Kinematics
The study of how objects move
We must start with a
question
Where are You?
Position
To find where you are you need the following:
Reference Point:
Zero location in a coordinate
system or frame.
Position:
Separation between
an object
andused
a to define
Frame
of Reference:
Coordinate
system
reference point. It needs motion.
a distance and direction.
i.e. 3.0 meters to the right.
You are Here
You are Here
You are Here
You are Here
You are Here
You are Here
You are Here
You are Here
Distance vs. Displacement
50 m
Distance is simply the total path length
traversed in moving from one location
to another
Distance is a SCALAR quantity. A measurement
with only magnitude, or size.
Distance vs. Displacement
10 m East
Displacement is the straight-line distance between
two points, along with the direction from the
starting point to the final position.
Displacement is a VECTOR quantity. A
measurement with magnitude, or size and
a direction
Walking to
Physics Lab
Difference between
Distance and
Displacement
Distance = 8 m
Displacement =+8m
= 8m right
Direction Convention :
Right - Positive
Left - Negative
Speed
Speed is the rate at which distance is traveled.
20 m
2 sec
Distance
20m
Speed 
=10 m/s
2 Time
sec
Speed is a scalar
quanity
Average vs.
Instantaneous Speed
Average speed is the distance d traveled, the actual path, divided
by the total time t elapsed in traveling that distance.
Distance Traveled
Average Speed 
Total Time
Instantaneous Speed is how fast an object is moving at a
particular instant of time.
Velocity
Velocity tells how fast something is moving and in what
direction. Velocity is a vector quantity.
Average Velocity is the displacement divided by the total
travel time
x x  xo
V

t
t  to
x
V
t
xo =initial position, commonly
zero
t o = initial time, commonly zero
When xo and to are zero
Speed
and
Velocity
Problem
A jogger jogs from one end to the other of a straight 300 meter track
(from point A to point B) in 2.50 min and then turns around and
jogs 100 m back toward the starting point (to point C) in another
1.00 min. What are the jogger’s average speeds and velocities in
going
A) From A to B
B) From A to C after turning around
Given:
xB = 300m - 0m = +300m (from A to B)
xC = 200m – 300m = -100m (from B to C)
tB = 2.50 min = 150 s
tC = 1.00 min = 60.0 s
A) Average speed and velocity from A to B
d 300m
avg speed  
 2.00 m s
t
150s
x B 300m
v

 2.00 m s
tB
150s
B) Average speed and velocity from A to C
d 300m  100m
avg speed  
 190
. ms
t
150s  60.0s
xB  xC 300m  ( 100m)
vc 

 0.952 m s
t B  tC
150s  60.0s
Two Components
of Velocity
Graphical look
at Motion
Acceleration
Position is the separation between an object
and a reference point.
Displacement is a change in position
Velocity is the time rate change of displacement
Acceleration is the time rate change of velocity
All four are vector quantities
Represented by a magnitude and direction
Kinematic
Equations
velocity
d d2  d1 d 2  d1


v
t
t2  t1
t
acceleration
v v2  v1
v

v
2
1
a


t
t 2  t1
t
v2  v1  at
d Total Displacement and total time
v
t
v1  v2 With a constant acceleration
v
2
d  12 (v1  v2 )t
Average Velocity
11 (v  v )t v  v  at
d

d  22 (v11  v22 )t 2
1
d
1
d
2
1
(v1  (v1  at ))t
2
(2v1 )t  2 at
1
d  v1t  2 at
1
2
2
v1  v2 v  v  at
v
2
1
2
v2  v1
t

2
2 a
v

v
2 v 1 v
v

v
d
1 2
2
1
d  vt

2
a
2
a
v  v  2ad
2
2
2
1
d
v
t
v
a
t
Kinematic
Equations
v2  v1  at
d  v1t  2 at
1
2
v  v  2ad
2
2
2
1
Problem:
18
.
m
s2
Ally the Alligator takes off from an airfield.
She starts from rest and obtains a velocity of
36 m/s at take off in 20.0 seconds.
v

v
v m 36

0
2
1
s
s
m
m
1
1 go


18
.

d
0

36
)
20
.
0
sec
2
How a
far
she
down
the
runway?
dddid
(
v

v
)
t
2(
s
s
2 360
m
s
1
2
20 tsec
t
m the runway?
m
What is her acceleration down