CAPM Unit 3 Summary

In this unit entitled CAPM, or Constant Acceleration Particle Model, we were finally able to learn about acceleration and how it relates to velocity,  a topic we had held out on all year. This unit, in a way, was a more advanced continuation of the CVPM Model we did at the beginning of the year, as we took the same concepts and added in more elements to fully connect everything together. We covered new topics such as acceleration, instantaneous velocity, acceleration vs. time graphs,  and finding displacement from a velocity time graph. 

CAPM.1: 

Instantaneous velocity by definition is the velocity of an object in motion at a specific point in time. So that means that one would find the instantaneous velocity at 2 seconds to find out how fast the object was going RIGHT AT 2 seconds, not at any other time. 

Instantaneous velocity can be calculated 2 ways:

1. when using a position vs. time graph, one can find the slope of the tangent at a given point

2. use the formula: v final= at+ v initial

EXAMPLES (as related to the methods above): 

1.In the position vs. time graph pictured below,  the object is shown with constant acceleration, meaning it is speeding up, and traveling in the positive direction (how I know this will be explained later).  Because this is a position not a velocity graph,  you must calculate the instantaneous velocity using the slopes of the tangent, which are the points on either side of the desired midtime. 

SOLVING:

In this particular example, I am calculating the instantaneous velocity at 2 seconds, which then becomes the midtime. I will then use the x,y      coordinates at 3 and 1seconds, plug it into the formula change in x divided by change in time and get an answer in m/s (meters per second), which will be the instantaneous velocity at 2 seconds. The work is shown in the photo below. Notice that the smaller number is always subtracted from the larger number.

2:  The second method that can be used to solve for instantaneous velocity is when you simply have to plug values into the formula:                     v (velocity)= (a {acceleration})(t {time}) + vi (initial velocity).                                                                                                                   THIS EQUATION CAN ONLY BE USED WHEN THERE IS A STRAIGHT LINE!!!!!!  

SOLVING: (Acceleration has not been described, nor has solving it been displayed yet in this blog, but for the sake of this problem it will              be briefly overviewed.) 

One has the same goal when using this formula: to find the specific velocity of an object at a given time. Using the graph pictured below, I will show how to find the instantaneous velocity at 6 seconds.

a. the first step is to solve for the acceleration of the object over the course of the whole period using the formula acceleration=                    change in v divided by change in t (time). Basically,  written out longhand it is final velocity- initial velocity/ final time- initial time.  Using the graph above as a reference and the picture below to see the work written out, the final velocity of the object was 0 m/s, the initial was -15 m/s, all over the final time which was 8 seconds - the initial time, 0 seconds.  This got me an answer of about 1.87 m/s^2, which are the units for acceleration. 

b. calculating the acceleration would have to be the hardest part, because now all that is left is to plug the given values into the formula v=at+vi. One must be careful to pay special attention to the exact value the problem is asking for because it is very easy to plug in the wrong time. For this problem, since we are finding the instantaneous velocity at 6 seconds, 1.87 is plugged into a, 6 is plugged into t, and -15 is added at the end as the initial velocity of the object. In all the formula would look like this: (1.87)(6)+(-15). I would get the answer of -3.78m/s as the instantaneous velocity at 6 seconds.

CAPM.2:  To continue our study on displacement, which is the shortest distance from the initial to the final position of an object, we learned 2 new methods when working with velocity time graphs. 

1. finding the area under a velocity vs time curve

2. using the mathematical model x= 1/2at^2 +(v initial)(time)

EXAMPLES (as related to the methods above): 

1: When using a velocity vs. time graph, one can calculate the area under the line in 2 ways using the formula l (length) x w (width), or the formula 1/2 ( b {base})(h {height}) + (length)(width).  One would only use just length times width when there is a straight line, meaning constant velocity on the graph, because that is the formula for area for a rectangle, which is the only shape made from a straight line. The other formula would be used when there is a slanted line, and a triangle, which used the formula 1/2bh to find the area, and a rectangle can be formed, which will be shown in both examples below. 

a. FINDING DISPLACEMENT USING L x W USING A RECTANGLE

The graph below shows a straight line, meaning the object is traveling at a constant velocity. As you can see, I have drawn a straight 

line down to the axis to show myself that only a rectangle is present, meaning that I only have to multiply length times width to find the displacement. The length of the rectangle is 5 and the length width is 2, multiplied together gives me 10 meters, which is the displacement.

b. FINDING DISPLACEMENT USING 1/2bh USING A TRIANGLE

This example is just like the one above except this time. only a triangle is formed under the curve because the line touches the X axis, so          only the formula 1/2bh needs to be used to find the displacement.  For this particular graph one would use 1/2(2)(5) which would get 5m for the displacement.

c. FINDING DISPLACEMENT USING 1/2bh + L x W USING A TRIANGLE AND RECTANGLE 

The graph to the right shows a slanted line, meaning that the object is not traveling in one constant velocity, it is speeding up. That means that I have to use the formula 1/2bh + L x W because both a rectangle and a triangle can be formed under the curve as you can see labeled. Now all that is left is finding the values to plug into the formulas. For the triangle, the formula would look like this: 1/2(5)(3), and for the rectangle it would look like this: (5)(1). Altogether, the formula would be: 1/2(5)(3) + (5)(1)= 12.5 meters which is the displacement. 

2. You can also use the formula 1/2(a{acceleration})(t{time})^2+(v initial)(time). This formula can be used when the acceleration is constant.

SOLVING: Using the graph to the right as a reference, the displacement will be found using the new formula. As overviewed earlier and will be overviewed again until acceleration is fully described later on, the first step to completing this problem is finding the acceleration of the object. Once again, use the formula final velocity- initial velocity divided by final time - initial time.  As can be seen in the work below, I used 0 -(-3) divided by 5-0 which is 3 over 5 which gave me .6m/s^2 as the acceleration. Now I need to plug that value into the rest of the equation along with the other numbers. I plugged the numbers in like this: 1/2(acceleration=.6)(total time=5)^2+(initial velocity=-3)(total time=5), without the explanations in the equation it would look like: 1/2(.6)(5)^2+(-3)(5). The answer is -7.5 which is the displacement.

CAPM.3: 

By definition, acceleration is the rate of change of velocity of an object. In this class we have looked at acceleration as whether or not an object is speeding up. There are a few ways to find acceleration based on graphs, but the main ways we have used in class include finding the slope of a v vs t graph which is basically the same as using the mathematical model that was used earlier which was r a=change in v (final velocity-initial)/ change in t(final time-initial).  The example below shows the work and the graph of a problem that asks for acceleration. 

Other ways of finding acceleration including rearranging the mathematical models that were used in the previous example: x=1/2at^2 +(vi)(t), and final velocity= (a)(t)+ v initial. You simply manipulate the equation so you are solving for acceleration. 

EXAMPLE: 

1. x=1/2at^2 +(vi)(t)- subtract (vi)(t) and move it to the other side, then divide both sides by 1/2(t)^2 so a is isolated

2. velocity= (a)(t)+ v initial- subtract v initial from both sides as well as divide both sides by t which leaves a alone on the right side

CAPM.4/CAPM.5:  These 2 sections have been combined because they both involve position-time graphs, velocity-time graphs, and acceleration-time graphs, and how they relate with each another. 

POSITION VS. TIME GRAPH: Below is a picture of all the different position time graphs we have learned about throughout the course of this unit and CVPM. The corresponding numbers will describe what the object is doing in each graph.

1. the object is at rest, there is no movement 

                                                          2. the object travels in the positive direction at a constant velocity

                                                          3. the object travels in the negative direction at a constant velocity  

                                                          4. the object is speeding up in the positive direction

                                                          5. the object is slowing down in the positive direction

                                                          6. the object is speeding up in the negative direction

                                                         7. the object is slowing down in the negative direction

A line with a positive slope on a position time graph is always going forward, one with a negative slope is going backward, and the steepness of the slope determines how fast the object is going with a horizontal line meaning the object is at rest.

REMINDERS:

• finding instantaneous velocity: use the tangent of the slope, or one point on either side of the desired midtimet using the formula: x. Finding average velocity is the same as finding the instantaneous velocity of the midtime using the same formula, change in x over change in t!!

VELOCITY VS. TIME GRAPH:  below is a picture of the different types of velocity time graphs.  Above the x axis represents the positive direction, and below the axis represents the negative direction, which corresponds with the slope of a position time graph.  As the object moves farther from the x axis, it begins to speed up, and when it moves closer to the x axis, it slows down, when the motion is constant the graph has a straight line. The very first graph below shows an object going in the positive direction speeding up, (moving away from the axis), the second one shows an object going in the positive direction slowing down, (going towards the axis), the the third dhows an object going in the negative direction slowing down, and the fourth is an object in the negative direction speeding up. Using this information, one can construct a velocity time graph that corresponds with the position time graph, but one can also construct a position time graph from the velocity time graph information as well, the key is simply to remember the rules from each graph. 

CONSTANT VELOCITY:  

REMINDERS: 

• determining acceleration: use the same formula as before, v final- v initial/ t final- t initial, but the only difference is, this time you can pull the velocities straight from the graph without having to calculate velocity like from the position time graph 
• determining instantaneous velocity: because this is a velocity time graph, there is no need to take the tangent of the line because there are already velocities on the graph, so if you would like to find the instantaneous velocity at 2 seconds, for example, you would just find the specific velocity at the corresponding y point on the graph
• determining displacement for a given time interval: one would use the formula 1/2(a)(t)^2+vi. After calculating the acceleration and the initial velocity, the time is the difference between the time intervals, for example, between 2 and 6 seconds would be 4 seconds. Once you have figured out all those values, you would then plug them into the formula as pictured below:

ACCELERATION  VS. TIME GRAPH:  acceleration-time graphs are drawn based off of velocity-time graphs and it tells us whether the acceleration is positive or negative. Once again, above the x axis is positive acceleration and below is negative acceleration, and where the line is drawn is based on the SLOPE of the velocity-time graph, not the direction or location of the line, which is a common mistake. The acceleration line is always straight, it will never be diagonal, and if the object is traveling at constant velocity or if it’s not moving at all, the line will be drawn on the x axis representing 0 acceleration. 

MOTION MAPS: The motion maps we completed for this unit were much less specific than the ones we have done in the past. These maps are more focused on the speed of the object represented by the length of the vectors, rather than the specific location the object is located. When constructing these graphs, however, it is important to label which side is positive and which is negative and draw the direction of the vectors specific to the graph. When the object has positive acceleration, you would draw the vectors going towards the positive end of the map getting longer as they go, and vice versa if there was negative acceleration. 2 example pictures are shown below. So far, we have only learned about constant acceleration, so the motion map for acceleration will always have equal vectors.

CAPM.6: Below are a few examples of how to construct position time, velocity time, acceleration time, and motion maps when given written information. 

1. The object travels forward in the positive direction at constant acceleration speeding up

2. The object travels backward at constant velocity

3. The object travels backward in the negative direction slowing down 

CAPM.7:  When you want to solve for the amount of time it took for an object undergoing constant acceleration using the formula x=1/2at^2 +(v initial)(time), you simply manipulate the formula to isolate t which would look like the picture below.