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

vvtva

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

-505

101520253035

0 1 2 3 4 5 6 7 8 9 10 11

Time (s)

Velo

city

(m/s

)

B

AC

D

What is the Acceleration at A, B, C, D?

Figure 2-10

tvaavg

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

txvavg

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

Disp

.

time

Disp

.

time

Disp

.

time

Disp

.

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

tva

v

t

v

t

No acc.

acc. = slopedisp = area

vavg

tva

tvvx fi )(21

No acc.d

t

d

t

Final velocity depends on initial velocity, acceleration, and time

tva

tvv

a if

if vvta

tavv if

tavv if tvvx fi )(21

ttavvx ii )(21

2

21 tatvx i

2

21 tatvx i

vi

vf

v

t0 tf

tvi

2

21)(

21 tavvt 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 )(21

Solve for ∆t

avv

t if Sub ∆t in this equation

avv

vvx iffi

)(

21

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