Kinematics
Describing how things move is called kinematics and it
has terms that are very specific. These terms include
position, displacement, distance, velocity, speed and
acceleration. Proper use of these terms first demands
that you understand each term’s definition, symbol and
any equations associated with it.
POSITION, DISPLACEMENT
Kinematics is a branch of physics that is concerned with
describing motion. The first concept of motion is
position. Think of your position at this instant. Another
way of asking this question is “Where are you?”
Position is defined as the distance and direction from a
reference point. A reference point may be in fact moving
(e.g. Earth). Reference points are arbitrary choices
made by us. Position is a vector quantity because it has
size or magnitude and direction. Scalar quantities have
only size or magnitude. The symbol for position is

d
Direction can be given above or below the horizon,
within a 2-D system (NEWS), or within a 3-D system
(on Star Trek).
After the examples on the next slide complete the
position worksheet.
For this class use the 2-D map system.
Record direction relative to north or south.
15.41 cm (measure)
11.6o (measure)
6.52 cm (measure)
r.p.
30.2o (measure)
6.52 cm [S11.6oW]
15.41 cm [N59.8oE]
DISPLACEMENT
Displacement is defined as change in position.



 d  d 2  d1
Displacement is a vector with magnitude and direction.
It is found in the difference between two position
vectors. Take note that it is always final position minus
initial position.
Displacement does not need a reference point.
Displacement remains the same even if the reference
point was changed.
Another equation
for displacement is




 d R   d1   d 2   d 3 ...
25.31 cm (measure)
initial point
Arrow points to final position!
r.p.
25.31 cm [S65.5oE]
final point
24.5o (measure)
Try worksheet
Distance is defined as the length of path
traveled. It is a scalar quantity and its
symbol is
d
If you walked 10 m [E] then 10 m [W] then
10 m [N] your distance walked is 30 m while
your displacement for the trip is 10 m [N].
Velocity and Speed
Are you moving?
Can you feel movement?
How do you know you are moving?
a vector divided by
a scalar yields a
vector
Velocity is defined as the change in position
with respect to time. It is a vector quantity.
Speed is defined as distance traveled with
respect to time. It is a scalar quantity.
Another equation
for velocity is



 d1   d 2   d 3 ...
v
t


d
v
t

d
v
t
Acceleration
Acceleration is defined as change in velocity with
respect to time.



v
a
t

v 2  v1
a
t

In everyday usage acceleration often seems to mean speed up
while decelerate means to slow down however technically this is
not correct. Acceleration is a vector quantity therefore has
magnitude and direction. In one-dimensional acceleration can
have a positive or negative value. The sign indicates the sign of
the velocity change since the change in time is always positive.
Physicists do not mention deceleration. Speeding up refers to
increasing magnitude of velocity while slowing down means the
opposite.
Graphing
Graphs are often used in kinematics to
represent position,velocity or
acceleration (all versus time). Other
quantities can be calculated from
analysis of these graphs.
Curve Sketching
Dave begins west of the school and
walks with a constant velocity and he
passes the school. ([E] is positive)
This means position on a graph. We will
never graph distance vs. time.
d
v
t
t
This means velocity on a graph. We will
never graph speed vs. time.
Tanita walks back to school with a
constant velocity, when she reaches the
school she begins running with a
constant velocity.
d
v
t
t
Karan runs away from school at a
constant rate for a time then slows down
until he stops.
d
v
t
t
a
t
A ball is thrown upwards and is caught
below where it began.
d
v
t
t
a
t
Sketch a position vs. time graph, a velocity vs. time graph and an acceleration
vs. time graph for the following situations.
a) A car travels forwards with a constant velocity of 50 km/hr [N].
b) A person walks backwards with a velocity of 4 m/s [N].
c) A person walks at 3 m/s for a period of time then jogs 6 m/s for the same
period of time.
d) A person walks at 3 m/s for a period of time then jogs home at 6 m/s in the
opposite direction.
e) A ball is thrown up out of a person’s hand; it rises and then returns to the
person’s hand.
i) Sketch the graph for the velocity of the ball when it is out of the
person’s hand.
ii) Sketch the graph for the velocity of the ball for the complete situation.
f) A hockey puck slides along the smooth ice and bounces straight off the
boards.
g) A hockey puck is shot at a goalie and the goalie catches it.
h) A person who slows down and speeds up without changing direction.
i) A person who slows down and speeds up who does change direction.
Constant Acceleration Kinematic Equations
Motion with a constant acceleration is an important type
of motion which bears further analysis. In this unit only
constant acceleration problems will be studied. The
kinematic equations about to be shown or developed are
only for constant acceleration!
The average velocity during a constant acceleration is
found with this equation.


v 2  v1
vav 
2



v 2  v1
 d  (
)t
2

So at this point when studying constant acceleration we




have two equations.


v 2  v1
v 2  v1
d  (
)t
a
2
t
Constant Acceleration Kinematic Equations
The two constant acceleration equations below lead to 3 other
kinematic equations.


v 2  v1
a
t


v 2  v1
d  (
)t
2


v 2  v1
a
t





v 2  v1
d  (
)t
2



a t  v 1  v 2

at  v 1  v 1
d  (
)t
2




at  2 v 1
d  (
)t
2


at 2
 d  v 1 t 
2

v2  v1  at
v2  v1
d  (
)t
2
a 2
d  v1t  t
2
a 2
d  v2 t  t
2
2
2
v
 v  2ad
2
1
ACCELERATION DUE TO GRAVITY
Objects that have a net force of gravity
acting upon them experience an acceleration due to gravity of 9.81 m/s2 [down].
g = 9.81 m/s2 [down]
In these situations the force of air
resistance is ignored.
This acceleration applies to objects going
up or down (or both).
(terminal velocity, coffee filter and motion sensor)
An object is thrown upwards
and returns to the same height.
position
time
slope = - 9.81 m/s2
velocity
time
- 9.81 m/s2
acceleration
time
at max. height v = 0
the velocity at this
time is negative of
the initial velocity
position
time
these two t’s
are equal
DOUBLE OBJECT PROBLEMS
-set up a direction system for each
object such that each object moves in a
positive direction
-there will always be a link between the
two objects so that one variable can be
eliminated
-extra information allows another
equation to be written