graph, the tops of the “hills” and the bottoms of the “valleys” where the slope is zero, have zero velocity.
external image curvedYvsTgraph.gif
In this graph, for example, the velocity is zero at points A and C, greatest at point D, and smallest at point B. The velocity at point B is smallest because the slope at that point is negative. Because velocity is a vector quantity, the velocity at B would be a large negative number. However, the speed at B is greater even than the speed at D: speed is a scalar quantity, and so it is always positive. The slope at B is even steeper than at D, so the speed is greatest at B.Velocity vs. Time GraphsVelocity vs. time graphs are the most eloquent kind of graph we’ll be looking at here. They tell us very directly what the velocity of an object is at any given time, and they provide subtle means for determining both the position and acceleration of the same object over time. The “object” whose velocity is graphed below is our ever-industrious ant, a little later in the day.
external image VvsTgraph.gif
We can learn two things about the object’s velocity by a quick glance at the graph. First, we can tell exactly how fast it is going at any given time. For instance, we can see that, two seconds after it started to move, the object is moving at 2 cm/s. Second, we can tell in which direction the object is moving. From t = 0 to t = 4, the velocity is positive, meaning that the object is moving to the right. From t = 4 to t = 7, the velocity is negative, meaning that the object is moving to the left.Calculating AccelerationWe can calculate acceleration on a velocity vs. time graph in the same way that we calculate velocity on a position vs. time graph. Acceleration is the rate of change of the velocity vector, external image phy.total80.gif, which expresses itself as the slope of the velocity vs. time graph. For a velocity vs. time graph, the acceleration at time t is equal to the slope of the line at t.What is the acceleration of our ant at t = 2.5 and t = 4? Looking quickly at the graph, we see that the slope of the line at t = 2.5 is zero and hence the acceleration is likewise zero. The slope of the graph between t = 3 and t = 5 is constant, so we can calculate the acceleration at t = 4 by calculating the average acceleration between t = 3 and t = 5:
external image NEWacc=vf.gif
The minus sign tells us that acceleration is in the leftward direction, since we’ve defined the y-coordinates in such a way that right is positive and left is negative. Att = 3, the object is moving to the right at 2 cm/s, so a leftward acceleration means that the object begins to slow down. Looking at the graph, we can see that the object comes to a stop at t = 4, and then begins accelerating to the right.Calculating DisplacementVelocity vs. time graphs can also tell us about an object’s displacement. Because velocity is a measure of displacement over time, we can infer that:
external image displac.gif
Graphically, this means that the displacement in a given time interval is equal to the area under the graph during that same time interval. If the graph is above the t-axis, then the positive displacement is the area between the graph and the t-axis. If the graph is below the t-axis, then the displacement is negative, and is the area between the graph and the t-axis. Let’s look at two examples to make this rule clearer.First, what is the ant’s displacement between t = 2 and t = 3? Because the velocity is constant during this time interval, the area between the graph and the t-axis is a rectangle of width 1 and height 2.
external image VvsTgraph_shade1.gif
The displacement between t = 2 and t = 3 is the area of this rectangle, which is 1 cm/sexternal image phy.total81.gifs = 2 cm to the right.Next, consider the object’s displacement between t = 3 and t = 5. This portion of the graph gives us two triangles, one above the t-axis and one below the t-axis.
external image VvsTgraph_shade2.gif
Both triangles have an area of 1 /2(1 s)(2 cm/s) = 1 cm. However, the first triangle is above the t-axis, meaning that displacement is positive, and hence to the right, while the second triangle is below the t-axis, meaning that displacement is negative, and hence to the left. The total displacement between t = 3 and t = 5 is:
external image 1cm-1cm.gif
In other words, at t = 5, the object is in the same place as it was at t = 3.Curved Velocity vs. Time GraphsAs with position vs. time graphs, velocity vs. time graphs may also be curved. Remember that regions with a steep slope indicate rapid acceleration or deceleration, regions with a gentle slope indicate small acceleration or deceleration, and the turning points have zero acceleration.Acceleration vs. Time GraphsAfter looking at position vs. time graphs and velocity vs. time graphs, acceleration vs. time graphs should not be threatening. Let’s look at the acceleration of our object at another point.
external image AvsTgraph.gif
Acceleration vs. time graphs give us information about acceleration and about velocity. In this graph, the object is accelerating at 1 m/s2 from t = 2 to t = 5 and is not accelerating between t = 6 and t = 7; that is, between t = 6 and t = 7 the object’s velocity is constant.Calculating Change in VelocityAcceleration vs. time graphs tell us about an object’s velocity in the same way that velocity vs. time graphs tell us about an object’s displacement. The change in velocity in a given time interval is equal to the area under the graph during that same time interval. Be careful: the area between the graph and the t-axis gives the change in velocity, not the final velocity or average velocity over a given time period.What is the object’s change in velocity between t = 2 and t = 5? Because the acceleration is constant during this time interval, the area between the graph and the t-axis is a rectangle of height 1 and length 3.
external image AvsTgraph_shade1.gif
The area of the shaded region, and consequently the change in velocity during this time interval, is 1 cm/s2 · 3 s = 3 cm/s to the right. This doesn’t mean that the velocity at t = 5 is 3 cm/s; it simply means that the velocity is 3 cm/s greater than it was at t = 2. Since we have not been given the velocity at t = 2, we can’t immediately say what the velocity is at t = 5.Summary of Rules for Reading GraphsYou may have trouble recalling when to look for the slope and when to look for the area under the graph. Here are a couple handy rules of thumb:
  1. The slope on a given graph is equivalent to the quantity we get by dividing they-axis by the x-axis. For instance, the y-axis of a position vs. time graph gives us displacement, and the x-axis gives us time. Displacement divided by time gives us velocity, which is what the slope of a position vs. time graph represents.
  2. The area under a given graph is equivalent to the quantity we get by multiplying the x-axis and the y-axis. For instance, the y-axis of an acceleration vs. time graph gives us acceleration, and the x-axis gives us time. Acceleration multiplied by time gives us the change in velocity, which is what the area between the graph and the x-axis represents.
We can summarize what we know about graphs in a table:
external image graphsummary.gif

The total area is then the area of the rectangle plus the area of the triangle. The calculation of these areas is shown below.

Area = base * height
Area = (10 s) * (5 m/s)

Area = 50 m
Area = 0.5 * base * height
Area = 0.5 * (5 s) * (10 m/s)

Area = 25 m
The total area (rectangle plus triangle) is equal to 75 m. Thus the displacement of the object is 75 meters during the 10 seconds of motion.

Check Your Understanding

1. Consider the graph at the right. The object whose motion is represented by this graph is ... (include all that are true):
  1. moving in the positive direction.
  2. moving with a constant velocity.
  3. moving with a negative velocity.
  4. slowing down.
  5. changing directions.
  6. speeding up.
  7. moving with a positive acceleration.
  8. moving with a constant acceleration

  1. external image U1L4a8.gif

Answers: a, d and h apply.
a: TRUE since the line is in the positive region of the graph.
b. FALSE since there is an acceleration (i.e., a changing velocity).
c. FALSE since a negative velocity would be a line in the negative region (i.e., below the horizontal axis).
d. TRUE since the line is approaching the 0-velocity level (the x-axis).
e. FALSE since the line never crosses the axis.
f. FALSE since the line is not moving away from x-axis.
g. FALSE since the line has a negative or downward slope.
h. TRUE since the line is straight (i.e, has a constant slope).

Here's a website with plenty of animations and other kinematic questions
Here is a website that has a lot of stuff on Kinematics.
It has the kinematic equations and examples with them, it also has graphs and explanations of the graphs.
There is alot of general information on things. more kinematics problems with solutions go to this website:

This PDF file has my (Adrian's) notes about different cases for Free-Fall Motion (not Projectile Motion).

This PDF file explains the 3 different cases of Projectile Motion.

Here are some pictures that relate to topics discussed above:
external image projectile.gif
picture showing change in velocity vector during ground to ground launch
animation demonstrating horizontal launch, notice how x component does not change

Step-by-step Workout to a "Ground-to-Ground" Problem
This problem works out a "Ground-to-Ground" case and explains the reasining behind each step very well. Note: As opposed to using (x) or (y), they use dx or dy. Also, as opposed to simply using "v" to denote final velocity, they use vf

Here are the knowns for that problem for x and y:
Finding: (t/2), then ∆y, then t, then ∆x

This figure shows a projectile motion in the "ground-to-ground" case at different points during the trajectory


This figure shows how far an object would travel in a projectile motion problem in relation to it's initial "launch angle"

Additional Example:
A bullet is shot horizontally with a velocity of 350 m/s from a height of 1.5 meters at a target 10 meters away. If the gun was aimed at the center of the target, how far does the bullet hit below the center?

The following PDF file has the entire problem worked out.

If you need more help with ranking tasks, this site is good: 2-Dimensional Kinematics (Ranking Tasks)EXAMPLE PROBLEMS:

external image 4a4f36a7-b7a7-4acb-a074-c4be2027f10f.gif

Rank the three from largest to smallest according to their time in flight.
(a) blue > green > red (b) red > green > blue (c) red = green = blue

Rank the three from largest to smallest according to their initial vertical component of velocity.
(a) blue > green > red (b) red > green > blue (c) red = green = blue

Rank the three from largest to smallest according to their initial horizontal component of velocity.
(a) blue > green > red (b) red > green > blue (c) red = green = blue

Rank the three from largest to smallest according to their initial speed.
(a) blue > green > red (b) red > green > blue (c) red = green = blue

Refer to the following information for the next five questions.
A ball was kicked at a speed of 10 m/sec at a 37º and eventually returns to ground level further downrange.

What were the horizontal and vertical components of the ball's velocity?

How much time did it spend in the air?

How far downrange did it land?

How high did it rise in the air?

Just watch this video. Good times. Shows a crazy video of an "impossible" basketball shot. Is this possible? Look at the analysis from "Popular Science" columnist Adam Weiner who is also the author of the book: Don't Try This at Home! The Physics of Hollywood Movies. Make your own decision!

(Project Excelsior was a series of high-altitude parachute jumps made by Colonel Joseph Kittinger Jr.)
New video link:
other link (longer video):
Kittinger talking through video:

Baumgartner article + video:
Baumgartner spin:
(uncontrollable spin)
Baumgartner videos and pictures:
article + videos (Baumgartner talking about drop):
During this project, there were three high altitude jumps accomplished from a balloon-supported gondola; the first from 76,400 feet; the second from 74,700 feet 25 days later; and on Aug. 16, 1960, from 102,800 feet, the highest altitude from which man has ever jumped. It was Kittinger who did the jumping.

In freefall for four and a half minutes, Kittinger fell at speeds over 700 mph, approaching the speed of sound. He experienced temperatures as low as -94 degrees Fahrenheit. Kittinger opened his parachute at 18,000 feet and landed safely in the New Mexico desert after a 13 minute 45 second descent. Project Excelsior successfully proved the new parachute system, the Beaupre Multi-Stage Parachute, would solve the problem of high altitude escape by crewmen.
In a 1960 National Geographic article, Kittinger describes the jump and the associated support and research efforts called "The Long, Lonely Jump"

In the early years of the United States space program, the United States Air Force (USAF) conducted a variety of pioneering tests to investigate the effects of high-altitude flight on the human body. In August 1960, less than a year before the Union of Soviet Socialist Republics (USSR) sent the first human into space, USAF Captain Joseph W. Kittinger, Jr., tested a new parachute design by jumping from a helium balloon at 31,354 m (102,867 ft).
Shows an astronaut dropping a feather and hammer on the moon, confirming that, in the absense of air resistance, the two objects would fall with the same acceleration.

(This link is for the last two pictures on the page -- the one showing a ground-to-ground case and the one talking about horizontal distance vs. vertical distance in relation to the initial launch angle)
(Video of the Project Excelsior high-altitude parachute jump made by Colonel Joseph Kittinger)
(Information about Col. Kittinger can be found here, at the US Air Force website)
(Ground-to-Ground problem)
(animations and practice problems with kinematics)
(example problems)
(graph explanations)