Pre AP Physics
One Dimensional
Motion
Velocity, Acceleration, Motion Graphs
Velocity
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Displacement divided by time interval
Velocity is the change in position divided by the
time interval during which the motion took place.
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Velocity is a vector quantity
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Speed is scalar and is distance divided by time.
Types of Velocity
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Average Velocity – the total displacement of motion
divided by the total time interval.
Instantaneous Velocity – the velocity in a specific
instant. Like a quick check on your speedometer.
Initial Velocity – the velocity at the start of the time
interval. When object starts at rest, Vi = zero.
Final Velocity – the velocity of the object at the end of
the time interval. For an object braking to a stop, Vf =
zero.
Terminal Velocity – got to wait for the force chapter,
sorry.
Average Velocity Equation
Vavg=
Manipulate to solve for d and t
Velocity and Direction
Practice Problems 2A p.44 1-5
ClassWork
1.
2.
3.
How far have you walked if your velocity is
2m/s West and you walk for 2 minutes?
How much time would it take to drive 2km if
your average velocity is 30m/s?
What is Mr. Crabtree’s average velocity if he
runs 3.2km in a time of 18 minutes and 20
seconds?
Distance – Time Graphs
Also called Position – Time Graphs
Position (m)
Time (sec)
Slope – what does it mean?
Constant Velocity
A special Case
● An unchanging Velocity
● Not speeding up, not slowing
down, not changing
direction.
● Its like perfect cruise control
on a level, straight highway.
●
Change in Velocity
speed up, slow down, or change in direction
ā= V2-V1
t2-t1
Units=
m/s2
m/s
s
ms-2
Acceleration is a vector quantity
(like velocity)
Direction makes a difference!
If you have (+) positive velocity and speed up
=(+)positive acceleration
(+) positive velocity and slow down
=(-)negative acceleration
An acceleration value tells you by
how much an object is changing its
For an object
accelerating
@ 3m/s2
Time
velocity
Velocity
0 sec
O m/s
1 sec
3 m/s
2 sec
6 m/s
3 sec
9 m/s
4 sec
12 m/s
5 sec
15 m/s
Example Problem
The velocity of a car increases from 2 m/s at 1 sec.
To 16 m/s at 4.5 sec. What is the car’s average
acceleration?
Given
Vi= 2m/s
Vf= 16m/s
T1= 1 sec
T2= 4.5 sec
Formula
a=
v
t
Solution
a= 16m/s – 2 m/s
4.5 sec- 1 sec
a= 14m/s
3.5 sec
= 4m/s2
Look at the following
Practice Problems
p.49
1,2,3
p.49
Sample
On A velocity-Time
Graph:
Slope= Acceleration
Sample 2B p.49
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A shuttle bus slows to a stop with an average
acceleration of –1.8m/s2. How long does it take
the bus to slow from 9m/s to 0m/s?
Practice
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What is the acceleration of a car that goes from 4m/s to
36m/s in a time interval of 4 seconds?
What is the acceleration of a car that goes from 36m/s to
15m/s in a time interval of 3 seconds?
What is the acceleration of a vehicle that goes from
–3m/s to 4.5m/s in a time of 2.5 seconds?
Solutions on the next 2 slides
#1.
Given
Vi= 4m/s
Vf= 36m/s
t= 4s
Formula
a = Vf-Vi
t
36 – 4
4
#2. Vi= 36 m/s
8 m/s2
a = Vf-Vi
Vf= 15 m/s
t= 3 sec
Solution
t
15
– 36
3
- 7 m/s2
#3. Vi= -3 m/s
Vf= + 4.5 m/s
t= 2.5 sec
a= Vf – Vi
t
3 m/s2
4.5 – (-3)
2.5
GREAT JOB!
Acceleration that does not change is called uniform or constant acceleration
(for simplicity sake, the problems we will solve will be in uniform acceleration)
2nd Formula–
Vf= Vi + at
Vi= Vf – at
t
t = Vf – Vi
a
a= Vf – Vi
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If a car with a velocity of 2 m/s accelerates at
a rate of (+)4 m/s2 for 2.5 sec., what is the
velocity at 2.5 sec?
Given
Formula
Vi= 2 m/s
Vf= Vi + at
A = 4 m/s2
Vf= 2 + 4(2.5)
T= 2.5 sec
Vf= ??
Do sample problem p. 55
Solution
Vf = 12 m/s
Sample Problem p.55
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A plane starting from rest at one end of the runway undergoes a uniform
acceleration of 4.8m/s2 for 15 seconds before takeoff. What is the speed
at takeoff?
 Displacement when velocity and time are known
d= ½ (Vf + Vi) t
*True
when object is accelerated uniformly
Displacement when acceleration and time are known
d= Vit + ½ at2
*recall
that an object starting from rest has Vi= 0 m/s
When starting from rest
d= ½ at2
Sample Problem
●
A plane starting from rest at one end of the runway undergoes a uniform
acceleration of 4.8m/s2 for 15 seconds before takeoff. How long must the
runway be for the plane to be able to take off?
2C
Page 53
practice problems 1-4
2D
Page 55 practice problems 1-4
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A certain airplane wants to takeoff from a runway
that is 1 kilometer long. The plane starts from rest
and must get to a speed of 71m/s in order to lift off
the ground. What is the minimum acceleration
the pilot has to achieve to order to takeoff?
** you are not given time.
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Answer is on the next slide.
Solving without time
Vf2= Vi2 + 2 ad
Vf= Vi2+ 2ad
a= Vf2 – Vi2
2d
Given
Formula
Vi= 0
a= Vf2- Vi2
Vf= 71 m/s
2d
d= Vf2 – Vi2
2a
Solution
a = 712 - 0
2(1000)
2.5 m/s2
d= 1000m
HW pages
2E
Page 58 practice problems 1-4
Velocity Time Graphs
Velocity
(m/s)
Time (sec)
V – t Graphs
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Slope equals ______________________
Area under the slope (curve, line) equals
________________
examples
Velocity (m/s)
Time (sec)
Class Work
Page 68 N and P
D-t and V-t graphs side by side.
Acceleration due to
gravity
 Earth’s
gravity pulls everything towards the center
of the earth
 All
things are pulled at the same rate. (atoms to
elephants- all the same) – so long as we neglect air
resistance, which we will most of the time.
 Gravity
is a uniform acceleration
 Gravity
on Earth can change as you go from place
to place. Gravity changes as you change the
distance from the center of the earth to your
location.
 EXAMPLE: on
a mountain gravity is less than at
sea level. also earth isn’t perfectly round gravity is
less at equator than poles
Gravity can vary from place to place;
but only at the same place are all
objects accelerated the same.
Every planet- (as well as the moon and
sun) has its own acceleration due to
gravity
size and mass determine gravity
acceleration due to gravity is a
vector quantity
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We will use an average value for gravity when solving problems on
Earth.
g= - 9.81 m/s2
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As an object falls w/out air on Earth
Time
Velocity
0s
0
1s
-9.8 m/s
-32 ft/s
2s
-19.6 m/s
-64 ft/s
3s
-29.4 m/s
-96 ft/s
4s
-39.2 m/s
-128 ft/s
Displacement During free fall
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Displacement = ½ a t2 for objects falling from
rest. a=accel due to gravity, -9.8m/s2.
After falling for 1 second, an object on Earth has
fallen -4.9m
After falling for 2 seconds, the object is -19.6m
below the drop position.
After falling for 3 seconds, the object is -44.1m
below the drop position.
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**overhead model**
as the ball goes up gravity takes 9.8 m/s off the
velocity for every second
as the ball goes down gravity puts on 9.8 m/s for
every second the ball falls
Example problems p.68U
Practice problem p.68U-V 3-7
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reaction timer page 68T
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Gravity is constant and therefore we
can “sub” it into ALL of our formulas
1.
g=
v
t
2. Vf = Vi + gt
4. D = Vit+ ½ gt2
5. Vf2 = Vi2 + 2 gd
*Class work*
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The time the sky screamer ride at Astro world is free falling is 1.5
sec, (A) what is its velocity at the end of the time? (B) How far
does it fall?
Given
g= - 9.8 m/s 2
Vi= 0 m/s
t = 1. 5 s
Formula
Solution
Vf = Vi + gt
Vf= 0 + (-9.8)(1.5)
d= Vit + ½ gt2
(A): Vf= - 15 m/s
d= 0 + ½ (-9.8)(1.5)2
(B): d= -11 m
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A brick falls freely from a high scaffold (A) what is the velocity after 4 seconds
(B) how far does the brick fall during these 4 seconds?
Given
g= - 9.8 m/s2
t= 4 sec
Vi= 0 m/s
Formula
Solution
Vf= Vi + gt
Vf= 0 + (-9.8)4
d= Vit + ½ gt2
(A): Vf= -39.2m/s
d= ½ (-9.8)(4)2
( B): d= - 78.4m
Formulas
a= Vf – Vi
Vi
Vf
a
t
Vf= Vi+ at
Vi
Vf
a
t
d= ½ (Vf + Vi)t
Vi
Vf
d= Vit + ½ at2
Vi
Vf2= Vi2 +2ad
Vi
Vf
g= Vf – Vi
Vi
Vf
g
t
Vf= Vi + gt
Vi
Vf
g
t
d= Vit + ½ gt2
Vi
d
g
t
Vf2= Vi2 + 2gd
Vi
d
g
t
d
t
d
a
d
a
t
t
Vf
What’s going on here …..?
What’s going on here…..?