Watch this movie a few times. What evidence do you see that indicates that the cart moves at (approximately) a constant velocity after it leaves my hand? Are there times when it doesn't move at a constant velocity? What evidence do you have of this?
(Unfortunately, your computer may have trouble showing this movie full speed. You might want to step through it frame by frame with your keyboard arrows.)
Now here's an animated look at that same motion. (The numbers don't match, but the motion is otherwise the same.)
Clearly, a constant velocity is indicated by a straight line position, time graph. Its slope is equal to the velocity. The Δx corresponding to each equal Δt is the same.
1. What will a velocity, time graph of this motion look like?
How will it indicate that the velocity is constant?
This time our cart illustrates constant acceleration. A fan on board provides a constant accelerating force. (Use your keyboard for better control.) How do you think a position, time graph of this motion will compare to the one above? How will it be different? Specifically, how will later Δx's compare to early Δx's?
Sketch your prediction of the shape of a position, time graph and a velocity, time graph for this motion.
It's a little hard to tell about this new motion. Let's look more closely using Frame Flipper.
After you collect the 10 data points, plot them with Graphical Analysis. Print out the graph and data table if you can.
The (average) velocity of an object equals Δx/Δt. Let's see what happens to the velocity as time passes.
The simplest way of describing what you've found is to say that the average velocity is increasing by about .2 m/s every .4 second or .5 m/s every second. This is what we call a constant acceleration.
This equation actually defines the average acceleration. In nearly all cases, we will consider only constant accelerations. If a body has a constant acceleration, it will change its velocity by a constant amount each second. A constant change sounds strange, but it's not new. An object has a constant velocity if its position changes by a constant amount each second.
Using our numbers above,
a = (v2 - v1)/Δt
= (.5 - .3)m/s / .4s
= .2m/s / .4 s
= .5 m/s/s
In words, the velocity changes by .5 m/s each second.
We can simplify the units as follows:
Square seconds do look a bit weird. We do this a lot with units–algebraically reducing them to their simplest form. We'll generally use the m/s2 form, but the concept of acceleration is best understood using the m/s/s form.
Also, don't confuse .5 m/s/s with .5 m/s. Discuss the meaning of each of these before you proceed.
5. What change in average velocity do you expect during the 4th .4-s interval? Check and see what it actually was?
The position, time graph of this constant acceleration is a parabola. There are many things we can determine from the shape of this graph. Let's see how the graph indicates the direction of motion and in what way the velocity is changing,
Look at the graph. The displacement, Δx and the time interval, Δt are just the rise and run for a chord connecting two points on the curve.
The slope of a chord connecting two points on a position, time graph is the average velocity for the time interval between those two points. |
So the average velocity between the times .6 s and 1.0 s is just
Note that this is exactly what we found from the data table when we first looked at this motion. Ordinarily we don't deal with data points after we draw our graphs, but our data is so good that the curve goes right through the data.
1. What's the average velocity between the instants 1.4 s & 1.8 s?
We've found that we could calculate the average velocity over some interval of time using the definition:
Is this the best we can do? In our cars the speedometer gives us the instantaneous velocity, that is, the velocity at a given instant. The speedometer doesn't do any calculations; it uses electrical effects. Could we use our average velocity equation to calculate instantaneous velocity?
How could we calculate Δx and Δt at an instant? We can't. An instant is not a time interval. This doesn't mean that there is no velocity at an instant. What it means is that our equation has met its match. It took the likes of Isaac Newton to get past this obstacle.
The following activity illustrates the mathematics necessary to describe instantaneous velocity. It also illustrates one of the most basic concepts in the calculus. Don't Panic! We won't be using calculus in this course. We just want to take a quick look at a different conceptualization of the problem we're working on and make clear that an instantaneous value for something that is defined in terms of a rate of change makes sense.
That was a lot to digest. The big ideas were:
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Instantaneous Velocity
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We need to be careful when referring to these two types of velocities - average and instantaneous. You should understand the difference between them by now. But you also need to be alert to the language. When we mean average, we say average. When we just use the term velocity we mean instantaneous velocity.
2. Under what circumstances is the instantaneous velocity at every instant equal to the average velocity?
In case you're thinking that finding the slope of a tangent is a pretty tedious & inaccurate way to find the instantaneous velocity, there are other methods that we'll find more dependable. We'll find later that we can describe these graphs with equations. These equations will give us the velocities we want with no need for graphs. For now, let's look at another way of finding it graphically.
Load the file "loacr.dat" into Graphical Analysis. It contains the data from the graph of our fan cart's motion. Try this with GA.
3. Click somewhere in the graph to "focus " on it. Click on the tangent icon 
in the GA menu bar to select it. (It looks lighter when it's selected.)
Funny thing about that, huh? Try it with another point. Just make your chord connect points an equal time interval either side of the point where you want to know the instantaneous velocity. For example, find the average velocity from 1.2 s to 1.6 s. You'll find it to be close to the instantaneous velocity at the mid-time - 1.4 s. For a perfectly constant acceleration it will be right on.
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For a constant acceleration, the average velocity over an interval is equal to the instantaneous velocity at the mid-time of that interval. |
Let's try a few calculations from our cart's motion. Just as with velocity, to calculate acceleration we need a pair of data points. In the data table you'll find some data for the fan cart. Look at how the data is arranged. The first pair of columns gives the cart's position at each instant.
1. Use this data to complete the velocity column. Use the relationship between average and instantaneous velocity you just learned. A sample value is given in the first box.
Do you notice an obvious trend that (almost) happens with this data? Look at how the velocity goes up by (almost) equal increments during each .2-s interval. That is, the velocity changes by about .1 m/s per .2 seconds. This is exactly what we mean by a constant acceleration. The cart changes its velocity by about .1 m/s per .2 second or .5 m/s per second.
(The amount of increase actually drops over time. We'll worry about why later in our study of forces.)
Since we now know the velocity at six different instants, we can calculate the acceleration between any pair of velocities. That is, since the (average) acceleration is the change in velocity per change in time, the difference in two velocities divided by the difference in the corresponding times is the acceleration.
2. Go ahead and calculate that acceleration to fill in the last 3 entries in the table.
You should have found that the acceleration was almost constant, about .5m/s2. It actually decreased a bit as we observed above.
In summary, we've found that a constant acceleration means that an object is changing its velocity by equal amounts during each equal increment of time. As we found with constant velocities, we have quantities that are changing at a constant rate and other quantities that have constant values.
Looking at graphs of each of these quantities is the best way to see these relationships. We'll do that next.
Run GA and load the file "loacr2.dat". It includes a text box that will give you directions to follow. Good-bye
GA Part 1
Hello again. You should have gotten a graph something like this.
1. What about this graph tells us that the velocity is not constant?
2. Is the velocity increasing or decreasing? How do you know?
3. Now think about a velocity, time graph of this motion. What do you expect it to look like?
Let's get GA to make us a velocity, time graph. Go back to "loacr2.dat" and follow the second set of instructions. Good-bye.
GA Part 2
Welcome back. Your new creation should should look something like this. We get a nice straight line during the middle section of the v,t graph where the acceleration is fairly constant.
The slope of this graph is Δv/Δt which is just the acceleration. The constant slope indicates that we have a constant acceleration during that interval.
4. If the acceleration is constant (in the central section), what would you predict an acceleration, time graph to look like?
Let's check it out. Go back to "loacr2.dat", part 3. Good-bye.
GA Part 3
Back so soon! Ok here's what you should have gotten. In the central region, we got just what we expected, a (sort of) horizontal line.
5. Let's get a little picky before we leave these graphs. Our v, t graph actually bends a bit at the start and finish. If the slope of the graph is the acceleration, what does the bending indicate about the actual acceleration of the cart? Be specific about each of the two bends.
This effect actually has more to do with the way GA does calculations that with our data. The curves get less and less straight because of lack of data, not because our acceleration is not constant. When we wanted to find instantaneous velocities by averaging over an interval we needed points before and after the point in question. GA is doing the same thing here. The problem is, it doesn't have data before t = 0 and after t = 1.8 s to work with. Forget about it. Too much detail.
A constant acceleration is indicated by:
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A look back at a table we started during our study of velocity would be useful. You should now be familiar with the first three entries.
1. How can you tell from the graph that the second run had a greater acceleration than the first? Be specific!
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The more quickly the position, time graph increases in steepness, that is the more sharply it curves, the greater the rate at which the velocity is changing, i.e., the greater the acceleration.
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What would velocity, time graphs for this pair of runs look like? What would be similar about them? What would be different? How would the v,t graphs indicate the accelerations of the carts?
Before we investigate these questions you need to think them through. Sketch quick graphs (no numbers) of your predictions of the x,t, v,t, and a,t graphs we'll find for each cart. (You've already seen the x,t graphs.)
We'll use GA to display our data. Run Graphical Analysis and load the file: "rt-spd.dat". You should see a data table and an x,t graph which shows both runs.
GA 1
2. What would a velocity, time graph for each cart look like? How should the velocity, time graphs compare?
Let's check our predictions. To make a v,t graph we'd need to calculate a lot of slopes. We'll let GA do all that.
GA 2
3. While you were in GA you found the (sort of) constant accelerations for the two trials. What were they?
| Accel. #1 | |
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| Accel. #2 |
There's just one step left. We need to make a graph of acceleration vs. time. You just found the (average) acceleration for each run.
| 4. What would a plot of this, almost constant, acceleration vs. time look like? |
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Go back to that predicted a,t graph that you drew at the start of this section and fix it up to reflect what you've learned. Draw it correctly, with the correct acceleration, for the interval from 2 to 3 seconds.
Now go to GA and have it create the a,t graphs. It works just like the velocity graph.
GA 3
Did you take all that in? Check So Far... for a run-down on this section. It's easy to get lost in the details.
How quickly we forget! Like displacement and velocity, acceleration is a vector. We've learned that the sign of a vector just indicates its direction. Let's use an animated version of the graphs we just created for our investigation. Use the movie controls at the bottom to move through time and collect any data you need to answer the questions below. (The ordinate labels x (or Px), Vx, and Ax represent the position, velocity, and acceleration in the x-direction, respectively.)
1. What are the positions, x1 & x2 at 1 & 3 seconds?
| Position, x1, at 1 s | |
| Position, x2, at 3 s |
2. What's the displacement between these two points?
| 3. What do the signs of x1, x2, and Δx indicate? |
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4. What are the velocities, v1 & v2 at 1 & 3 seconds?
| Velocity, v1 at 1 s | |
| Velocity, v2 at 3 s |
| 5. What do the signs of v1, v2 indicate? |
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Now the plot thickens. It's time to investigate the sign of the beast, the acceleration. Beware, this is not for the weak of heart. First we need to look at Δv.
What does the positive sign of this quantity indicate? Most likely, you answered "it's speeding up." You can't actually answer the question that simply. Acceleration is a bit more complicated than displacement and velocity.
Look at the figure below. Δv is calculated by subtracting a small positive v1 (.5 m/s, on the left) from a larger positive v2 (1.5 m/s, on the right.) The result, Δv is +1.0 m/s.
. . . 
Note that this is not a velocity. It's a change in velocity, a more abstract idea. It has a positive direction. Since the acceleration is just Δv/Δt, the acceleration is also positive. Did you get that? The acceleration always has the same sign as Δv!
So, what does a positive acceleration mean? For this case we can say that the sign indicates that the cart is speeding up in a positive direction.
Notice that there are two parts to this answer - which direction the object is going and what's happening to its speed. This gives us four possibilities for an object accelerating along a line. It can speed up or slow down, and it can do it while traveling to the right or left.
The activity below is designed to help you explore these alternatives. The whole show is run from the table below. The left column lets you watch movies of the 4 possible motions. Watch each motion and then use the buttons in the columns to its right to identify the signs and relative magnitudes of the important quantities. Each large or small sign is a button. Just click on your choice. If you choose incorrectly, you'll get a red response. Click on it to try again until you get a green response. As you get to each acceleration, think about how you got the sign and how you might interpret it in terms of the two criteria mentioned above.
All graphs are position vs. time using the UMD which is located off the screen to the left. The subscripts 1 & 2 refer to earlier and later instants in the motion.
| Δx | ||||||
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Notice that you got two positive and two negative accelerations.
| 7. Does a positive acceleration indicate an increase in speed? Does a negative acceleration indicate a decrease in speed? |
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A look at the velocity vectors in each case should make things clearer. Here they are, in the same order as in the table. The graphics in the center column are buttons.
| to right |
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+a |
| to right |
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-a |
| to left |
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-a |
| to left |
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+a |
We can sum up our results as follows:
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Sign of the Acceleration
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The table below pairs up movies of moving carts with animations of the same motions. Your task is to create position, velocity, and acceleration graphs to match the motions in the animations. These are just sketches. You don't even need to use numbers. You'll notice that the animations do not start at x=0. All accelerations are constant.
You figure out the last one, the area under the acceleration, time graph. A good way to do this would be to use the four "animations with graphs" above. Here are some buttons to bring them back.
| Graph 1 | Graph 2 | Graph 3 | Graph 4 |
1. How does the a,t graph provide the data that would let you draw a corresponding v,t graph? (You'd have to know the initial velocity.)
2. How does the a,t graph determine whether the v,t graph is rising or falling?
| 1 |  |
| 2 |  |
We're going to build three more with the help of something old (average velocity) and something new (constant acceleration.) Watch this movie of a cart on a ramp and a ball on a level table and note how each type of motion is illustrated. Use the movie controls to check out all the details.
The ball is given an initial velocity which it maintains for the whole trip. The cart starts out at the same point, at rest. Its acceleration is just enough to let it catch the ball at the end. (We'll ignore the fact that the cart goes a little farther than the ball due to the diagonal orientation of the ramp.)
| 1. How do the average speeds compare? |
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2. Sketch x,t, v,t, and a,t graphs showing both objects on each graph. You don't need to include numbers, but carefully consider when the two have the same & different values for the three variables. Also, what lines should be straight, curved, level, etc.
Here is a movie of the same motions, with graphs & numbers added. Compare it to your graphs. Fix yours up and add numerical labels to your graphs.
You can clearly see that both objects have the same average velocities since they travel equal distances in equal times.
| 3. What is the average velocity of the two objects? |
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Let's use the v,t graphs of our blocks to find the relationship between the average velocity and the initial and final velocities of an object with constant acceleration. Here's the logic.
5. Although we created our equation with a motion that started from rest, the equation is valid for any constant acceleration. Using our new equation along with equation 1, calculate the displacement of the blue block between 1 & 4 seconds. Compare your results with the values in the movie.
We now have three equations that we can use for calculating the motions of objects. We'll generally refer to them by number, in the order listed below.
| 1 | |
| 2 | |
| 3* |  |
Our three equations can be used to solve many kinematics problem, but we'll often find problems that don't have the same mix of variables as any one of these three. Let's make a couple more equations to help us with some typical problem situations. Consider the following problem.
A car, initially traveling at 10.0 m/s, accelerates at 2.0 m/s2 over a distance of 100.0 m. How fast was it traveling at the end of this distance?
| We're given: | vo = 10.0 m/s |
| a = 2.0 m/s2 | |
| Δx = 100.0 m | |
| We want: | vf |
Only equation 1 has Δx, but, to use that equation, we need vavg and the time. We need something new.
Our three equations have the four variables we want to work with spread out among them. If we do a little algebra now, we can create a new equation that includes just what we need: vo, a, Δx, and vf. Here's how it works out.
Substituting the left side of equation 3 for vavg in equation 1 will get rid of vavg and replace it with velocities we're interested in:
Δx = vavg t = ((vo + vf)/2) t
To eliminate the time we'll solve equation 2 for t and substitute what we get into the equation we have so far.
a = (vf - vo) / t, so
t = (vf - vo) / a
Δx = ((vo + vf)/2) t
= ((vo + vf)/2) ((vf - vo)/a)
= ((vo + vf) (vf - vo))/2a
Δx = (vf2 - vo2)/2a
We usually do one more step and write our new equation this way:
Let's try it out on our accelerating car problem.
| We're given: | vo = 10.0 m/s |
| a = 2.0 m/s2 | |
| Δx = 100.0 m | |
| We want: | vf |
Our new equation is already solved for vf.
vf2 = (10.0 m/s)2
+ 2 (2.0 m/s2) (100.0 m)
= 100 m2/s2 + 400 m2/s2
= 500 m2/s2
vf = +/- 22.4 m/s
We can reason that the positive result must be the correct one since the car is initially traveling in the positive direction with a positive acceleration.
We now have 4 equations. Since equation 4 was derived using equation 3, it also requires a constant acceleration.
| 1 | |
| 2 | |
| 3* | |
| 4* | |
Let's find one more equation. Consider this problem.
A car traveling at 5.0 m/s begins to accelerate at 4.0 m/s2. How far will it travel during the next 6.0 seconds?
| We're given: | vo = 5.0 m/s |
| a = 4.0 m/s2 | |
| t = 6.0 s | |
| We want: | Δx |
Again, we find that no single equation has the right blend of variables. We could solve our problem as we did before by deriving another equation algebraically. Let's use our graphing skills instead.
Watch this animation to see the step-by-step derivation. Be sure to justify each step. The displacements come from the areas within the lower rectangle and the upper triangle. Note that the height of the triangle is vf - vo. Do you see why?
Let's use our new equation to solve our problem.
| We're given: | vo = 5.0 m/s |
| a = 4.0 m/s2 | |
| t = 6.0 s | |
| We want: | Δx |
Our new equation is already solved for Δx, so let's put the numbers in.
Δx = (5.0 m/s) (6.0 s)
+ 1/2 (4 m/s2) (6.0 s)2
= 30 m + 72 m = 102 m
Let's animate this problem to check it out. Notice (with your imagination) the small 30-m rectangle at the bottom and the much larger 72-m triangle above. (Note that Vf = 29m/s.)
Here's our complete list of kinematics equations. They are extremely versatile at solving problems. We'll find them useful throughout the course.
| 1 | |
| 2 | |
| 3* | |
| 4* | |
| 5* |  |
| 1. What would happen if we made the ramp steeper? |
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If we tried successively steeper ramps, we'd find successively greater accelerations. Is there a limit? Well, you can't get any steeper than a vertical ramp, that is, a vertical drop.
A body falling straight down moves only under the influence of gravity and air resistance. For fairly short falls by massive objects, the effect of air resistance is negligible. We'll ignore friction for now.
How about a body thrown straight up? What sort of motion would it have?
You've got the tools. You figure it out. Your task is to investigate free fall - the motion of a body under the influence of gravity alone. Here's a movie of the motion you'll work with. Be sure to use your left and right arrow keys to step through this movie.
1. Use Frame Flipper to get some position, time data (y,t where y is the vertical position). Amazingly, the blurry image does not cause much of a problem. Use GA to analyze the data. You'll need graphs of y,t, v,t, and a,t. Look back at part 3a if you have problems.
2. Show that the acceleration is (approximately) constant? How do you know, based on your v,t and a,t graphs.?
3. What is the acceleration?
4. What does the negative sign of the acceleration indicate?
Part II - An Object Thrown Upward
Toss an object up and down, catching it at the bottom. How does the trip up compare to the trip down?
5. Our falling object had a negative acceleration. What would be the sign of the acceleration while an object is going upward? Remember, it's slowing down.
6. Let's find out, but first, make some predictions. Watch this movie of an object thrown up & down. Sketch y,t, v,t, and a,t graphs for this object. Ignore the time while the object is in contact with the hand.
7. Use Frame Flipper to get a set of data for the complete trip. Use GA to produce a set of three graphs as you did before.
8. How does the acceleration while rising compare to the acceleration while falling? Explain your answer using the v,t graph and the a,t graph.
9. The ball is certainly at rest at the top of its motion. (How do the y,t & v,t graphs indicate this?) Does this mean that its acceleration is zero there? Look at your graphs. What do they say?
What we've found is that, in the absence of friction, vertical motion is motion at a constant acceleration. Note that the term vertical motion does not imply motion in any particular direction, up or down. The acceleration due to gravity, g is 9.8 m/s2 which we'll sometimes round off to 10 m/s2. Its direction is always downward, in the negative direction.
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g = -9.8 m/s2 |
1. Using GA, look at your y,t graph with the tangent tool turned on. Compare the slope of the graph at some height on the left side (rising) with that on the right side (falling). Repeat for some other heights. (You won't be able to get identical heights. Just get close.) State your observations.
2. Now pick a couple of vertical positions on your y,t graph. Turn on the examine tool in GA. Note the time the ball takes to rise between the two chosen positions. Now compare this to the time it takes to fall between these two positions. (Again, this will just be approximate.) State your observations.