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Oct. 30, 2012 AGENDA: 1 – Bell Ringer 2 – Kinematics Equations 3 – Exit Ticket Today’s Goal: Students will be able to identify which kinematic equation to apply in each situation Homework 1. Pages 4-6

Oct. 30, 2012

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Oct. 30, 2012. AGENDA: 1 – Bell Ringer 2 – Kinematics Equations 3 – Exit Ticket. Today’s Goal: Students will be able to identify which kinematic equation to apply in each situation Homework 1. Pages 4-6. CHAMPS for Bell Ringer. C – Conversation – No Talking - PowerPoint PPT Presentation

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Page 1: Oct.  30,  2012

Oct. 30, 2012AGENDA:1 – Bell Ringer2 – Kinematics

Equations3 – Exit Ticket

Today’s Goal:Students will be able to identify which kinematic equation to apply in each situationHomework1. Pages 4-6

Page 2: Oct.  30,  2012

CHAMPS for Bell RingerC – Conversation – No Talking H – Help – RAISE HAND for questionsA – Activity – Solve Bell Ringer on

binder paper. Homework out on desk

M – Materials and Movement – Pen/Pencil, Notebook or Paper

P – Participation – Be in assigned seats, work silently

S – Success – Get a stamp! I will collect!

Page 3: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 4: Oct.  30,  2012

4 MINUTES REMAINING…

Page 5: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 6: Oct.  30,  2012

3 MINUTES REMAINING…

Page 7: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 8: Oct.  30,  2012

2 MINUTES REMAINING…

Page 9: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 10: Oct.  30,  2012

1minute Remaining…

Page 11: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 12: Oct.  30,  2012

30 Seconds Remaining…

Page 13: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 14: Oct.  30,  2012

BELL-RINGER TIME IS

UP!

Page 15: Oct.  30,  2012

October 30th (p. 13)Objective: Students will be able to identify which kinematic equation to apply in each situation

Bell Ringer:1. How many quantities did

we underline in each problem?

2. How many known variables are you given in each problem?

3. How many unknown variables are you asked to find in each problem?

4. How do you decide what equation to use?

5. What do the equations mean to you?

Page 16: Oct.  30,  2012

Shout OutsPeriod 5 – ChrisPeriod 7 – Latifah, Shawn

Page 17: Oct.  30,  2012

Oct. 30, 2012AGENDA:1 – Bell Ringer2 – Kinematics

Equations3 – Exit Ticket

Today’s Goal:Students will be able to identify which kinematic equation to apply in each situationHomework1. Pages 4-6

Page 18: Oct.  30,  2012

Week 8Weekly AgendaMonday – Kinematic Equations ITuesday – Kinematic Equations IIWednesday – Kinematic Equations

IIIThursday – ReviewFriday – Review

Unit Test next week!

Page 19: Oct.  30,  2012

What are equations?Equations are relationships.

Equations describe our world.

Equations have changed the course of history.

Page 20: Oct.  30,  2012

CHAMPS for Problems p. 4-6C – Conversation – No Talking unless

directed to work in groupsH – Help – RAISE HAND for questionsA – Activity – Solve Problems on Page

4-6M – Materials and Movement –

Pen/Pencil, Packet Pages 4-6P – Participation – Complete Page 4-6S – Success – Understand all

Problems

Page 21: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 22: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4)

1. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. What is the final velocity of the Road Runner?

vi = 0 m/sa = 3 m/s2

Δt = 10 secondsvf = ?

Page 23: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 24: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4)

1. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. What is the final velocity of the Road Runner?

vi = 0 m/s a = 3 m/s2 Δt = 10 seconds vf = ?vf = vi + aΔt

Page 25: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4)

1. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. What is the final velocity of the Road Runner?

vi = 0 m/s a = 3 m/s2 Δt = 10 seconds vf = ?vf = vi + aΔtvf = 0 m/s + (3 m/s2)(10 s) =

Page 26: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4)

1. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. What is the final velocity of the Road Runner?

vi = 0 m/s a = 3 m/s2 Δt = 10 seconds vf = ?vf = vi + aΔtvf = 0 m/s + (3)(10) = 30 m/s

Page 27: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4

2. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. How far does the Road Runner travel during the ten second time interval?vi = 0 m/s a = 3 m/s2 Δt = 10 seconds Δx = ?Δx = viΔt + aΔt2

2

Page 28: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4)

2. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. How far does the Road Runner travel during the ten second time interval?vi = 0 m/s a = 3 m/s2 Δt = 10 seconds Δx = ?Δx = viΔt + aΔt2

2Δx = (0)(10) + (3)(10)2

2

Page 29: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4)

2. Starting from rest, the Road Runner accelerates at 3 m/s2 for ten seconds. How far does the Road Runner travel during the ten second time interval?vi = 0 m/s a = 3 m/s2 Δt = 10 seconds Δx = ?Δx = viΔt + aΔt2

2Δx = (0)(10) + (3)(10)2

2Δx = 0 + 150 m = 150 m

Page 30: Oct.  30,  2012

Solving Problems: THE EASY WAY (p. 4

3. A bullet starting from rest accelerates at 40,000 m/s2 down a 0.5 m long barrel. What is the velocity of the bullet as it leaves the barrel of the gun?vi = 0 m/s a = 40,000 m/s2 Δx = 0.5 m vf = ?

Page 31: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 32: Oct.  30,  2012

Solving Problems (p. 4)3. A bullet starting from rest accelerates at 40,000 m/s2 down a 0.5 m long barrel. What is the velocity of the bullet as it leaves the barrel of the gun?vi = 0 m/s a = 40,000 m/s2 Δx = 0.5 m vf = ?vf

2 = vi2 + 2aΔx

Page 33: Oct.  30,  2012

Solving Problems (p. 4)3. A bullet starting from rest accelerates at 40,000 m/s2 down a 0.5 m long barrel. What is the velocity of the bullet as it leaves the barrel of the gun?vi = 0 m/s a = 40,000 m/s2 Δx = 0.5 m vf = ?vf

2 = vi2 + 2aΔx

vf2 = (0)2 + 2(40,000)(0.5)

Page 34: Oct.  30,  2012

Solving Problems (p. 4)3. A bullet starting from rest accelerates at 40,000 m/s2 down a 0.5 m long barrel. What is the velocity of the bullet as it leaves the barrel of the gun?vi = 0 m/s a = 40,000 m/s2 Δx = 0.5 m vf = ?vf

2 = vi2 + 2aΔx

vf2 = (0)2 + 2(40,000)(0.5)

vf2 = 40,000

vf = 200 m/s

Page 35: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?

Page 36: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 37: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 38: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?vf = vi + aΔt

Page 39: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?vf = vi + aΔt0 = 20 + 4a

Page 40: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?vf = vi + aΔt0 = 20 + 4a

0 = 20 + 4a

Page 41: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?vf = vi + aΔt0 = 20 + 4a

-20 + 0 = 20 + 4a + -20-20 = 4a

Page 42: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?vf = vi + aΔt0 = 20 + 4a

-20 + 0 = 20 + 4a + -20-20/4 = 4a/4

Page 43: Oct.  30,  2012

Solving Problems (p. 4)4. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. What is the acceleration of the car?vi = 20 m/s vf = 0 m/s Δt = 4 seconds

a = ?vf = vi + aΔt0 = 20 + 4a

-20 + 0 = 20 + 4a + -20-20/4 = 4a/4

a = -5 m/s2

Page 44: Oct.  30,  2012

Solving Problems (p. 5)5. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. How far does the car travel before coming to a stop?vi = 20 m/s vf = 0 m/s Δt = 4s Δx = ?

Page 45: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 46: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 47: Oct.  30,  2012

Solving Problems (p. 5)5. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. How far does the car travel before coming to a stop?vi = 20 m/s vf = 0 m/s Δt = 4s Δx = ?

Δx = (vf + vi)Δt

2

Page 48: Oct.  30,  2012

Solving Problems (p. 5)5. A car traveling at 20 m/s applies its brakes and comes to a stop in four seconds. How far does the car travel before coming to a stop?vi = 20 m/s vf = 0 m/s Δt = 4s Δx = ?

Δx = (vf + vi)Δt = (0 + 20)(4) = 40 m

2 2

Page 49: Oct.  30,  2012

Solving Problems (p. 5)6. The USS Enterprise accelerates from rest at 100,000 m/s2 for a time of four seconds. How far did the ship travel in that time?

Page 50: Oct.  30,  2012

Solving Problems (p. 5)6. The USS Enterprise accelerates from rest at 100,000 m/s2 for a time of four seconds. How far did the ship travel in that time?vi = 0 m/s a = 100,000 m/s2 Δt = 4s Δx = ?

Page 51: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 52: Oct.  30,  2012

Notes: Kinematic Equations

The Four Kinematic Equations:vf = vi + aΔt

Δx = viΔt + aΔt2

2vf

2 = vi2 + 2aΔx

Δx = (vf + vi)Δt 2

Page 53: Oct.  30,  2012

Solving Problems (p. 5)6. The USS Enterprise accelerates from rest at 100,000 m/s2 for a time of four seconds. How far did the ship travel in that time?vi = 0 m/s a = 100,000 m/s2 Δt = 4s Δx = ?

Δx = viΔt + aΔt2 =

2

Page 54: Oct.  30,  2012

Solving Problems (p. 5)6. The USS Enterprise accelerates from rest at 100,000 m/s2 for a time of four seconds. How far did the ship travel in that time?vi = 0 m/s a = 100,000 m/s2 Δt = 4s Δx = ?

Δx = viΔt + aΔt2 = (0)(4) + (100,000)(4)2

2 2

Page 55: Oct.  30,  2012

Solving Problems (p. 5)6. The USS Enterprise accelerates from rest at 100,000 m/s2 for a time of four seconds. How far did the ship travel in that time?vi = 0 m/s a = 100,000 m/s2 Δt = 4s Δx = ?

Δx = viΔt + aΔt2 = (0)(4) + (100,000)(4)2

2 2Δx = 800,000 m

Page 56: Oct.  30,  2012

Solving Problems (p. 5)7. At the scene of an accident, a police officer notices that the skid marks of a car are 10 m long. The officer knows that the typical deceleration of this car when skidding is -45 m/s2. What can the officer estimate the original speed of the car?

Page 57: Oct.  30,  2012

Solving Problems (p. 5)7. At the scene of an accident, a police officer notices that the skid marks of a car are 10 m long. The officer knows that the typical deceleration of this car when skidding is -45 m/s2. What can the officer estimate the original speed of the car?

Page 58: Oct.  30,  2012

Solving Problems (p. 5)7. At the scene of an accident, a police officer notices that the skid marks of a car are 10 m long. The officer knows that the typical deceleration of this car when skidding is -45 m/s2. What can the officer estimate the original speed of the car?Δx = 10 m a = -45 m/s2 vf = 0 m/s vi = ?