 |
Copland CollegeYear 11 - Physics |
Linear
Motion including Speed, Velocity and Acceleration Reference Material: Your text has some excellent material to help you with this topic. Chapter 4 - Graphical Analysis of Motion: Complete the practice exercises on pages 60 to 62. Answers to these will be found at the back of the text. You should also work through the Questions and Applying the Concepts sections at the end of the chapter. Answers will be provided for these on the notice board at school and on this site. Our work on the equations of motion is based on Chapter 3- Motion in a Straight Line. Information on which questions you should do will be posted here soon. Graphing is probably one of the most important mechanisms to examine relationships between sets of physical measurements. Most of the graphing that we will do is done using Excel. By now most of you have completed a number of graphs using this program. If you have lost your data for the trolley practical you can download it by clicking here. Graphing so far: With the trolley prac that you are now analysing you have complete the graphs of Displacement vs Time. You would have noted that in general a curved line results up to the time that the mass struck the floor. At this point the graph continued as a straight line. This indicated that the object has ceased accelerating and continues along at more or less constant velocity. Of course the graph does some very strange things after the trolley reaches the end of the table. Conversion of a Displacement vs Time Graph to a Velocity vs Time graph. Typical data and the associated graph are shown below.
Your task was to recognise that up to the point where the mass strikes the floor the trolley is accelerating. Beyond this point the trolley moves at constant velocity. The slope of the line on a displacement vs time graph represents the velocity of the object. Our task is to determine the slope of the graph at various time intervals. In the graphic below you will see that the slope of the various parts of the curve is represented by the arrow E in triangle EFG. As you can see the slope of this line has changed, increased, for each time interval shown. To determine this slope we use the Excel spreadsheet and the formula for the slope of a straight line, i.e., Slope = (y2 – y1)/(x2-x1) Remember it does not matter which point is identifies as 1 or 2.
In the example shown below the points at 0.1 second is defined as x2 and y2 and the point at time 0.0 is defined as x1 and y1
The problem now is that the slopes that you have calculated refer not to the slope at time 0.0 or time 0.1 but at some intermediate point. This point we can assume is about 0.05 seconds. For each subsequent calculation the same will apply. See below.
Our task is now to plot Time (adj) vs Slope. Of course the slope that we have just calculated for each interval is really the average velocity for the time intervals 0.0 to 0.1, 0.1 to 0.2 etc. The graph generated looks something like the graph shown below.
The graph shown above has a small problem in that the early data shows a line which is not straight. Nevertheless this may not be the case with your data. What the line shown above indicates is that the trolley has some initial velocity at t = 0. This is of course not the case and is probably associated with some uncertainties associated with the measuring technique that we used. The trend line for the graph is generated in the graph as shown below. From this we can determine the slope to be 240.71. The slope of the line on a velocity vs time graph represents the acceleration. The acceleration of this trolley is 240.71 cm/s/s. This tells us that the trolley increases its velocity by 240.71 cm/s every second. Note here that we have ignored the fact that the equation shows us that there must have been some initial velocity.
You will note that the last three points on this graph are essentially a horizontal line. This indicates that the velocity remains constant after the mass strikes the floor. That is that the acceleration is zero, zero slope. An Alternative Approach. The analysis of ticker tape can be very tedious indeed if we follow the approach shown above. An alternative approach is to consider one of the basic equations of motion. This equation is shown below, d = 1/2 at2 This equation which will be discussed in great detail in class relates the displacement an object will travel knowing its acceleration and the time. In our situation we are more concerned with determining the acceleration given a known acceleration and a known time. Upon rearranging the equation becomes, a = 2d/t2 This equation has much similarity to the equation for the slope of a straight line, that is slope = vertical rise divided by horizontal run. If we plot 2d vs t2 the a straight line of positive slope should arise if the motion is uniformly accelerated motion. Go back to your spreadsheet data and create a 2d column and a t2 based on the original displacement and time data. See diagram below.
The graph generated by this data is shown below,
Note that a trend line has been generated using the same points as for the velocity vs time graph. Here the slope is found to be 319.53. This represents the acceleration of the object. The value previously determined was 240.71 cm/s/s. Although the agreement could be better at least they are reasonably close. Your data may be better. Area beneath a Velocity vs Time graph. The final activity relates to the area beneath a velocity vs time graph. What physical significance does this have. If we consider an object moving at constant velocity. In the following example the object moves at a constant velocity of 4 m/s for the time interval shown. Note here that the initial velocity is also 4 m/s
Looking at the graph we see that the slope is zero indicating that there is zero acceleration. If an object travels at 4 m/s than after 1 second it has travelled 4 metres, after 2 seconds it has travelled 8 metres and so on. Top find out how far it travels after 8 seconds we just multiply 4 by 8, that is 32 metres. Multiplying 4 by 8 is really multiplying the height by the length. This is the area beneath the line. The same analysis can be applied to lines that have a slope other than zero and even to curves. Your task now is to take your velocity vs time graph for your trolley and determine the area beneath that part of the line that has a constant slope. Remember there are two parts to this, see diagram below.
To calculate the area of A you will need to use the area of a triangle formula and for area B the area of a rectangle formula. This value is then compared to the actual distance travelled by the trolley in this time. For the graphs shown here the total area is approximately 90, that is the trolley has travelled 90 cm. The actual value was about 120 cm. Again your values may show a higher correlation. Summary: From this experiment you have found that the slope of a displacement time graph represents the velocity of an object. The slope of a velocity vs time graph represents the acceleration. Also the slope of a 2d vs t^2 graph represents the acceleration as well. The area beneath a velocity vs time graph represents the displacement of the object. Your task is to ensure that all graphs have been completed. A report will be required to be pasted into your prac book. In this reports you will have an aim, a brief introduction outlining the background an theory of the experiment. Then you will need a method concentrating on diagrams. In your results/discussion section you will need to include all graphs and outline what you have found. In your report you will also need to include answers to the Further Activities section that is shown below. Further Activities: Answers to these activities need to be included in your report of the Trolley Practical Position vs Time: The following simulation will allow you to examine the shape of a position vs time graph for an object and its relationship to the type of motion the object is undergoing. The "type of motion" refers to whether the object is stationary or moving at constant speed in a straight line either towards or away from the starting point, or whether the object is speeding up, that is accelerating or slowing down, decelerating. Click on the diagram below to open the simulation. The position, in relation to the starting point, is plotted on the vertical axis and the time is plotted on the horizontal axis. The horizontal line represents the origin. You are able to start the bike either before the origin or after the origin.
Once you have opened the simulation do some experimenting and then see if you can interpret the following graphs. You need to describe what is happening in each section of the graph that has a different slope. Ignore slight irregularities. Motion 1:
Motion 2:
Motion 3:
Motion 4:
Velocity vs Time: In the following simulation you can examine the effect of altering three factors associated with the straight line motion of an object. In the first graph the Xo refers to the objects initial position relative to some origin defined by the vertical axis. You can alter this position up or down. The Vo refers to the velocity or speed of the object. The ao refers to the objects acceleration. All quantities can be either positive or negative.
Questions: Answer the following questions relating to this simulation. Before answering these questions set the value of Vo and ao to zero. 1) Describe the effect on each graph of altering the value of Xo from a negative value through zero to positive values. Focus particularly on the slope or gradient of the graph. 2) What happens on each graph as you alter the value of Vo from negative through zero to positive values? What is happening to the object when the value of Vo is negative? 3) The starting acceleration of the object is represented by ao. What effect does the change of acceleration from negative through zero to positive have on each graph? What is negative acceleration? To see how good you are at interpreting position vs time graphs click on the following diagram.
Your task is to determine which graph correctly describes the motion of the car. Work through the task as described on the web site. You will need to use the step buttons so that you are able to record numerical data. Holding the cursor at any point over the graph and pressing the left mouse button will give values for the y and x coordinates. Remember to determine the velocity of an object you need to recognise that velocity is described as change of position divided by change of time. The Greek letter delta is used to represent change. Click on the diagram below to find out more about the use of delta. "Change" always represents the final position minus the initial position.
Change in time and change in velocity (speed in a straight line) are shown in the following formulas.
To see how good you are at interpreting velocity vs time graphs click on the following diagram.
As with the position vs time graph your task here is to determine which graph represents the motion of the car. Acceleration is defined as change in velocity divided by change in time. Change of (delta) is always final minus initial situation. In the following diagram you will notice that you can alter the initial position, velocity and acceleration of the object. You will also notice that graphs of these three quantities vs time are shown. You will also notice that a new term appears. This is the term VECTOR. In this case the vector of velocity or the vector of acceleration can be displayed. The vector is represented by an arrow that is attached to the car. The length of this arrow represents the magnitude of the velocity or acceleration and the way the arrow points represents the direction of this the velocity or acceleration.
|
