KINEMATICS (PowerPoint download) by dffhrtcv3

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									Motion in 1-D
KINEMATICS
   The Major Player
Isaac Newton 1642-1727
                         Develops Calculus to explain the
                                theory of Mechanics




                                       
                                   F  ma
                                          Kinematics
                  Dynamics                We begin here in 1-D
 Note that vx0 is the same throughout the
  motion of the object



                              vx 0  v0 cos
                              v y 0  v0 sin 
KINEMATICS to describe “motion”

   Kinematics-describes motion
   Dynamics-concerned with the cause of motion
   Measurable quantities needed to describe
    motion are
     Displacement:

      Dx=xf-x0 measured in units of distance
      (e.g., meters)
     Time or time interval, Dt=tf-t0

      measured in sec
                Dx = xf-xi in fact
                this is a Vector
                Two possible
                directions +/-

Coordinate space and time


x(t) vs t


for armadillo
   Put them together!!!!
   Get a rate of change
   Average Velocity
   Slope of x versus t
   graph
         xt1   xt0  x1  x0
vavg                   
            t1  t0       t1  t0
Freedom to choose “origin of time”
   If initial time is chosen to be
    t=0

                          xt1   x0  x1  x0
                 vavg                  
                                t1          t1




   So let t1=t !!!                 xt   x0
                           vavg   
                                         t
Sample problem 2-1
      “Calculus 101” the derivative graphically
    “Average velocity at an instant” let the time interval be come very very

    small      Dt  dt
   Instantaneous velocity, v(t)
    slope of the displacement
    curve at a point                           Tangent at P is
                                               derivative of curve
                                               evaluated at t1


                                               Dxt  dxt 
                                 vt   lim         
                                          t 0  Dt     dt
Sample problem 2-2      Graphical representation
                               of motion
                        x(t)-position
                        v(t)-velocity
                       a(t)-acceleration
                     Relationship btwn
                     x and v: Calc. 001
                     Area under velocity plot is the
                     displacement!

                     Dx = vav Dt
                         =Area ??!!!
                     Will Generalize……….
                                       t
                           x  x0   vt  dt 
                                       0
          Average Acceleration
   acceleration rate of
    change of velocity
      units are
        distance/time2
   average acceleration is
    the average slope on a
    velocity vs. time graph
                                         vt1   v0  Dvt 
   average acceleration is
    the rate of change of the   aave                  
    slope on a displacement                 t1  0       Dt
    vs. time graph
Instantaneous Acceleration
 If the time interval
  between velocity
  measurements becomes
  vanishingly small,
 the acceleration is
  instantaneous
 Instantaneous
  acceleration is the slope
  of the velocity curve at a
  point                                      Dvt  dvt 
                               at   lim         
                                        t 0  Dt     dt
Example 2-4
   x(t)=4 - 27t + t3

a) v(t) ?

b) Is there a time, t when v(t) = 0 ?

    c) Plot results !
Example, constant acceleration:
Example, constant acceleration “g”
MOTION IN ONE DIMENSION :
  CONSTANTLY ACCELERATED
          MOTION

                            Equations based on
                            definition

                                       v ( t ) + v0
                              vavg =                  (1)
                                             2

             vt   v0
    aave                   vt   v0  a t        (2)
                   t
             xt   x0     xt   x0  vavgt
    vavg                                             (3)
                  t
MOTION IN ONE DIMENSION (3):
CONSTANTLY ACCELERATED MOTION



                 x  x0  v0t  1 at 2
                                2          (4)


                 2a  x  x0   v2  v0
                                       2
                                           (5)


                   See blackboard for
                   derivation !!!
MOTION IN ONE DIMENSION
    Special considerations
       Positive and negative values have meaning

        (use a coordinate axis or number line)
       Problems can have as complex a variability

        as necessary

       a
          dv
               implies   v = ò a ( t ) dt   and
          dt

                         x = ò v ( t ) dt
          dx
       v      implies
          dt
                     Calculus 102 – Equations of Motion
                                                   constant Acceleration
          dv ( t )
a (t) =                                  (Eq. 1) Instantaneous acceleration
            dt
dv ( t ) = a dt                          (Eq. 1b), re-write (Eq. 1) in differential form
                         v( t )
v ( t ) - v ( 0) =   ò ( ) dv
                         v0
                                             Integrate LHS of Eq. 1b

                     ò       a ( t ' ) dt ' = a ò dt ' = a t
                         t                         t
v ( t ) - v ( 0) =                                               Integrate RHS of Eq. 1b
                         0                         0

v ( t ) = v ( 0 ) + a t º v0 + a t                 (Eq. 1c)


          dx ( t )
v (t) =                                  (Eq. 2)
           dt
dx ( t ) = v ( t ) dt                    (Eq. 2b), re-write (Eq. 2) in differential form
                             x(t )
x ( t ) - x ( 0) =   ò ( ) dx
                         x 0


                     ò v (t )
                         t
x ( t ) - x ( 0) =       0
                                     '
                                           dt ' = (see black board)
                          1
x ( t ) - x ( 0 ) = v0 t + at 2
                          2
                        1
x ( t ) = x0 + v0 t + at 2 (Eq. 3)
                        2

          Work this out
MOTION IN ONE DIMENSION (4):
CONSTANTLY ACCELERATED MOTION
         SPECIAL PROBLEM CLASS:
           FREE FALL
         The acceleration of gravity is a
           constant when near the
           surface of the earth
         g = 9.8 m/s2 toward the center
           of the earth
Exercise
   What is the instantaneous velocity of an object
    at t = 3 s, if its displacement is described by
    x(t) = 7.8+9.2t – 2.1 t3 ?
   What is velocity at t=3.5 s ?
   What is the acceleration at t=3.0 s?
       EXAMPLE


A ball is initially thrown upward
   with a velocity of 20.0 m/s at
   the edge of a 50.0 m tall
   building.
a. How long does the ball take
   to hit the ground?
b. What is the ball’s velocity
   when it strikes the ground?

								
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