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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
ideasnotes
Acceleration def.
Acceleration direction
Graph representation
Equations
• AccelerationAcceleration
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?”
Acceleration = (change in velocity) / time
change in VELOCITY occuring over TIME
Acceleration is a Vector quantity
The unit for acceleration is
if
ifavg tt
vv
t
va
m/s2
• 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
Positive Velocity
Positive Acceleration
Speeding up in + direction
Negative Velocity
Negative Acceleration
Speeding up in - direction
Positive Velocity
Negative Acceleration
Slowing downEventually speeds up in – direction!
Negative Velocity
Positive Acceleration
Slowing downEventually speeds up in + direction!
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!!!
demo
+v
+a
The car will…
-v
-a
The car will… The car will… The car will…
+v
-a
-v
+a
Conceptual challengePage 50
The slope of the v-t graph describe the object’s acceleration
Velocity vs. Time
-20
-15
-10
-5
0
5
10
15
20
25
30
35
0 1 2 3 4 5 6 7 8 9 10 11
Time (s)
Vel
oci
ty (
m/s
)
B
AC
D
What is the Acceleration at A, B, C, D?
Figure 2-10
t
vaavg
A: +10 m/s2
B: zeroC: -5 m/s2
D:-15 m/s2
What is happening at point D?
Conceptual challenge
• Page 50
Constant acceleration• demoConstant acceleration - the velocity is changing by a constant amount - in each second of time.
= vf + vi
2∆x∆t
∆x = ½ (∆t)(vf + vi)
The displacement equals to the area under the velocity vs. time graph
2fi
avg
vvv
t
xvavg
a = slope
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 EXCEPTa. 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 paperShow name of assignment
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 ticker tape
Constant motion vs. accelerated motion vector diagrams
• displacement vs. time (constant motion)
disp
lace
men
t
time time
Constant motion - constant speed with constant direction
disp
lace
men
t
Alert: in displacement vs. time graph, displacement can be positive or negative
Slope represent velocity
At restConstant, positive velocity
When velocity is constant
• velocity vs. time (constant motion)ve
loci
ty
time time
velo
city
Alert: in velocity vs. time graph, velocity can be positive or negative
Slope represent acceleration
Constant negative velocity, zero acceleration
Constant, positive velocity, zero acceleration
• displacement vs. time (accelerated motion)di
spla
cem
ent
time time
Speeding up in positive direction
Slowing down in positive direction
disp
lace
men
t
disp
lace
men
t
time
disp
lace
men
t
time
Speeding up in negative direction
Slowing down in negative direction
When velocity is changing
• velocity vs. time (constant accelerated motion)ve
loci
ty
time time
Speeding up in positive direction at constant rateAcceleration is constant, positive
Slowing down in positive direction at constant rate, acceleration is constant, negative
velo
city
velo
city time
velo
city time
Speeding up in negative direction at constant rateAcceleration is constant, negative
Slowing down in negative direction at constant rate, acceleration is constant, positive
Alert – v vs. t and d vs. t must matchve
loci
ty
time time
velo
city
velo
city
time
velo
city
time
Dis
p.
time
Dis
p.
time
Dis
p.
time
Dis
p.
time
Determine acceleration
From 0 s to 4 s:From 4 s to 8 s:
Determine acceleration - Example 1
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 notes
Acceleration def.
Acceleration direction
Graph representation
Equations
Speed up: a and v have same signSlow down: a and v have opp. Sign
t
va
v
t
v
t
No acc.
acc. = slopedisp = area
vavg
t
va
tvvx fi )(2
1
No acc.d
t
d
t
Final velocity depends on initial velocity, acceleration, and time
t
va
t
vva if
if vvta
tavv if
tavv if tvvx fi )(2
1
ttavvx ii )(2
1
2
2
1tatvx i
2
2
1tatvx i
vi
vf
v
t0 tf
tvi
2
2
1)(
2
1tavvt if
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
tavv if tvvx fi )(2
1
Solve for ∆t
a
vvt if Sub ∆t in this equation
a
vvvvx iffi
)(
2
1
xavv if 222
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 - review1. 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.
HomeworkCastle learning, must show work for incorrect answers to get full credit
In one dimensional motion:
∆x = d
• 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?
Example
84 m
∆x = ½ (vf + vi)(∆t)
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?
– How long does it take the train to reach its final velocity?
vf2 = vi
2 + 2advf
2 = 0 + 2(1.20 m/s2)(500. m)vf = 34.6 m/s
vf = vi + at34.6 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?
– What would his acceleration need to be to not get a ticket?
vf2 = vi
2 + 2advf
2 = (85. mph)2 + 2(-500. mph2)(2.00 mi)vf = 72.3 mph *YES*
vf2 = vi
2 + 2ad(55 mph)2 = (85. mph)2 + 2a(2.00 mi)
a = -1050 mph2
Class work
• Problem workbook 2A, 2B, 2C, 2D, 2E