Transcript Outline

Kinematics
Position, Velocity , and Acceleration Graphs
Overview
 Kinematics: A Description of Motion
 Position
 Velocity
and Displacement
 Average
 Instantaneous
 Acceleration
 Average
 Instantaneous
 Graphing
The topics we
will discover on
this powerpoint
A Story
We will introduce Kinematics
at motion
along a Questions by using a both
Ifby
welooking
can answer
the following
Imagine
walking
fromwe
your
house
to visit1-dimensioal
an ice
single line.
This away
restricted
Algebra
and
Graphing,
then
know
we GROK
cream
store is
that
is located
perspective
usually
called 2 km away. You walk at a
kinematics:
constant
speedMotion.
and you arrive at the store in 30 minutes.
1-Dimensional
spend
10 minutes
at the store eating your ice cream.
You
How
far did
you walk?
thenwhere
leave your
the store
and
walking
awayjourney?
from
You
What
speeds
atcontinue
any location
on your
your normal
pace forat
5 minutes.
You realise that
home
Whatatwhere
your velocities
any location?
should
heading
home
and and
immediately
walk back
you
What
was be
your
average
Speed
Average Velocity?
home
atDisplacement?
twice your normal pace. At a distance
towards
What was
your
theaccelerations
ice cream store
andthe
yourjourney
home you
halfway
Tell mebetween
about your
during
stop for 10 minutes at a library (good students always
visit libraries). You then continued back home at your
increased speed.
Position vs Time Plots

Gives us the location, x, at any time, t.
x
3
For example:
1
t
Position at t=3, x(3) = 1
3
-3
Graph the Journey
Get out some graph paper and Sketch a Position-Time
graph of our story. Don’t forget to use those graphing
skills you learned in Math class.
Imagine walking away from your house to visit an ice
cream store that is located 2 km away. You walk at a
constant speed and you arrive at the store in 30 minutes.
You spend 10 minutes at the store eating your ice cream.
You then leave the store and continue walking away from
home at your normal pace for 5 minutes. You realise that
you should be heading home and immediately walk back
towards home at twice your normal pace. At a distance
halfway between the ice cream store and your home you
stop for 5 minutes at a library (good students always
visit libraries). You then continued back home at your
increased speed.
You
Imagine
You
stopthen
spend
realise
forwalking
5
leave
continued
10
minutes
that
minutes
the
away
you
store
at
back
should
from
at
a library
the
home
and
your
be
store
continue
at
heading
(good
house
your
eating
students
walking
increased
to
home
visit
yourand
away
an
ice
always
speed.
immediately
ice
cream.
from
cream
visit
home
libraries).
store
walk
at that
your
backis
normal
towards
located
pace
home
2 km
for
Compare
your
graph
to
mine.
Let’s
do
some
algebra
to
5
away.
atminutes
twice
Youyour
walknormal
at a constant
pace. Atspeed
a distance
and you
halfway
arrive between
at the store
the ice
in 30
cream
minutes
store and your
calculate
distances
and
speeds
at
each
line
segment
of
home you stop
the graph.
Kinematic Definitions
• Position (x) – where you are located
• Distance (d) – how far you have traveled, regardless of
direction
• Displacement (x = xf-xi) – where you are in relation to
where you started, need initial and final position.
•Displacement is a vector: It has both, magnitude and direction!!
•For one dimensional motion a +ve sign means the displacement is
toward the right, a -ve sign means the displacement is toward the
left.
•Distance is a scalar, only magnitude.
Kinematic Definitions
• Speed (v) – distance divided by time. It is always a
positive quantity and the direction of motion is
irrelevant. Speed is a scalar quantity.
• Velocity (v) – is displacement divided by time. Since
displacement depends only on your starting and ending
points, velocity (v) is x divided by t and thus a
vector. If displacement is negative, then the velocity
will also be negative. Similarly if displacement is
positive, then the velocity will be positive.
v avg , x 
x
t

x f  xi
t f  ti
average speed 
total distance
total tim e
Graphical Interpretation of Average Velocity

Velocity can be determined from a position-time graph
v average 
x
t

 13.3
 40 m
3.0 s
m
s

Average velocity equals the slope of the line joining the
initial and final positions (A and D)
The average velocity you travelled at on your walk
between home and the ice cream store is:
v avg , x 
x
t

x f  xi
t f  ti

2 km  0 km
30 m in  0 m in

2 km
30 m in

1km
 0.06667
15 m in
We also notice that the
total distance travelled by
you is also 2km and the
total amount of time it took
you was also 30 min. So your
average speed is also
0.0667 km/min
2km
30 min
km
m in
The average velocity you travelled when sitting at the
ice cream store eating your ice cream:
v avg , x 
x
t

x f  xi
t f  ti

2 km  2 km
30 m in  40 m in

0 km
10 m in
0
km
m in
We also notice that the total
distance travelled by you is
also 0 km and the total amount
of time it took you was also 10
min. So your average speed is
also 0 km/min
10 min
The average velocity you travelled for 5 minutes away
from the ice cream store:
v avg , x 
x
t

x f  xi
t f  ti
2

1
km  2 km
1
km
1 km
km
3
 3

 0.0677
45 m in  40 m in 5 m in 15 m in
m in
1/3 km
5 min
We also notice that the total
distance travelled by you is
also 1/3 km and the total
amount of time it took you was
also 5 min. So your average
speed is also 0.0677 km/min
The average velocity you travelled when heading back
home:
v avg , x 
x
t

x f  xi
t f  ti
1km  2

1
km
1
1
km
2 km
km
3
 3

  0.1333
55 m in  45 m in
10 m in
15 m in
m in
1 1/3 km
10 min
We also notice that
the total distance
travelled by you is
also 1 1/3 km and
the total amount of
time it took you was
also 10 min. So your
average speed is
also 0.1333 km/min
Note: Since we are returning home, look what happens to the
average speed and average velocity from home to this location
:
x f  xi
x
1km  0 km
1km
km
v avg , x 



 0.01818
t
t f  ti
55 m in  0 m in
55 m in
m in
1 km
55 min
We notice that the
total distance
travelled by you is
2 km +1/3 km + 1
1/3 km = 3.667 km
and the total
amount of time it
took you was also
55 min. So your
average speed is
now 0.0667 km/min
The average velocity during your library:
v avg , x 
x
t

x f  xi
t f  ti

1km  1km
60 m in  55 m in

0 km
5 m in

0 km
15 m in
0
km
m in
We also notice that
the distance
travelled by you is
also 0 km and the
total amount of time
it took you was also
5 min. So your
average speed is
also 0 km/min
5 min
The average velocity final walk home:
v avg , x 
x
t

x f  xi
t f  ti
0 km  1km

67
1
2
v ave 
1 km
7.5 m in
 0.1333
km
m in
m in  60 m in
1km

7
1

2 km
  0.1333
15 m in
m in
km
m in
2
We also notice that
the distance
travelled by you is
also 1 km and the
total amount of time
it took you was also
7.5 min. So your
average speed is
also 0.1333 km/min
1 km
7.5 min
The average velocity for the whole journey:
v avg , x 
x
t

x f  xi
t f  ti
0 km  0 km

67
1
2
v ave 
67
1
m in
0
km
m in
2
We also notice that
the distance
travelled by you is
now 4.667 km and
the total amount of
time it took you was
also 67.5 min. So
your average speed
is 0.06913 km/min
4.667 km
67.5 m in
 0.06913
m in  0 m in
0 km

km
m in
67.5 min
Understanding
An object (say, car) goes from one point in space
to another. After it arrives to its destination, its
displacement is:
1.
2.
3.
4.
5.
either greater than or equal to
always greater than
always equal to
either smaller or equal to
either smaller or larger
than the distance it traveled.
Example:
Suppose that in both cases truck covers the distance in 10 seconds:
v 1 average 
 x1
t

80 m  10 m

10 s
 70 m
10 s
 7 m s
v 2 average 
 x2
t

 6 m s
20 km  80 km
10 s

 60 m
10 s
Instantaneous Velocity

Instantaneous velocity is defined as the limit of the average
velocity as the time interval becomes infinitesimally short, or
as the time interval approaches zero
vinst

x f  xi dx
x
 lim
 lim

t 0 t
t 0
t
dt
The instantaneous velocity indicates what is happening at
every point of time
This notation tells us that Instantaneous Velocity is a Derivative
of position with respect to time. I invented the mathematics for
this, and it is called Calculus. Why? Because I need two points to
determine a slope, but I have only one point available:
Graphical Interpretation of Instantaneous
Velocity

Instantaneous velocity is the slope of the
tangent to the curve at the time of interest
The slope of the
tangent line drawn
at B, is the
Instantaneous
Velocity at B

The instantaneous speed is the magnitude of
the instantaneous velocity
Average vs Instantaneous Velocity
Average velocity
Instantaneous velocity
Understanding
The graph shows position as a function of time for two
trains running on parallel tracks. Which of the
following is true:
1.
2.
3.
4.
5.
at time tB both trains have the same velocity
both trains speed up all the time
both trains have the same velocity at some time before tB
train A is longer than train B
all of the above statements are true
position
A
B
Note: the slope of curve B is parallel to
line A at some point t< tB
tB
time
Average Acceleration


A velocity that changes indicates that an
acceleration is present
Average acceleration is the rate of change
of the velocity Note: Velocity also
changes when the speed
remains the same, but
the direction changes.
f
i
This is because velocity
average
is a vector.
a

v v  v


t
t
Average acceleration is a vector quantity (i.e.
described by both magnitude and direction)
Average Acceleration
When the sign of the velocity and the
acceleration are the same (either positive
or negative), then the speed is increasing
When the sign of the velocity and the
acceleration are the opposite (either
positive or negative), then the speed is
decreasing
Units
SI
Meters per second squared (m/s2)
CGS
Centimeters per second squared (cm/s2)
US Customary
Feet per second squared (ft/s2)
Instantaneous and Uniform
Acceleration

Instantaneous acceleration is the limit of
the average acceleration as the time
interval goes to zero
ainst

v f  vi
v
 lim
 lim
t 0 t
t 0
t
dv

dt
Yep, that
Calculus
thing again.
When the instantaneous accelerations are
always the same, the acceleration will be
uniform. That is, the instantaneous
accelerations will all be equal to the
average acceleration
Graphical Interpretation of
Acceleration


Average acceleration is the
slope of the line connecting
the initial and final
velocities on a velocity-time
graph. That is, the slope of
secant line PQ
Instantaneous acceleration
is the slope of the tangent
to the curve of the velocity
- time graph, That is, slope
of tangent line at t=b
b
Velocity vs Time Plots





Gives velocity at any time by just looking at the
height of the graph at any time t.
Net area gives displacement (how far you have
moved from initial position to final position)
Total area gives how far you have travelled
Slope gives acceleration (rise over run).
Speed s = | v |
If the shapes
you are
calculating the
area of is not
simple, you can
calculate the
area using my
Calculus.
v (m/s)
9-1=8
9+1=10
3
4
-3
t
Let’s Draw a Velocity Time Graph based on out
previous ice cream trip’s Position Time graph. Then we
will use that to determine how far we walked.
Slope (Velocity) is
0.06667 km/min
Slope (Velocity) is
-0.1333 km/min
Slope
(Velocity) is
0.06667
km/min
Slope (Velocity)
is 0 km/min
Slope (Velocity)
is 0 km/min
Slope (Velocity) is
-0.1333 km/min
Hey,
walked
2 km Time
+ 1/3 Graph
km= 2 based
1/3 kmon
away
Let’s you
Draw
a Velocity
out from
home
andice
1 1/3
km+trip’s
1 km Position
= 2 1/3 km
back
toward
home.
previous
cream
Time
graph.
Then
we
So
distance
travelled
0km.
the total
willour
usenet
that
to determine
howisfar
weWhile
walked.
distance travelled was 2 1/3 km+ 2 1/3 km = 4 2/3 km
2 km
1/3
km
1 1/3
km
1 km
Acceleration vs Time Plots
Gives acceleration at any time.
 Area gives change in velocity

The change in
velocity
between t=4
and t=1 is then
6+(-2)=4 m/s.
a (m/s2)
3
6
24
-3
t
Let’s Check your Understanding
Is it possible for an object to have a positive
velocity at the same time as it has a negative
acceleration?
1 - Yes
2 - No
If the velocity of some object is not zero, can
its acceleration ever be zero ?
1 - Yes
2 - No
Velocity Understanding
If the average velocity of a car during a trip along a
straight road is positive, is it possible for the
instantaneous velocity at some time during the trip
to be negative?
A - Yes
B - No
Drive north 5 miles, put car in reverse and
drive south 2 miles. Average velocity is
positive while the instantaneous velocity is
negative when the car is going backwards.
.
Graphical Relationships
x (t )
vx 
ax 
dx
dv x
dt
dt
2

d x
dt
2
• Relating Graphs
x
t
v
t
a
t
Understanding
A ball is dropped from a height of two meters above the ground.
y
x
Draw vy vs t
9
v
A
-6
9
0.5
v
B
9
t
9
0.5
-6
v
-6
0.5
C
t
9
D
v
v
0.5
-6
E
t
-6
0.5
t
t
Understanding
x
A ball is dropped for a height of two meters above
the ground.
t
v
Draw v vs t
t
Draw x vs t
Draw a vs t
a
t
Tossed Ball
A ball is tossed from the ground up a height of two meters above the
ground. And falls back down.
y
Draw v vs t
9
v
x
A
-6
1
9
v
B
9
t
9
1
-6
v
-6
1
C
t
9
D
v
1
-6
v
E
t
-6
1
t
t
Tossed Ball
x
A ball is tossed from the ground up a height of two
meters above the ground. And falls back down.
t
v
Draw v vs t
t
Draw x vs t
Draw a vs t
a
t
Understanding
A ball is thrown straight up in the air and returns to its
initial position. During the time the ball is in the air,
which of the following statements is true?
A - Both average acceleration and average velocity are zero.
B - Average acceleration is zero but average velocity is not zero.
C - Average velocity is zero but average acceleration is not zero.
D - Neither average acceleration nor average velocity are zero.
Vave = Y/t = (Yf – Yi) / (tf – ti) = 0
aave is not 0
since Vf and
Vi are not the
same !
aave = V/t = (Vf – Vi) / (tf – ti)
Example
x (meters)
100
0
•
•
•
•
-100
-200
position vs. time
-300
0
v (m/s)
5
10
t (seconds)
15
20
• Where is acceleration zero?
• Where is acceleration positive?
20
0
-20
-40
-60
-80
-100
Where is velocity zero?
Where is velocity positive?
Where is velocity negative?
Where is speed largest?
velocity vs. time
0
5
10
t (seconds)
15
20
Final Check



x(t)
Slope of x vs t gives v
Area under v vs t gives x!
4s
x
Which plot best represents v(t)
t
5s
v(t)
v(t)
t
t
v(t)
t
t
Matching Graphs