Transcript Kinematics in One Dimension

Kinematics in One
Dimension
Position Vector
Displacement Vector, Distance
Velocity Vector
Acceleration Vector
Kinematics
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Geometrical and algebraic description
of motion
No regard to the causes of motion
(forces)
Makes use of the mathematical
concept of coordinate system
attached to a point in space or to
objects called ‘frames of reference’.
Coordinate system is used to describe
position relative to the origin.
Location of origin is arbitrarily
selected.
One Dimensional Coordinate System
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When dealing with a single dimension the coordinate
system is a straight line.
Points on the line are located relative to an arbitrary
point called the origin.
Distance on one side of the origin is positive; negative
on the opposite side.
A point in 1D space corresponds to a point on the line; a
positive or negative number with units of length.
Positive and negative axis directions are arbitrarily
selected; usually the positive direction is chosen to the
right. Thus a point can be described by a single signed
coordinate value with unit.
Sample point is 1.5 m on the positive side of the origin.
+1.5 m
x (m)
-2
-1
O
+1
+2
Position as a Vector
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A point can also be located by a position
vector.
Position vector has tail at the origin and tip
at the point.
The length of the vector is the distance
(coordinate value) of the point from the
origin.
The vector direction corresponds to the sign
of the coordinate value.
+1.5 m
x (m)
-2
-1
O
+1
+2
Displacement Vector - describes
the change in position of a moving
object
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Vector pointing from an object’s initial
position to its final position.
In one dimension position and
displacement, although vectors, can
be unambiguously described by
scalars.
x 0  initialposition vector
x  final position vector
Displacement Vector
x  x  x 0
Displacement has unit of length
such as m.
x 0  initial positioncoordinate
x  final positioncoordinate
x  x  x0  displacement
Sign of x indicatesdirectionof
displacement as eitheralong
positiveor negativex - axis.
Positive and Negative
Displacement
Distance
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Magnitude of displacement if motion is in
one direction only (no reversal of direction).
Distance is always positive because it is has
magnitude only with no direction.
If motion consists of a sequence of positive
and negative displacements, the distance is
the sum of distances for each of the
segments of the displacement. 4.0 m
4.0 m
2.0 m
x = + 4.0 m
d = 4.0 m
x = + 2.0 m
d = 4.0m + 2.0 m = 6.0 m
Distance versus Displacement
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What if you travel around the running track?
What is your displacement?
What is your distance?
Average Speed - a scalar quantity
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A stage of the Tour de France from Melun to
Paris has a distance of 140 km.
Australian, Robbie McEwen, won the stage
by cycling the distance in 3h, 30 min and 47
s. What was his average speed?
 1h 
 1 min  1 h 
t  3 h  30 min
  47 s 
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
60
min
60
s
60
min
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

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t  3.513h
d 140 km
vspeed  
 39.851 39.9 km/h
t 3.513h
Speed has units of length per unit time such as m/s.
v speed
d

t
Average Velocity - a vector
quantity
A man taking a leisurely walk takes 15 minutes to
walk 100 m in an eastward direction. He then walks
50 m back (west) in 5 minutes. What is his average
speed and velocity in m/s?
AverageSpeed
d
100 m  50 m
v spee d  
t
15 min  5 min 60 s 
 1 min 
v spee d  0.12 m/s
T akeeastwarddirectionas positiv e.
x  100 m - 50 m  50 m (east)
t  20 min  1200s
x 50 m
v

 0.042 m/s (east)
t 1200s
100 m
W
50 m
E
Velocity has units of length per unit time such as m/s.
x
v
t
Instantaneous Velocity - average
velocity when the elapsed time approaches
zero
Instantaneous Velocity or simply velocity
x
v  lim
t 0 t
Instantaneous Speed or simply speed - the magnitude
of the velocity.
If v is constant, the x-t graph is that of a straight line. The average and instantaneous
velocities are the same and is equal to the slope of the line. If v is not constant the
instantaneous velocity at a time t is the slope of the curved x-t graph. Instantaneous
and average velocities are not necessarily the same (although they can be).
Small intervals give you instants
Meaning of the sign of velocity
Positive velocity means direction of v is towards
positive direction of x axis, i.e., displacement is
positive. Slope of x-t graph will be positive.
Negative velocity means direction of v is towards
negative x axis, i.e., displacement is negative.
Slope of x-t graph will be negative.
The sign of v does not indicate whether speed is
increasing or decreasing. It, however, indicates
what is happening to the displacement.
Meaning of the sign of velocity
•You can consider the world to have a positive
direction and a negative direction.
•Are you going towards POSIWORLD?
•Or NEGILAND?
Acceleration - a vector quantity
Average Acceleration
a a vg
Unit: length/time2 such as m/s2
v

t
Instantaneous Acceleration - average acceleration
when the elapsed time approaches zero.
v
a  lim
 t  0 t
Acceleration and Velocity
Whether an object is speeding up or slowing down
does not depend on the sign of a but on its direction
relative to the direction of v.
a
v
Result (assume +
)
v is increasing in magnitude
and becoming more positive;
speeding up
v is decreasing in magnitude
and become less negative;
slowing down
v is decreasing in magnitude
and becoming less positive;
slowing down
v is increasing in magnitude
and becoming more negative;
speeding up
a and v in the same direction
v
t
v  v f  v i  v  v 0
a
v  a t  v  v 0
v  v 0  a t
v  v 0  at if init ial t imeis assumed 0
a and v in opposite directions
t = 3 s
v
t
v  v f  vi  v  v 0
a
v  at  v  v 0


v  v 0  at  28 m/s  - 5 m/s2 3 s 
v  28  15  13 m/s
v  v 0  at if init ial t imeis assumed 0
a and v in opposite directions
t = 3 s
When acceleration and velocity ‘compete’ then the object is slowing down
Kinematic Equations for Constant
Acceleration
We already have one kinematic equation (previous slide)
v  v0  at
The graph of v vs. t if a is constant is a straight line.
y  b  mx
v
vavg
v0
slope = a
t
v  v0
vavg  v 
2
 x x  x0
v

t t  t 0
Usually, we assume initial time t0 = 0 and initial position x0 = 0.
 v  v0   v0  at  v0 
x  vt  
t  
t
2
 2  

1 2
x  v0t  at
2
Kinematic Equations . . .
v  v0
v  v0  at or t 
a
 v  v0   v  v0  v  v0 
x
t  


 2   2  a 
v 2  v02
x
2a
v 2  v02  2ax
v 2  v02  2ax
Kinematics Equations
Summary of Kinematic Equations in 1 D
constant a
Variables
Equation
x
a
v
v0
t
v  v0  at
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x  12 v0  vt
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x  v0t  12 at2
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v2  v02  2ax
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
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Remember, kinematic variables are vector quantities. Their signs are
important. Also x is a displacement from the initial position which
was assumed to be zero. Initial time was also assumed zero.
Freely Falling Bodies
Example of motion with constant
acceleration.
Motion is vertical with acceleration of
gravity,
g = 9.81 m/s2 always downward.
Coordinate axis will be vertical (you
can call it x or y). Choose positive
direction up or down but be sure the
sign of a is correct.
If you choose upward is positive, then
a = -g.
If you choose downward as positive,
then a = +g.
Be sure to get study packet and
we will work some problems
out in class.