9/23 do now – on a new sheet
• Which of the following is the expression for
average velocity?
Homework questions?
Motion in one dimension
2.2
Acceleration
Lesson Objectives:
1. Describe motion in terms of changing velocity.
2. Compare graphical representations of accelerated
and nonaccelerated motions.
3. Apply kinematic equations to calculate distance,
time, or velocity under conditions of constant
acceleration.
Do Now - Notes
Key terms and
ideas
Acceleration def.
Acceleration direction
Graph representation
Equations
notes
Acceleration
• Acceleration
change in VELOCITY occuring over TIME
Acceleration = (change in velocity) / time

aavg
  
v v f  vi


t t f  ti
The unit for acceleration is
Acceleration is a
Vector quantity
m/s2
Questions:
1. “If an object has a large velocity, does it necessarily have a
large acceleration?
2. If an object has a large acceleration, does it necessarily have a
large velocity?”
• Anytime an object's velocity is changing, the
object is said to be accelerating; it has an
acceleration.
• What are the three ways to accelerate your
car?
Gas pedal, break, steering wheel
Example – finding time of acceleration
• A shuttle bus slows to a stop with an average
acceleration of -1.8 m/s2. How long does it take the
bus to slow from 9.0 m/s to 0.0 m/s?
Example – finding time of acceleration
• Find the acceleration of an amusement park ride that
falls from rest to a speed of 28 m/s in 3.0 s.
Do now
1. In addition to displacement, which of the
following must be used for a more complete
description of the average velocity of an
object?
2. finish Class work Page 49 #1-5
The Direction of the Acceleration Vector
Negative
Positive Velocity
Velocity
Negative
Positive Acceleration
Acceleration
Speeding
Speeding
Slowing
up
up in
indown
+- direction
direction
Eventually speeds up in +
– direction!
demo
The Direction of the Acceleration Vector
• The direction of the acceleration vector depends on whether
the object is speeding up or slowing down
• If an object is speeding up, then its acceleration is in the same
direction of its motion.
– moving in the + direction, the acceleration + direction
– moving in the - direction, the acceleration - direction
• If an object is slowing down, then its acceleration is in the
opposite direction of its motion.
– moving in the + direction, the acceleration - direction
– moving in the - direction, the acceleration + direction
• The direction of velocity and acceleration do not have to be
the same!!!
+v
-v
+v
-v
+a
-a
-a
+a
The car will…
The car will…
The car will…
Conceptual challenge
Page 50
The car will…
The slope of the v-t graph describe the object’s
acceleration
Figure 2-10
aavg
Velocity (m/s)
Velocity vs. Time
35
30
25
20
15
10
5
0
-5 0
-10
-15
-20
B
C
What is the
Acceleration at A,
B, C, D?
A
D
1
2
3
4
5
v

t
6
Time (s)
7
8
9
10 11
A: +10 m/s2
B: zero
C: -5 m/s2
D:-15 m/s2
What is happening
at point D?
Conceptual challenge
• Page 50
Constant acceleration
• demo
Constant acceleration - the velocity is changing by a
constant amount - in each second of time.
vavg 
vf + vi
2
vi  v f
2
=
∆x
∆t
vavg
x

t
a = slope
∆x = ½ (∆t)(vf + vi)
The displacement equals to the
area under the velocity vs. time
graph
Determining the Area on a v-t Graph
In velocity versus time graphs, the area bounded by the
line and the axes represents the displacement.
The shaded area is representative of the
displacement
area = base x height
area = ½ base x height
area = ½ base x ( height1 + height2)
or
area = big triangle – small triangle
example
• Determine the displacement of the object during the time
interval from 2 to 3 seconds (Practice A) and during the first 2
seconds (Practice B).
25 m
40 m
Example – 2C
• A racing car reaches a speed of 42 m/s. It then begins a
uniform negative acceleration, using its parachute and
breaking system, and comes to rest 5.5 s later. Find how far
the car moves before stopping.
120 m
Class work
• Page 53 – practice 2C
Do now
Velocity measures all of the following EXCEPT
a. the speed of an object.
b. the total displacement of an object.
c. the direction of an object’s motion.
d. the displacement for each time interval.
Lesson Objectives:
1. Describe motion in terms of changing velocity.
2. Compare graphical representations of accelerated
and nonaccelerated motions.
3. Apply kinematic equations to calculate distance,
time, or velocity under conditions of constant
acceleration.
Homework:
Castle learning
Must show work for incorrect answer after one
try on a separate paper
Show name of assignment
Constant motion vs. accelerated motion
ticker tape
constant velocity: displacement is changing by a constant amount - in each
second of time.
constant acceleration: displacement is increasing in each second of time. Velocity is
changing by a constant amount - in each second of time
Constant motion vs. accelerated motion vector
diagrams
When velocity is constant
time
At rest
displacement
displacement
• displacement vs. time (constant motion)
time
Constant, positive
velocity
Constant motion - constant speed with constant direction
Slope represent velocity
Alert: in displacement vs. time graph, displacement can be positive or negative
time
Constant negative velocity, zero
acceleration
velocity
velocity
• velocity vs. time (constant motion)
time
Constant, positive velocity, zero
acceleration
Slope represent acceleration
Alert: in velocity vs. time graph, velocity can be positive or negative
When velocity is changing
Speeding up in positive
direction
Slowing down in
positive direction
displacement
displacement
• displacement vs. time (accelerated motion)
Speeding up in negative
direction
time
time
Slowing down in negative
direction
displacement
displacement
time
time
Speeding up in positive
direction at constant rate
Acceleration is constant,
positive
velocity
velocity
• velocity vs. time (constant accelerated motion)
time
time
Speeding up in negative direction
at constant rate
Acceleration is constant, negative
velocity
velocity
time
Slowing down in positive
direction at constant rate,
acceleration is constant,
negative
time
Slowing down in negative
direction at constant rate,
acceleration is constant,
positive
time
time
time
time
time
time
Disp.
velocity
time
Disp.
velocity
Disp.
velocity
Disp.
velocity
Alert – v vs. t and d vs. t must match
time
Determine acceleration - Example 1
Determine acceleration
From 0 s to 4 s:
From 4 s to 8 s:
0 m/s2
2 m/s2
Do now
• Sketch displacement vs. time and velocity vs.
time graphs to indicate an object is
accelerating at a constant rate.
Lesson Objectives:
1. Describe motion in terms of changing velocity.
2. Compare graphical representations of accelerated
and nonaccelerated motions.
3. Apply kinematic equations to calculate distance,
time, or velocity under conditions of constant
acceleration.
review
Key terms and ideas
Acceleration def.
Acceleration direction
notes
a
v
t
Speed up: a and v have same sign
Slow down: a and v have opp. Sign
v
No acc.
d
No acc.
acc. =
disp = area
v
vavg
d
Graph representation
t
Equations
a
v
t
t
x 
1
(vi  v f )t
2
t
t
Final velocity depends on initial velocity,
acceleration, and time
v
a
t
a
v f  vi
t
at  v f  vi
v f  vi  at
v f  vi  at
1
x  (vi  v f )t
2
1
x  (vi  vi  at )t
2
1
x  vi t  at 2
2
1
x  vi t  at 2
2
v
vf
1
1
t (v f  vi )  at 2
2
2
vi
vi t
0
tf
t
In velocity versus time graphs, the area bounded by
the line and the axes represents the displacement.
Example 2D
• A plane starting at rest at one end of a runway
undergoes a uniform acceleration of 4.8 m/s2
for 15 s before takeoff. What is the its speed at
takeoff? How long must the runway be for the
plane to be able to take off?
Class work
• Page 55, practice 2D
Time and be found from displacement
and velocity
v f  vi  at
1
x  (vi  v f )t
2
Solve for ∆t
t 
v f  vi
Sub ∆t in this equation
a
v f  vi
1
x  (vi  v f )
2
a
v f  vi  2ax
2
2
Example – 2E
• A person pushing a stroller starts from rest,
uniformly accelerating at a rate of 0.500 m/s2.
what is the velocity of the stroller after it has
traveled 4.75 m?
Class work
• Page 58 – practice 2E #1-6
In one dimensional
motion:
∆x = d
Class work
• Section Review Worksheet 2-2
Do now
• A car travels 90. meters due north in 15 seconds. Then the car
turns around and travels 40. meters due south in 5.0 seconds.
What is the magnitude of the average speed and average
velocity of the car during this 20.-second interval?
• Objectives - review
1. Describe motion in terms of changing velocity.
2. Compare graphical representations of accelerated and non
accelerated motions.
3. Apply kinematics equations to calculate distance, time, or
velocity under conditions of constant acceleration.
Homework
Castle learning, must show work for incorrect answers to get full credit
In one dimensional
motion:
∆x = d
Example
• A bicyclist accelerates from 5.0 m/s to a
velocity of 16 m/s in 8 s. Assuming uniform
acceleration, what displacement does the
bicyclist travel during this time interval?
∆x = ½ (vf + vi)(∆t)
84 m
Example
• A train starts from rest and leaves Greenburg
station and travels for 500. meters with an
acceleration of 1.20 meters per second2.
– What is the train’s final velocity?
vf2 = vi2 + 2ad
vf2 = 0 + 2(1.20 m/s2)(500. m)
vf = 34.6 m/s
– How long does it take the train to reach its final
velocity?
vf = vi + at
34.6 m/s = 0 + (1.20 m/s2) t
t = 28.8 s
Example
• A driver traveling at 85. miles per hour sees a police
car hiding in the trees 2.00 miles ahead. He applies his
brakes, decelerating at -500. miles per hour2.
– If the speed limit is 55 mph, will he get a ticket?
vf2 = vi2 + 2ad
vf2 = (85. mph)2 + 2(-500. mph2)(2.00 mi)
vf = 72.3 mph *YES*
– What would his acceleration need to be to not get a ticket?
vf2 = vi2 + 2ad
(55 mph)2 = (85. mph)2 + 2a(2.00 mi)
a = -1050 mph2
Class work
•
Problem workbook 2A, 2B, 2C, 2D, 2E