We call any quantity that specifies both magnitude and direction a vector.
Suppose you asked two small children "which of you is the faster runner?" Their response would undoubtedly be a race. If you followed with "are you a faster runner than you were last year?" you'd probably have them stumped.
Of course, this is simple to older children and adults. In competitive races, records are kept of the time each runner takes to run a fixed distance. Shorter times indicate faster speeds. It all seems really straight-forward, but notice that distance and time individually have nothing to do with speed. It is their ratio which expresses exactly what we need. Speeddistance/timeis the rate at which distance is covered.
In this set of activities we will assume that everyone has a good understanding of ideas like distance and speed. Our goal is to learn about a more concise description that allows us to describe any motion, no matter how complex. We'll start with a very simple system and build on it. Pay attention to the rules and nuances of this system. As it is built up, you'll find that what seemed obvious at first becomes quite important later.
We want to start with a general look at how we describe the locations and motions of objects relative to an arbitrary coordinate system.
Let's start out with something that's near and dear to everyone's heart, Pizza!
"Let's see, that's two mediums with pepperoni and one small veggie and you're 2 blocks away.
That'll be 15 minutes or we pay for the cheese."
Just click on the Bob's Pizza sign to start the delivery guy on his way. (Remember, always run movies and animations immediately unless you're told to wait.)
And of course, they delivered 3 larges with double anchovies! Sigh.
Something must have been missing from their directions. I guess that's why they call them directions.
To prevent the problem of having to deliver a pizza to every house that's 2 blocks away you might have specified 2 blocks north or 2 blocks east. In this way you are specifying both the magnitude (the distance of 2 blocks) and direction (north) of the trip to your house.
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We call any quantity that specifies both magnitude and direction a vector. |
This particular vector involving distances is called a displacement vector. Velocity, force, and momentum are other quantities that we'll find are vectors. Distance, time, and energy are examples of scalars, which are quantities that have magnitudes, but no direction. We'll learn about vectors a little at a time on an as-needed basis.
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Any quantity that can be described entirely in terms of its magnitude is called a scalar. |
We make a lot of use of vectors in physics because, well, we want our pizza to be hot. There are of course other situations that are almost as important - finding our way to Mars to see if there really are fossils there, etc. So let's work out a system that is equally reliable in pizza delivery and Space Shuttle navigation. Our goal is a system that can describe the location of any object and the trip you'd have to take to get there from any point.
Relative to Bob's place, Laura's apartment building has a position of 2 blocks north, and 2 blocks, west. Since the position requires both magnitude and direction to describe it, we see that position is a vector quantity.
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The displacement is the distance and direction from some starting position to some ending position using the shortest straight-line path.
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In the previous animation, Bob's trip to Laura's took an indirect route because of the street layout. He could have gotten there by other routes. (Subterranean delivery! Cool!) But, regardless of the route, the displacement is the same and it's equal to about 2.8 blocks, northwest.
Stating the displacement as 2.8 blocks, northwest in words is one way of completely specifying it. In many cases it is useful to specify displacement graphically. We use an arrow drawn from the starting point to the destination. The length of the arrow indicates the distance and the direction it points indicates, you guessed it, the direction. We often label such vector arrows with names, usually letters. The following animation illustrates this technique.
The big red arrow is a graphical representation of the vector 
. It's magnitude is indicated by the scale of the drawing (see red rings) and its direction is indicated by the direction the arrow points. The map rose on the right indicates the orientation of the map. Note also the little arrow over the . Such arrows indicate that the quantity named "A" is a vector. Another convention is to use a bold-faced letter for the vector name and a regular letter for the magnitude of the vector. In these activities we will use both conventions.
Ex. A = 2.8 blocks, northwest is a displacement vector.
Ex. A = 2.8 blocks is the magnitude of the vector , that is, the distance.
(To be honest, we get a bit lazy about this. The bold type or arrow means "treat this quantity as a vector." Once you get the hang of it, you need less reminding & we only use the symbols for emphasis.)
In physics we usually don't use city blocks for our unit of distance; we'll use meters. Also, for the time being, we'll stick with one-dimensional motion, motion along a line. Let's try a few instructive examples of some positions and displacements in 1-D.
In the following animation you'll see that we use an x to represent position. We've chosen to arbitrarily set the x=0 point at Bob's pizza place. Any point to the left will have a negative position, any point to the right has a positive position. Go ahead and work through all four pages of the animation.
So far we've found that the displacement is a vector from the starting point to the ending point of some motion. We calculate it in one dimension using the equation: Δx = x2 - x1.
(The Greek letter Δ, delta, is the symbol we use for the change in a quantity. We always take the final value minus the initial value.)
The displacement has a magnitude equal to the length of that straight line path and a direction determined by the direction you'd go from the start to the finish in a straight line.
For motion along a line–always the case in this chapter–we can specify direction using sign conventions: If you go to the right, your displacement has a positive direction. A displacement to the left has a negative direction.
We've also found that different routes can result in the same displacement. The actual route taken is immaterial.
Try these four examples to make sure that all is clear.
One last thing about vectors before we move on. The use of the sign with the displacement to indicate direction is not an isolated case.
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The sign of a vector always indicates its direction.
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For example, a force can't have a magnitude of less than zero, but a force vector can be have a negative value, where the sign indicates that the force is acting in a negative direction. Don't forget that vectors don't always choose to point in the + or - direction. We saw one example of that with our pizza delivery.
Let's get some experience with some real motions. We'll be using a tool called an Ultrasonic Motion Detector (UMD) that works very much like a bat's echo location system. It emits very high frequency sounds and receives the reflection of these sounds after they bounce off of objects in front of it. Using the time it has to wait for the sound to return and the speed of sound (about 340 m/s) it calculates the distance to the object. If the object moves closer, the wait time is shorter, etc. Plotting the calculated position of the object vs. time provides a vivid description of the object's motion.
(When a bat uses this technique, it has to shut down its hearing equipment while it emits the sound to prevent damage to its hearing. Our UMD has to do the same thing. As a result, it can't "see" things close by because the reflections return while it's not listening. The closest distance that we can get to the UMD and get results is about .5 m.)
In this movie you see a hand moving toward and away from the UMD. (Get a load of that hair! I think I've seen this guy on TNN. I think that's George Jones!) You also see the graph produced by the UMD. Can you see how the location of the hand at each instant is matched by the height of the graph?
| 1. Notice that the graph only shows positive positions. This will always be the case. Why do you think it works this way? |
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| 1. How is the position of the hand indicated in the graph? Explain. | |
| 2. How is the direction of motion indicated in the graph? Explain. |
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| 3. How is the speed indicated in the graph? Explain. |
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Another tool for making graphs of motions is a piece of software called Interactive Physics. It lets us set up arrangements of objects and set them into motion. The program produces nice graphs of the motions of the objects. Let's try it with a little matching puzzle.
4. Here are four animations involving the collision of a pair of blocks on a frictionless surface. The red block is the object to watch. The blue block is just there to change the red block's velocity. Ignore it.
To the right of the animations are four position, time graphs. Each one matches the motion of one of the animations. Your task is to match them up. Write down your answers (Movie 1 = Graph c, Movie 2 = Graph d, etc.) and when you have them all, try the multiple-choice questions below to see how you did.
You'll find that this is a good place to try clicking and holding on the 
and 
icons to speed up the animations. OK, have at it. Take your time. Discuss all the details. And get all the answers before you try your hand at the multiple-choice answers.
| 1 = a | 2 = b | 3 = c | 4 = d | |
| 1 = d | 2 = c | 3 = b | 4 = a | |
| 1 = b | 2 = d | 3 = c | 4 = a | |
| 1 = d | 2 = a | 3 = c | 4 = b |
| 5. We haven't seen a graph yet for a stationary object. How would a position, time graph indicate that an object is at rest? Explain. |
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| 6. Thoroughly describe the motions of the toy car as indicated by the graph shown. Be sure to include positions, times, and directions in your explanation. Just indicate speeds using fast, slow, and at rest. | |
In your answer to the previous question you should have noted that the car traveled faster during the 4th second than during the 2nd second. You could have even quantified its speed using numbers from the graph. The speed is just a measure of how far it travels in a given amount of time. Another way of saying this is that speed is the rate at which the car moves. There are two aspects of our description of the speed of an object that need a little more work. The first involves the direction of motion.
| 1. From the graph, calculate the average speed of the car for each one-second interval, just using the distance traveled and the time required. |
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During the 2nd second, the distance from 4 m to 8 m, is 4 m. The time involved is 1 second. Thus the average speed is 4 m/s.
Our definition of speed does fine as far as it goes, but what it leaves out is the direction of motion. Clearly the car reversed direction. But the speed fails to take this into account. Speed is a scalar.
How could we modify our definition so as to include direction, that is, create a vector version of speed? Try making the calculations for the 2nd and 4th seconds in the graph using displacement instead of distance. Recall that for linear motion the displacement is the final position minus the initial position.
| 2. You should have found +4 m/s and -8 m/s for these two motions. How is the direction of motion indicated by these numbers? |
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We define the average velocity of an object as the displacement per second.
In the form of an equation this would be:
Suppose you got A's in English and History, and C's in Math and Physics. Your average is a B. What does this average say about you as a student? Does it sum up your academic ability somehow? Actually, in this case, it doesn't say much at all. It would be better to say that you do great in English and History and not so great in Math and Physics.
When we state the average velocity of something, we have only a vague notion of what really happened during its motion. Let's look at some motions that are similar overall, but different in the details.
Watch this animation. You may need to watch it several times to catch all the details. You may also need to re-watch it to answer the follow-up questions. The next frame 
and previous frame 
buttons may be particularly helpful.
| 1. From the animation, calculate the average velocity of each car for each 1-second interval. |
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Let's look at the 2nd second for each car. We got the same average velocity for each car for this interval they had equal displacements in equal times, but they all have very different motions. So just what does the average velocity tell us?
For the red and green cars, it just gives us a general idea of the motion. You certainly couldn't use this value to predict where the cars were at say, 1.5 s. Actually, all three cars were at different points at this instant.
The orange car, however, traveled at a constant velocity of 4 m/s during this second. This means that if it maintained that velocity it would travel 8 m in 2 s, 2 m in .5 s, etc. Be careful with this distinction. The average velocity says nothing about the details of motion, unless the object has a constant velocity.
2. At what constant velocity should a fourth car travel so as to start beside the other cars at t = 0 s, and finish beside them at t = 4 s?
Hint: If it is to have the same displacement for the same 4-s time interval, it must have the same average velocity. Find that velocity. It should be the same for each car. Now visualize the motion that that velocity describes.
OK, let's watch it.
| 3. For which car would the -1 m/s value be able to predict its location at any time during the 4-s period? |
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Let's return to our position, time graph.
1. Suppose that during the 4th second the car took twice as long to travel the same distance. What change would this indicate about its motion?
Be specific - use numbers.
2. Add a line to the graph indicating this motion.
3. Calculate the slopes of the original and new lines.
Hopefully, it's obvious by now that he slope of a position, time graph is equal to the velocity.
The Slope of the Position, Time Graph
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This makes perfect sense when you compare the equation for the slope of the graph to the equation for the average velocity. The rise is just Δx, and the run is just Δt. So both the slope and the average velocity are found from Δx/Δt.
| 4. What slope (velocity) would be indicated if the car took half as long to travel the last leg of its motion? Add a line to the graph indicating this motion. |
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This graph describes the motion of a rolling ball.
| 1. Where was the ball after
0 s, 1 s, 2 s, 3 s, 4 s, 5 s? |
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| 2. Look at this horizontal number line. Imagine the location of the ball at each of the times in the table. |
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3. Describe the motion of this ball. (Solution: 
)
| 4. Describe a possible situation that might involve such a motion. |
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| 5. What about the shape of the graph seems to indicate that the ball is speeding up? |
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6. Describe the motion indicated by this graph. Be sure to mention when it's going fast and slow or stopped. You should also include directions.
7. Describe a situation where you might find such a motion.
In the previous two examples you've looked a body with a non-constant velocity. The smooth curves indicate smooth changes in velocity. So at any given time the object has some velocity. A moment earlier and later it could have had different velocities. The velocity at some instant is called the instantaneous velocity. Your car's speedometer is quite good at telling you your instantaneous velocity. For us, in the lab, it will be somewhat more difficult. We'll return to this later. For now, just be aware that we can identify a velocity at an instant.
| 8. A car has a constant velocity of 40 mph, east for 5 seconds. What was its instantaneous velocity 1.5 s into this period? |
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Solving Vavg = Δx/Δt for Δx produces
In our everyday travels this just simplifies to distance = speed x time. On a trip you might want to call ahead for reservations. Deciding that you want to drive for 8 hours on one leg of your journey at an average speed of 50 mph, you'd want to choose a motel about 400 miles ahead.
Notice that the word "ahead" doesn't imply a specific direction. The road could be quite curvey so that our direction is changing. The 400 miles is just the distance along that road. This is typical of our everyday experience where our careful attention to direction just isn't needed. We'll assume that you're already skilled at such problems.
In this class we'll concern ourselves with situations that are not so familiar, where attention to direction is crucial. We'll take direction into account by using the vector version of speed, the velocity.
In all these problems you must be careful to consider what type of speed or velocity is being provided, constant, average, or instantaneous.
For example, could you predict where you'll be on the trip mentioned above after exactly 2 hours of driving? No. The average speed is for the whole trip. As the car commercials don't say: "actual speed may vary." If, however, you could travel at a constant speed of 50 mph, then you would be correct in assuming that you'd travel 100 miles during this time. You could also claim that your instantaneous speed at a given time was 50 mph.
Let's look at the two forms of this equation graphically. There's a lot here. Take your time. Also note that in this animation "dropped hints", the falling green buttons, are not used to suggest your next click. Just keep moving to the next frame until the next frame button disappears.
| 1. Sketch our x,t graph, then add a v,t grid below it as shown here. Plot the graph of velocity vs. time that you think corresponds to the car's motion. (Solution: )
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| 2. A position, time graph indicates constant speed by a constant slope. How is constant speed indicated by the v,t graph? |
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| 3. A position, time graph indicates the direction of motion by a rising (positive velocity) or falling (negative velocity) line. How is the direction of motion indicated by the v,t graph? |
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| 4. A position, time graph indicates a zero velocity with a horizontal line. How is a zero velocity indicated by the v,t graph? |
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| 5. A position, time graph indicates that an object is speeding up or slowing down by getting steeper or less steep. What would you expect a v,t graph to look like for an object that is speeding up, slowing down? | |
The following three problems should clarify these points. Sketch a position, time and a velocity, time graph for each motion. Let all velocity changes be instantaneous. Also, all velocities are constant.
(The answers are in this animation. Each click on the brings up the next answer.)
6. A low-friction cart starts at x = 2 m and travels in the positive direction at a steady speed of 1 m/s for 6 s. It then stops for 1 s. Finally, it travels 12 m in the negative direction in three seconds. 7. A low-friction cart starts at x = -2 m and travels in the positive direction at a steady speed of 1 m/s for 6 s. It then stops for 1 s. Finally, it travels 12 m in the negative direction in three seconds. 8. A low-friction cart starts at x = 2 m and travels in the negative direction at a steady speed of -1 m/s for 6 s. It then stops for 1 s. Finally, it travels 12 m in the positive direction in three seconds. |
This lab should help you confirm what you've learned so far and extend it a bit. You'll notice that the real world situations are not quite so well-behaved as our animations. In particular, be on the lookout for curving lines.
A car travels at three different constant velocities during three successive time intervals. (Velocity changes are instantaneous.) Both the graph and the table below contain exactly the same record of this motion.
| Time Interval (s) | Velocity (m/s) |
| 0-1 | 10 |
| 1-2 | 20 |
| 2-4 | -10 |
| 1. Using the equation Δx = v Δt, and data from the table, calculate the car's displacement, during each of the three time intervals? |
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Here's an important variation on this calculation.
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The area under a velocity, time graph for some time interval is equal to the displacement during that interval.
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Here's what we have so far.
| Time Interval (s) | Velocity (m/s) | Displacement (m) |
| 0-1 | 10 | 10 |
| 1-2 | 20 | 20 |
| 2-4 | -10 | -20 |
| 2. Now that we have all the displacements, (10 m, 20 m, -20 m), drawing a position, time graph should be no problem, right? | |
So, given, the information above along with the initial position we should have everything we need to go from velocity, time information to a graph of position, time. Let's try it with the data in our table.
Let's assume that the car started at x = -20 m.
Our table isn't quite right since it uses time intervals rather total elapsed time. What we need is a table of position vs. time.
The car starts at -20 m. After 1 s it has been displaced by +10 m. So it's now at x = -10 m. During the next second it's displaced by +20 m from -10 m to +10 m. Continuing for the rest of the data we get the following table. Make sure you can account for each position value .
| Time (s) | Position (m) |
| 0 | -20 |
| 1 | -10 |
| 2 | +10 |
| 4 | -10 |
Here's our pair of graphs. Take some time to discuss how each graph can be used to produce the other.
Before you try some examples, let's see what we've learned about constructing one graph from another.
First, here's a table that sums up the information that each graph provides. It goes a little farther than we've gotten yet. The first two entries are what we've seen so far. Feel free to be inquisitive and look at the rest.
In summary, here are the steps you can use to create one graph from the other. Looking at this as a magical list of steps won't get you anywhere. You really need to understand how it works.
v,t -> x,t for constant velocity
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Now for some practice. The following four problems give you the opportunity to practice producing graphs of both types. Feel free to look back at the example above.
Don't panic! This is not easy. You are probably a bit in the dark at this point because you've been on the sideline watching. It's time to get your hands dirty.
| 1. Run the movie. Create a position vs. time graph for the red block from 0 s - 5 s. The collision occurs at 1.4 s. The block's velocity is 2.0 m/s before the collision and -.7 m/s afterwards. It starts at x = 0 m.
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| 2. Run the movie. Create a velocity vs. time graph for the red block from 0 s - 5 s. The collision occurs at (1.6 s, 2.5 m). The block starts at x = 5.7 m and ends at 4.8 m at t = 5 s.
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| 3. Run the movie. Create a position vs. time graph for the red block from 0 s - 5 s. The collision occurs at 3.4 s. The block's velocity is -.5 m/s before the collision and 1.1 m/s afterwards. It starts at x = 5 m.
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| 4. Run the movie. Create a velocity vs. time graph for the red block from 0 s - 5 s. The collision occurs at (3.5 s, 1.8 m). The block starts at x = 0.0 m and ends at .2 m at t = 5 s.
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The most familiar case of velocity addition involves walking up and down moving escalators. Let's try a couple of examples.
Hopefully you understood these examples without feeling like you were using any special mathematical technique. The techniques we want to develop are based on the same intuitive techniques you used, but by being completely general they will work for any situation, even those that are more foreign to us.
The first thing we have to realize is that the effect of the two motions is cumulative, that is, they can be added together, somehow taking direction into account. In both cases, the man's velocity relative to the store is equal to his velocity relative to the escalator plus the velocity of the escalator relative to the store. You may object and say that you added one time and subtracted the next. Our method will relieve you of having to make this distinction. And it will always work, even in two and three dimensions.
In vector terminology, the man's net velocity is:
Man relative to Building
= Man relative to Steps
+ Steps relative to Building
or
The arrow over each term indicates that it is a vector. This also informs us that this is vector addition not scalar addition which means that we have to take direction into account somehow.
Also, note the arrangement of the terms in the equation so that the subscripts in the sum, (VMB) are the first and last subscripts in the terms being added (VMS + VSB). This also works with more than two terms.
In one-dimensional situations we've found that we can just use ordinary addition if we include signs to indicate direction. This method is called analytical addition of vectors.
Let's try our two problems using this method. Let's let the "upward" direction be positive and the "downward" direction be negative.
1.
= (+2 m/s) + (+1 m/s) = +3 m/s
The man's speed, (the magnitude of his velocity) relative to the building, is 3 m/s. His direction is up the escalator. This is indicated by the positive sign of the result.
2.
= (+2 m/s) + (-1 m/s) = +1 m/s
His speed this time relative to the building is 1 m/s. His direction is still up the escalator.
Notice a couple of important points.
There are many other types of vectors that we will need to add togetherforce, momentum, electric fields, etc. Our "add with signs" method will still work, but only in one dimension.
For example, our pizza man went 2 blocks north then 2 blocks west. The resulting displacement, a vector, was the sum of these two individual displacements. We found the result to be 2.8 blocks northwest. We obviously can't get this value by adding 2 and 2.
What we need is a method that will also work in two and three dimensions. It's called graphical addition of vectors. It works by representing the vector quantities as arrows drawn to a scale such that the length represents the magnitude and the direction the arrow points indicates the direction of the vector. The arrows are then connected together in a special way to form a vector sum.
We'll just work here with one dimension. We'll apply our method later to more dimensions.
We might represent the first problem above like this. The length of the blue vector defines the scale of the drawing. The green vector must be twice as long.
It looks like connecting the vectors and measuring the length of the two together gives us the answer we expected. Is this the whole story? Let's try again, this time with our second example (2 m/s, up plus 1 m/s, down).
What would be different this time. If we just connect them together again, they'll be the same length. We have to make a refinement to our method. We don't just connect them, we connect them so that the tail of the second vector is connected to the head of the first.
The sum is a vector from the tail of the first to the head of the last.
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Graphical Addition of Vectors
To add several vectors together, connect them tail to head, in any order. The vector sum or Resultant is the vector from the tail of the first to the head of the last. The Resultant of several vectors is a vector indicating the net effect of those vectors. |
The Frame Flipper allows you to take measurements of an object's position at different instants. As you'll learn, you can determine a lot about an object's motion from just this much information. For our first use of the Frame Flipper we'll just be finding the velocity of a toy car.
There are short instructions in a window on the Frame Flipper screen. But since you've not used Frame Flipper before, select Help in the menu bar and do the practice lesson now, then come back to this lesson. You might want to print out the help screen for further reference.
Imagine the motions you'll see when the car runs
What would the motion of the car look like in each case when observed by someone riding on the belt, someone standing beside the belt?
Consider how these motions would vary when the car travels faster and slower than the belt.
In all these situations, the resultant velocity is the vector sum of the individual velocities. That is, the car moves relative to the belt, the belt moves relative to the store. The combined effect, the resultant of these, is the motion of the car relative to the store.
We'll refer to three different velocity vectors:
They're related by the following vector sum:
Let's match these terms up with the motions they represent. At the bottom of this animation we see the car as it would move on a stationary belt. At the top we see the car moving on a moving belt. (Note the light gray seam which shows the belt's motion.)
Now let's look at the real thing.
1. In this movie, watch the motion of the seam in the belt. Its motion is bs.
Since the car's motor is turned off (cb = 0), we'd get the same value for cs.
2. In this movie, the belt is at rest (bs = 0), but the car's motor is running, so the motion we see is cb.
What other vector is this velocity equal to?
| bs |
cs |
3. In this movie, which vector would have the largest magnitude (size)?
| cb |
bs |
cs |
4. Find the car's velocity relative to the belt with Frame Flipper. The Frame Flipper tells you x and t at any point. You just need two data points. That is, you need an early x,t and a later x,t.
Using Vconst = Δx/Δt you can calculate the velocity. Choose your points as far apart as possible. Why?
Don't forget to close Frame Flipper when you're finished with it.
5. Now find the belt's velocity relative to the store. You'll note that the car is at rest relative to the belt. Since it's hard to pick out a point on the belt to follow with FF, just use the car as the point to follow since it has the same velocity as the belt in this case.
//About .18 m/s
6. What should we find for cs?
Check it out with Frame Flipper. You'll notice that both the car and the belt move this time.
7. If your predictions don't fit fairly well with Frame Flipper's value, review your work and find the problem. You might also have noticed that there's a lot of error due to parallax. These movies were made 8 years before the birth of Frame Flipper. //About .95 m/s. About .04 off.
8. OK, let's try it with the car going the other way. What should you get for cs if the car is run to the left? Don't forget the sign.
Check it out with Frame Flipper. //Get about - .6 m/s, about .03 off.
9. Did your result match your prediction?
| 10. What was the meaning of the sign of the answer? |
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11. Go back to FF and measure that last velocity again, but this time with a completely different origin. That is, if you chose the origin on the left side of the screen, move it to the right, or visa versa.
| 12. The location of the origin has no effect on the position values. |
true
false |
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| 13. The location of the origin has no effect on the displacement. | true false |
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| 14. The location of the origin has no effect on the velocity. | true false |
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Here's a puzzler to finish with. Watch the top movie to get a good reminder of the speed of the belt, bs.
Now, check out the bottom movie. Did you notice that the strange-looking car in this movie was moving at an odd velocity? What must be going on?
15. Use Frame Flipper to find cs.
So, if bs = +.18 m/s and cs = +.08 m/s, then what is cb?
Note that cb is not +.26 m/s since it's not the resultant of the other two velocities. Recall that the equation is
cs = cb + bs.
16. So what do you expect to get? If +.18 plus something equals +.08, then what must that something be?
As long as we're dealing with collinear vectors we can do ordinary algebra with our equations as long as we keep signs to indicate direction. The Analytical Subtraction of collinear vectors works just like Analytical Addition. The previous problem becomes:
cs = cb + bs
cb = cs - bs
= +.08 m/s - (+.18 m/s) = -.1 m/s
where the sign of cb indicates that the car was moving to the left relative to the belt.
Unfortunately a problem arises if we try to do the same with Graphical Subtraction. Recall that we add vectors graphically by connecting them together tail to head... with the resultant from the tail of the first to the head of the last. It's not really simple to rewrite this for subtraction although you can do it. Another method works better. Let's use our conveyor problem as an example.
Rather than subtracting a vector, we add its opposite, a vector of the same length but opposite direction.
Here's an example that we'll need when we study acceleration.
Suppose a car is initially traveling at +5 m/s. It gradually speeds up to +15 m/s. Its new velocity, 2, can be written as
2 = 1 + Δ
That is, the final velocity equals the initial velocity plus the change in velocity.
1. What is the magnitude and direction of Δ?
Using analytical subtraction:
Δ =
2 -
1
Δ = (+15 m/s) + (-5 m/s) = +10 m/s.
The car's change in velocity is 10 m/s in the positive direction. (Don't worry about the meaning of the direction. We'll have plenty of time for that later.)
Here's a graphical solution..