Experiment E1010 Constant Velocity in One Dimension

Figure 1. A bicyclist rides from left to right across the video screen. Discussion The Model Marking the Video Graphical Analysis

Discussion

Figure 1 shows a picture of a bicycle moving from left to right across the video screen. This is the positive x direction.

An object's position is located in the video using an x-y coordinate system. When you move the mouse from left to right in the video, you will notice the x values increasing at the bottom of the screen. When you move the mouse from bottom to top in the video, you will notice the y values increasing at the bottom of the screen. The x position is zero at the left side of the video and the y position is zero at the bottom of the video.

If an object is moving right in the video, then we say it is moving in the positive x direction. If an object is moving left in the video, then we say it is moving in the negative x direction. If an object is moving up in the video, then we say it is moving in the positive y direction. If an object is moving down in the video, then we say it is moving in the negative y direction.

When you play the video, or step through the video, notice the time changing at the bottom of the screen. When you mark an object in the video, the computer records its position and time.

The computer uses SI units for time, length and mass. In the International System of Units (SI), the second (s) is the unit of time, the meter (m) is the unit of length, and the kilogram (kg) is the unit of mass.

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The Model

Definitions:

1. t is time, in seconds.
2. x is the x position, in meters, at time t.
3. x_o is the x position, in meters, at t = 0. The subscript, o, is an identifier and means this is a special x, the x value when time is zero.
4. D = x - x_o is the definition for displacement in terms of position. In words, displacement is a measure of where the object is located with respect to its location at t =0, or how far it has been displaced (moved) from its original position.
5. v = Dx/Dt is the definition for velocity in terms of position and time. In words, the velocity is the slope of a position-versus-time graph or the slope of a displacement-versus-time graph.
6. v_o is the velocity, in m/s, at t = 0. Again, the subscript, o, is an identifier and means this is a special velocity, the velocity when t = 0.
7. m is the object's mass in kg.
8. p = mv is the object's momentum, in kg^.m/s. The momentum is the product of the object's mass and its velocity.

Using these definitions, the one dimensional equations for constant velocity in the x direction are:

1. x = x_o+ v_ot
2. D = x - x_o or D = v_ot
3. v = v_o
4. p = mv_o

Notice that if you know x_o, v_o, and m, then you can write all the equations as functions of time. Also, notice that v = v_o when the velocity is constant. Constant means, no change, does not change, zero change, stays the same, etc. Therefore, v and v_o are interchangeable in the equations above. This will not be true when the velocity is not constant.

In Equation 2, the displacement is also written as:

D = v_ot

This is the area under a velocity-versus-time graph. It is a rectangle with the height equal to v_o and the width equal to t. The area of this rectangle is (height)(width) = v_ot with units of (m/s)(s) = m. Therefore, the displacement can also be defined as the area under a velocity-versus-time graph.

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Marking the Video

1. Play the video and observe the motion.
2. Starting at time zero, put the mouse cursor on the front bicycle wheel's center and press the left mouse button. A red circle will appear.
3. Step the video forward by pressing the right mouse button while the cursor is in the video window, or by pressing the left mouse button on the STEP button. Put the mouse cursor on the center of the bicycle wheel and press the left mouse button again. Another red circle will appear. Continue marking the bicycle wheel until the video does not step any more.
4. What are your observations? Is the bicycle moving in the positive x, negative x, positive y, or negative y direction(s)? Describe the spacing of the circles. Select the Lines menu option in the Menu bar above the video, then select Vertical. You should see vertical lines passing through all the data points. Your video should now look similar to Figure 1 above. Describe the spacing of the lines.
5. Press the Circles menu item and then the Vectors-Direction of v sub item. Lines are drawn from the data circles in the direction of the velocity. Notice that all the lines go in the same direction and that they are all the same length. This is a characteristic of constant velocity. Now use the Circles menu item and the Vectors-Direction of p sub item. This is the direction of the momentum. Notice that it has the same direction as the velocity.

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Graphical Analysis

Note: To review linear plots and constant velocity graphs, run the Graph Tutor software that came on this CD.

1. Press the GRAPHS Button and the Data Analysis Choices will appear. Pick Option 1, then click the Next button.
2. Direction of Motion? Pick x since the bicycle is moving in the positive x direction. Click the Next button.
3. From the plots options, select position, velocity and momentum. Click the Next button.
4. The first thing you will see is a cover sheet. Write down the mass. Using the Graph>> and <<Graph buttons, you can step through the various graphs. Using the Play, Step and Back buttons you can control the video. Notice that stepping through the video causes a cursor to move in the video along with a line on the graph. This is to help you correlate the motion in the video with the data on the graph.
5. Look at the position-versus-time graph. It should look like a straight line with a positive slope. The slope of a position-versus-time graph is velocity. The units of the slope are meters/second (m/s) and these are the SI units of velocity. The variable used for velocity is v. You should note that the slope is positive and constant. Since the slope of this plot is velocity, then this means the velocity is positive and constant. How does this correlate with the motion in the video?
6. Move the mouse around on the graph and notice the numbers changing above the graph. Find two points on the line, then find the slope of the position-versus-time graph. Write your answer as v = (your answer) m/s.
7. Now find the vertical intercept. This is where the line crosses the position (vertical) axis. Since this is a position, it has units of meters (m). Write your answer as x_o= (your answer) m. The subscript, o, is an identifier. It tells you that this is a special position, the position of the bicycle at time = 0 s. You now have a slope and an intercept. In math, you probably wrote this as y = mx + b, where m = Dy/Dx is the slope and b is the y intercept. For this graph the slope is v = Dx/Dt and has SI units of m/s and the intercept is x_o in meters. Write the equation for this graph as x = x_o+ vt with your numbers for x_o and v. Note that the product of vt has units of m. Does this makes sense? Also note that physicists write this equation with the intercept first on the right side.
8. Check your equation by substituting at least two times off the graph. Do your answers match the graph at these times? You can also check your equation with the computer. Move the mouse close to the first data point on the graph, then click the LEFT mouse button. Next, move the mouse close to the last data point on the graph, then click the RIGHT mouse button. You should see the slope and intercept below the graph. How do these numbers compare to yours?
9. Click the Graph>> button and the displacement-versus-time graph will appear. Displacement is how far the object has moved and has SI units of meters. Notice that the displacement is zero when time is zero. The slope of this graph is also velocity, however, the intercept is zero. Therefore, the equation for this graph is D = vt. Using your value of v from above, calculate the displacement for at least two times off the graph. Do your answers match the graph at these times?
10. Click the Graph>> button and the velocity-versus-time graph will appear. Does the velocity appear fairly constant? Is the value close to your calculation of v from the position-versus-time graph? Note how this graph looks with respect to the position and displacement graphs. Also note how this graph looks for the motion in the video.
11. Estimate the area under the velocity-versus-time graph for two times. What are the units of this area? Use your equation for displacement (D = v_ot) to find the displacement at the same two times. How do your results compare? What is the area under a velocity-versus-time graph? You can also use the Area button to calculate the area under a graph.
12. Click the Graph>> button and the momentum-versus-time graph will appear. Does this graph have the same shape as the velocity graph? Does this graph have the same numbers as the velocity graph? Momentum is the product of mass and velocity and has SI units of kg^.m/s. Physicists use the variable, p, for momentum and write it as p = mv. If you multiply the velocity graph by the mass, then you will get the momentum graph. If you divide the momentum graph by the mass, then you will get the velocity graph. Try this. If you forgot to record the mass, then use the <<Graph button to return to the Cover Sheet.
13. Click the Graph>> button and you will see a data table similar to Table 1 below.

First notice that x = 0.247 m when t = 0 s in the table. This is x_o, so you can write x_o = 0.247 m. Next notice that D = 0 m when t = 0 s. Also notice that D = x - x_o in the table. For example:

At t = 0 s, D = x - x_o = 0.247 - 0.247 = 0 m.

At t = 0.4 s, D = x - x_o = 1.379 - 0.247 = 1.132 m.

At t = 0.7 s, D = x - x_o = 2.264 - 0.247 = 2.017 m.

Table 1. A data table for constant velocity in one dimension.

The last calculation gives D = 2.017 m, while the data table shows 2.016 m. This difference is due to formatting in the table. The computer uses more significant digits than are shown. So, while you may not get the same numbers as shown in the table, you should be reasonably close.

Since velocity is the slope of position versus time, v = Dx/Dt, the computer calculates velocities using position and time data pairs. However, to get a better velocity at a specific time, the computer uses values before and after that time. For example:

At t = 0.1 s, v = (0.808 - 0.247)/(0.2 - 0) = 2.805 m/s.

At t = 0.4 s, v = (1.676 - 1.093)/(0.5 - 0.3) = 2.915 m/s.

At t = 0.6 s, v = (2.264 - 1.676)/(0.7 - 0.5) = 2.94 m/s.

Again, while you may not get the same numbers as shown in the table, you should be reasonably close.

Finally, the computer calculates momentum using p = mv. The mass in this experiment is 40 kg so:

At t = 0.1 s, p = 40(2.8) = 112 kg^.m/s.

At t = 0.4 s, p = 40(2.91) = 116 kg^.m/s.

At t = 0.6 s, p = 40(2.94) = 118 kg^.m/s.

Is the bicycle's velocity reasonable? Convert the velocity to mi/h and to km/h using these conversions:

1 mi = 1609 m

1 km = 1000 m

1 h = 3600 s

Here h is the variable used for hour.

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