Oct. 30, 2012
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Transcript of 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
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!
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?
4 MINUTES REMAINING…
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?
3 MINUTES REMAINING…
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?
2 MINUTES REMAINING…
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?
1minute Remaining…
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?
30 Seconds Remaining…
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?
BELL-RINGER TIME IS
UP!
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?
Shout OutsPeriod 5 – ChrisPeriod 7 – Latifah, Shawn
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
Week 8Weekly AgendaMonday – Kinematic Equations ITuesday – Kinematic Equations IIWednesday – Kinematic Equations
IIIThursday – ReviewFriday – Review
Unit Test next week!
What are equations?Equations are relationships.
Equations describe our world.
Equations have changed the course of history.
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
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
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 = ?
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
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
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) =
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
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
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
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
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 = ?
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
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
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)
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
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 = ?
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
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
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
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
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
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
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
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 = ?
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
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
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
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?
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 = ?
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
Notes: Kinematic Equations
The Four Kinematic Equations:vf = vi + aΔt
Δx = viΔt + aΔt2
2vf
2 = vi2 + 2aΔx
Δx = (vf + vi)Δt 2
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
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
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
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?
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?
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 = ?