Kinematic Graphs by HC121001133815

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									            Kinematic Graphs

             Graphically exploring
            derivatives and integrals




10/1/2012                        Dr. Sasho MacKenzie - HK 376   1
                          Slope
            Y

                                    (4,8)
            8




                (0,0)                       X
                                   4
  Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 =     2
          run X      X2 – X1   4–0     4


10/1/2012                                     Dr. Sasho MacKenzie - HK 376   2
                Velocity is the slope of
                    Displacement
                   Y


                   8
                                              (4,8)
    Displacement
         (m)



                       (0,0)                            X
                                Time (s)      4

        Average
        Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m =        2 m/s
                   run    t   t2 – t1   4–0      4s

10/1/2012                                     Dr. Sasho MacKenzie - HK 376   3
     1. The displacement graph on the previous
        slide was a straight line, therefore the
        slope was 2 at every instant.

     2. Which means the velocity at any instant
        is equal to the average velocity.

     3. However if the graph was not straight
        the instantaneous velocity could not be
        determined from the average velocity


10/1/2012                           Dr. Sasho MacKenzie - HK 376   4
             Instantaneous Velocity
        • The average velocity over an infinitely
          small time period.
        • Determined using Calculus
        • The derivative of displacement
        • The slope of the displacement curve




10/1/2012                              Dr. Sasho MacKenzie - HK 376   5
            Instantaneous Acceleration
        • The average acceleration over an
          infinitely small time period.
        • Determined using Calculus
        • The derivative of velocity
        • The slope of the velocity curve




10/1/2012                           Dr. Sasho MacKenzie - HK 376   6
             Average vs. Instantaneous
                   Y         The average velocity does not accurately
                             represent slope at this particular point.
                 8
                                                    (4,8)
    Displacement
         (m)



                     (0,0)                                    X
                                Time (s)           4

        Average
        Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s
                   run    t   t2 – t1   4–0      4s

10/1/2012                                           Dr. Sasho MacKenzie - HK 376   7
                      Derivative
        • The slope of the graph at a single point.
        • Slope of the line tangent to the curve.
        • The limiting value of D/ t as t
          approaches zero.




10/1/2012                              Dr. Sasho MacKenzie - HK 376   8
            Infinitely small time period (dt)
                                  •Tangent line
                                  •Instantaneous Velocity
                       dt
    Displacement




                            t3



                                  t2


                                        t1

                            Time
10/1/2012                                        Dr. Sasho MacKenzie - HK 376   9
                  Velocity from Displacement
                        The graph below shows the vertical displacement
                        of a golf ball starting immediately after it bounces
                        off the floor and ending when it lands again.
  Displacement




                                          Velocity = 0

                 Velocity > 0
                                                                           Velocity < 0




                                               Time
10/1/2012                                                        Dr. Sasho MacKenzie - HK 376   10
                                Graph Sketching
                            0
                    Slope
                                 Differentiation
                       +        Displacement           Velocity
                0
        0
                                         OR



                                 Velocity        Acceleration

            0



10/1/2012                               Dr. Sasho MacKenzie - HK 376   11
                        Slope       Graph Sketching
                          0          Differentiation
                    +           _
                                        Displacement           Velocity
                0                   0
        0
                                                 OR



                                         Velocity       Acceleration
            0



10/1/2012                                      Dr. Sasho MacKenzie - HK 376   12
                Going the other way: area
                    under the curve
            Y

            4                   (2,4)




            (0,0)
                                        X
                            2

 Area under curve = Height x Base = Y x X = 4 x 2 = 8


10/1/2012                                   Dr. Sasho MacKenzie - HK 376   13
            Displacement is the Area
            Under the Velocity Curve
               Y

               4                        (2,4)

    Velocity (m/s)



               (0,0)
                        Time (s)                   X
                                    2

      Displacement = V x t = 4 x 2 = 8 m

10/1/2012                                  Dr. Sasho MacKenzie - HK 376   14
    What if velocity isn’t a straight line?
                               (2,4)
             4
                                        This would be
                                        an over
                                        estimate of
  Velocity                              the area under
                                        the curve



            (0,0)
                    Time        2




10/1/2012                       Dr. Sasho MacKenzie - HK 376   15
                     Integration
        • Finding the area under a curve.
        • Uses infinitely small time periods.
        • All the areas under the infinitely small
          time periods are then summed
          together.
        • D = Vdt = Vt



10/1/2012                               Dr. Sasho MacKenzie - HK 376   16
            Infinitely small time periods
              The area under the graph in these infinitely
              small time periods are summed together.


  Velocity




                              Time
   •D = Vdt = V1t1 + V2t2 + V3t3 + ……
   Instantaneous    Infinitely small
   Velocity         time period
10/1/2012                                    Dr. Sasho MacKenzie - HK 376   17
                                                   Graph
               Constant slope of 2               Sketching
                                                Integration
                                          Displacement               Velocity
        0

                                                      OR

        2

                                            Velocity          Acceleration
        0    t     t      t

            Area under curve increases by the same amount
            for each successive time period (linear increase).
10/1/2012                                            Dr. Sasho MacKenzie - HK 376   18
                                                   Graph
               Exponential curve
                                                 Sketching
                                                Integration
                                          Displacement               Velocity
        0

                                                       OR




        0      t    t     t              Velocity          Acceleration

            Area under curve increases by a greater amount for
            each successive time period (exponential increase).
10/1/2012                                            Dr. Sasho MacKenzie - HK 376   19
      Indicate on the acceleration graph below the location of the
      following point(s). Place the letter on the graph.

      A. Zero velocity      B. Zero acceleration C. Max velocity
      D. Min velocity       E. Max acceleration F. Min acceleration

     2




    0


    -1


                      1s             2s             3s
10/1/2012                                            Dr. Sasho MacKenzie - HK 376   20
      Indicate on the velocity graph below the location of the following
      point(s). Place the letter on the graph.

      A. Zero velocity        B. Zero acceleration     C. Max velocity
      D. Min velocity         E. Max acceleration      F. Min acceleration
      G. Max displacement     H. Min displacement




            0




10/1/2012                                                Dr. Sasho MacKenzie - HK 376   21

								
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