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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

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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