1 2012 ppt semester 1 graphing review
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1 2012 ppt semester 1 graphing review

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  • Welcome to the physics review for semester one. Hello everybody, this is Dr. Fiala. The tutorial you are about to begin is comprehensive. Material reviewed includes Kinematics, force, projectiles, momentum and impulse. To receive the most benefit from this video there are two important points that should be noted. <br />
  • Point one. I have color coded the slides so you can more easily practice skills with which you have difficulty. This is a good way for you to roll up your sweatpants. Word problems are presented on red slides. Data tables can be found on olive slides. Green slides work you graph reading skills and tan slides match concepts. Solutions are presented for each problem on the following slide. I will give you an audio cue that the slide will be changing. That way if you want to stop the video before you see the solution you can do so at that time. You may choose to view the entire video or simply move to the slide color representing your weakness. <br />
  • Point two. You must have all of the formulas handy while going though the semester on physics review. If you have them written down in my order it would be helpful because I refer to the formulas as formula one, formula two, formula three and so on according to the order in which they appear on my card. Because all students must use the formula given out on final exam day, it is important that you are skilled at finding any variable in any formula. You can stop this video at any time and I encourage you do do so. I suggest that you complete this video review and tutorial when you can think about what we are talking about without distraction. Stay positive . You can do this! <br />
  • Lets take a look at our first graph of the review. Eventually you should be able to shut off the sound on this video and talk your way through the graph as we have learned to do in class. Okay, here we go. Horizontal position is plotted on the y-axis and is measured in meters. Time is plotted on the x-axis and is measure in seconds. <br />
  • The y-intercept indicates an object started 3 meters to the left of the reference point. There is no x-intercept. The area (rise times run) of this graph is not useful to us because its units are meters times seconds. The slope of this graph represents the velocity of the object and is measure in m/s. As you can see there are four different slopes on this graph. Slope is equal to rise divided by run. To find the magnitude of the rise of the first slope we take our final position, which is -6 meters and subtract our initial position, which was -2 meters. The magnitude of the rise is -4 meters. To find the magnitude of the run of the first slope we take our final time, which is 2 seconds and subtract our initial time, which is 0 seconds. The magnitude of the run is 2 seconds. So the magnitude of the first slope is -4 meters divided by 2 seconds or -2 m/s. Again we are learning procedure so find the remaining three slopes using the same procedure. <br />
  • So now we can analyze a kinematic graph for all of its useful information. Remember to keep reading slopes out loud. The first slope on this graph would be read as negative (because of its downward slope) constant (because it’s a straight line) velocity (because that is the physics quantity this slope represents). In this case we also know its magnitude is -2 m/s. Reading the remaining slopes out loud yields constant neutral which means 0 m/s, constant positive 1 m/s, followed by constant neutral velocity. <br />
  • Try reading this position graph of a projectile on your own. Remember the six important factors when reading kinematic line graphs. Identify the quantities plotted on the x and y axis. Identify if there is an x or y- intercepts and what those intercepts represent. Identify the physics quantity of the slope by its units. You get the units of slope by dividing rise units by run units. Identify the physics quantity of the area by its units. You get the units of area by multiplying rise units by run units. Don’t worry about the area of a parabola. Remember the slope of this parabola is expressed by Yf = Yi + Vi t + ½ gt2. As you know, I often refer to this equation as the beast. Where can we get more information from this graph? What do you know about the 120 meter point? What is the time at that point? What is the velocity at this point? Good for you if you are KUEing the information as we go. <br />
  • The 120 meter point is important for several reasons. We recognize that the 120 meter point is the maximum vertical position (ΔY) for the object. We know that time to reach the object’s maximum height is 5 seconds. With that information we can apply the beast which is the formula that describes the slope to find initial velocity. This calculation yields 48.5 m/s. We also know that at its maximum height a projectiles vertical velocity (Vf) is 0 m/s. Using the formula final velocity equals initial velocity plus gravity times time we find the initial velocity to be 48.5 m/s. So as you can see graph reading yields much information. <br />
  • Please pause if you want to practice your graph reading before I begin mine. You may want to pause after each segment. Okay, here goes…Horizontal velocity is plotted on the y-axis and is measured in meters per second. Time is plotted on the x-axis and is measure in seconds. The y-intercept indicates an object had an initial velocity of 15 m/s. There is no x-intercept. The units for the area (rise times run) of this graph is meters, indicating the change in position of the object. The area is a rectangle so we simply multiply base times height to find it. 15 m/s times 2 seconds is 30m. Thirty meters is the objects change of position in 2 seconds. The slope of this graph represents the acceleration of the object and is measure in m/s2. As you can see there are four different slopes and therefore four different areas on this graph. Slope is equal to rise divided by run. To find the magnitude of the rise of the first slope we take our final velocity, which is 15 m/s and subtract our initial velocity, which was 15 m/s. The magnitude of the rise is 0 m/s. To find the magnitude of the run of the first slope we take our final time, which is 2 seconds and subtract our initial time, which is 0 second. The magnitude of the run is 2 seconds. So the magnitude of the first slope is 0 m/s divided by 2 seconds or 0 m/s2. Again we are learning procedure so find the remaining three slopes using the same procedure. <br />
  • Check your answers to see how successful you were at graph reading. <br />
  • Notice first of all that we are reading a velocity graph and predicting what the position graph will look like. So look at the velocity graph on top of the page and match one of the position graphs to it. For the first two seconds the object has constant positive velocity. This does not rule out any position graphs since they are all in the first quadrant at the two second mark. From two seconds to five second the object is still moving with positive velocity. The value at the five second mark is still positive. Now if we look back at the position graph choices we can see that the graph on the upper right and lower right cannot be correct. Both of those show a negative slop between two and five seconds indicating movement to the left. The next slope of the velocity graph shows no change in the positive velocity so the position graph on the lower left can be ruled out. <br />
  • The corresponding position graph would be read as constant positive velocity followed by positive changing velocity from fast to slow, followed by positive constant velocity followed by positive changing velocity from slow to fast. <br />
  • The quantity plotted on the y-axis is acceleration measured in m/s2. The quantity plotted on the x-axis is time measured in seconds. He y-intercept indicates the object had an initial acceleration of 6 m/s2. There is no x-intercept. The slope of an acceleration versus time graph is jerk measure in m/s3. The area between the slope and the x-axis is the change in velocity measured in m/s. So we need to calculate the slopes and the areas give magnitudes to each part of the graph. The first slope has a rise of 0 m/s and a run of 3 seconds. The magnitude of the first slope is 0 m/s divided by 3 seconds which equals 2 m/s3. The first area has a rise of 6 m/s and a run of 3 seconds. The magnitude of the first area is 6 m/s times 3 seconds which equals 18 meters. <br />
  • Now continue with the procedure you have just confirmed concerning slope and areas to complete the remaining two slopes and two areas. <br />
  • An acceleration graph is important for giving acceleration, time, jerk and the change in velocity. <br />
  • Remember that you now know the acceleration generated by the gizmo ever time it fires. Use this information to help complete the data table. <br />
  • We ended up using Newton’s 2nd law of motion to determine the masses of the students. <br />
  • Vertical velocity is plotted on the y-axis and is measured in meters per second. Time is plotted on the x-axis and is measure in seconds. The y-intercept indicates the projectile had an initial vertical velocity of 30 m/s. The x-intercept indicates that around 3 seconds the projectile had no vertical velocity and changed direction. We can find the exact seconds mathematically. The slope of this graph indicates constant negative acceleration. The area between the slope and the x-axis is the change in position of the projectile. <br />
  • The area between the slope and x-axis describes the same change of position as the position graph would. The beast describes the slope of the position graph and should result in the same calculation as the area of the triangle. We can get the magnitude of the x-intercept by using Vf = Vi + gt. <br />
  • This completes this comprehensive review for the semester one final exam. Thank you for taking the time to review with me. If you have studied hard you deserve to do well. Good luck and have a nice day. <br />

1 2012 ppt semester 1 graphing review 1 2012 ppt semester 1 graphing review Presentation Transcript

  • The slides used in this video are color coded. If you are experiencing difficulty with one aspect of your understanding than another you might find this coding… useful! Slides with Slides with tan red backgrounds backgrounds involve involve word matching Slides with green problems. concepts. backgrounds Slides with olive involve backgrounds graphing. involve reading data tables.
  • #1 #3 #2 #4
  • 2m /s m/s 1 0 m/s 0 m/s
  • Vy = 0 m/s Yf = Yi + Vi t + ½ gt2 Vi = 48.5 m/s ty = 5 s Vf2 = Vi2 + 2g Δy Vi = 48.5 m/s
  • 0 m/s2 2.5 30 m 2 m /s 2 .93 m/s 75 0 m/s2 33.75 m 15 m 90 m
  • Velocity (m/s) 16 14 12 10 8 6 4 2 0 0 2 4 6 8 10 12 14 16 Time (s)
  • Velocity (m/s) 16 14 12 10 8 6 4 2 0 0 2 4 6 8 10 12 14 16 Time (s)
  • 0 m/s3 18 m/s
  • 0 m/s3 -1 18 m/s m/ s3 17.5 m/s 0 m/s3 2 m/s
  • Mass (kg) Force (N) 0 0 42 46.2 47.56 44 48.4 48.65 49.25 49.51 45.45 50
  • Mass (kg) Force (N) 0 0 42 46.2 43.25 47.56 44 48.4 44.23 48.65 44.77 49.25 45.01 49.51 45.45 50
  • Vy = 0 m/s a= 9.8 m Vf = Vi + 2g Δt tf = 3.06 s - /s 2 Yf = Yi + Vi t + ½ at2 Yf = 45.9 m AREA = ½ Base x Height