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

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

EXAMPLE

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

(A)

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

(B)

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

(C)

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

(D)

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

(E)

“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.
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
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, is the change in the y direction and 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. 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. 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: 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.
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 Bis 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 Bis greater even than the speed at D: speed is a scalar quantity, and so it is always positive. The slope at Bis 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.
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, , 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:
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:
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.
The displacement between t = 2 and t = 3 is the area of this rectangle, which is 1 cm/ss = 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.
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:
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.
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.
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:

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.

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:

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

Rectangle

Triangle

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):

moving in the positive direction.

moving with a constant velocity.

moving with a negative velocity.

slowing down.

changing directions.

speeding up.

moving with a positive acceleration.

moving with a constant acceleration

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).

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

Here are some pictures that relate to topics discussed above:

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

X

Y

üa=0

üa=-9.8

üVox=vocosΘ

üVoy=vosinΘ

Vx=vox

üVy=-vosinΘ

∆x=?

∆y=?

T=?

T=?

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.

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?

DefinitionsDisplacement:shortest distance between starting point and ending point of a motion, including difection.Average Velocity:displacement covered in a time intervalAverage Speed:total distance traveled in a period of time (never negative!!!!)Average Acceleration:the rate of change of velocity; how fast the velocity changesEXAMPLEB.Ais an example of average velocity,Cis an example of instantaneous speed, and bothDandEare 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.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 aty= 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, then the velocity of the ant isyIf you recall, the slope of a graph is a measure of rise over run; that is, the amount of change in the

ydirection divided by the amount of change in thexdirection. In our graph, is the change in theydirection and is the change in thexdirection, sois a measure of the slope of the graph.vFor any position vs. time graph, the velocity at timetis equal to the slope of the line att.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 fromt= 3 tot= 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 betweent= 2 andt= 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.Average VelocityHow about the average velocity betweent= 0 andt= 3? It’s actually easier to sort this out with a graph in front of us, because it’s easy to see the displacement att= 0 andt= 3, and so that we don’t confuse displacement and distance.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: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.In this graph, for example, the velocity is zero at points

AandC, greatest at pointD, and smallest at pointB. The velocity at pointBis smallest because the slope at that point is negative. Because velocity is a vector quantity, the velocity atBwould be a large negative number. However, the speed atBis greater even than the speed atD: speed is a scalar quantity, and so it is always positive. The slope atBis even steeper than atD, so the speed is greatest atB.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.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 tot= 4, the velocity is positive, meaning that the object is moving to the right. Fromt= 4 tot= 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, , which expresses itself as the slope of the velocity vs. time graph. For a velocity vs. time graph,the acceleration at timetis equal to the slope of the line att.What is the acceleration of our ant att= 2.5 andt= 4? Looking quickly at the graph, we see that the slope of the line att= 2.5 is zero and hence the acceleration is likewise zero. The slope of the graph betweent= 3 andt= 5 is constant, so we can calculate the acceleration att= 4 by calculating the average acceleration betweent= 3 andt= 5: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 att= 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: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 thet-axis, then the positive displacement is the area between the graph and thet-axis. If the graph is below thet-axis, then the displacement is negative, and is the area between the graph and thet-axis. Let’s look at two examples to make this rule clearer.First, what is the ant’s displacement betweent= 2 andt= 3? Because the velocity is constant during this time interval, the area between the graph and thet-axis is a rectangle of width 1 and height 2.The displacement between

t= 2 andt= 3 is the area of this rectangle, which is 1 cm/ss = 2 cm to the right.Next, consider the object’s displacement betweent= 3 andt= 5. This portion of the graph gives us two triangles, one above thet-axis and one below thet-axis.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 thet-axis, meaning that displacement is negative, and hence to the left. The total displacement betweent= 3 andt= 5 is:In other words, at

t= 5, the object is in the same place as it was att= 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.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 tot= 5 and is not accelerating betweent= 6 andt= 7; that is, betweent= 6 andt= 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 thet-axis gives thechangein velocity, not the final velocity or average velocity over a given time period.What is the object’s change in velocity betweent= 2 andt= 5? Because the acceleration is constant during this time interval, the area between the graph and thet-axis is a rectangle of height 1 and length 3.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 att= 2. Since we have not been given the velocity att= 2, we can’t immediately say what the velocity is att= 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:- The slope on a given graph is equivalent to the quantity we get by dividing the
- The area under a given graph is equivalent to the quantity we get by multiplying the

We can summarize what we know about graphs in a table:y-axis by thex-axis. For instance, they-axis of a position vs. time graph gives us displacement, and thex-axis gives us time. Displacement divided by time gives us velocity, which is what the slope of a position vs. time graph represents.x-axis and they-axis. For instance, they-axis of an acceleration vs. time graph gives us acceleration, and thex-axis gives us time. Acceleration multiplied by time gives us the change in velocity, which is what the area between the graph and thex-axis represents.The total area is then the area of the rectangle plus the area of the triangle. The calculation of these areas is shown below.

RectangleTriangleArea = (10 s) * (5 m/s)

Area = 50 m

Area = 0.5 * (5 s) * (10 m/s)

Area = 25 m

1. Consider the graph at the right. The object whose motion is represented by this graph is ... (include all that are true):Check Your UnderstandingAnswers:

a, d and happly.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

http://www.physicsclassroom.com/Class/1DKin/

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.

http://id.mind.net/~zona/mstm/physics/mechanics/kinematics/kinematics.htmlFor more kinematics problems with solutions go to this website: http://www.solvephysics.com/problems_kinematics.shtml

Here are some pictures that relate to topics discussed above: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

XYThis figure shows a projectile motion in the "ground-to-ground" case at different points during the trajectoryThis 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:

(a) blue > green > red (b) red > green > blue (c) red = green = blue

(a) blue > green > red (b) red > green > blue (c) red = green = blue

(a) blue > green > red (b) red > green > blue (c) red = green = blue

(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.

WEBSITEShttp://www.popsci.com/entertainment-amp-gaming/article/2009-09/real-or-fake-worlds-longest-basketball-shot

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!