PHYSICS 11 TODAY: Acceleration

Grading Scale for Physics11 (it might change) 2013/2014 Assignments (worksheets, labs, homework, pop-quizzes) 15% Quizzes (around 25 of them) 20% Unit Tests (7 of them – 5% each) 35% Projects (posters, ppt presentations, research etc..) 10% Final Exam 20%

Letter Grades for Physics11 (it might change) 2013/2014 A 100% – 86% B 85% – 73% C+ 72% – 67% C 66% – 60% D 59% - 50% F 49% - under

2.2 ACCELERATION

WHAT HAPPENS TO ITS VELOCITY?? 2.2 ACCELERATION Imagine a train speeding up or slowing down… WHAT HAPPENS TO ITS VELOCITY?? Whenever velocity changes:

2.2 ACCELERATION

3.2 ACCELERATION

ACCELERATION SI Units: meters per second square (m/s2) It has DIRECTION and MAGNITUDE

from “rest” means (vi = 0 m/s) “comes to rest” means (vf = 0 m/s) Find the acceleration of an amusement park ride that falls from rest to a speed of 28.0 m/s in 3.0 s HINT: from “rest” means (vi = 0 m/s) “comes to rest” means (vf = 0 m/s) 9.3 m/s2

Speed Direction Velocity has two aspect to it: Acceleration can occur under 3 conditions:

3. Speed and Direction change When… 1. Speed changes 2. Direction changes 3. Speed and Direction change

vf > v0 vf < v0 1. Speed changes The object is speeding up slowing down vf > v0 vf < v0

1. Speed changes The object is speeding up

1. Speed changes The object is slowing down

If direction changes, velocity changes because:

3. Speed and Direction change

The car is slowing down… Practice Problems: 2.2.1 (pg. 49) - 15.0 m/s2 The car is slowing down…

Negative ACCELERATION Is an object speeding up OR slowing down?

Homework Textbook: pg. 49 – PRACTICE B (1 – 5)

What can you say about displacement and velocity? Magnitude and Direction of Acceleration Imagine a train moving to the right… What can you say about displacement and velocity? STATION 1 STATION 2 displacement is positive velocity is positive

velocity is not changing, therefore Δv is 0 m/s Magnitude and Direction of Acceleration Case #1: Imagine a train going with no speeding up, or stopping = at a constant velocity What can you say about velocity and acceleration? STATION 1 STATION 2 velocity is not changing, therefore Δv is 0 m/s acceleration is zero because velocity is constant

Magnitude and Direction of Acceleration Case #2: Imagine a train speeding up as it moves to the right … STATION 1 STATION 2

What can you say about velocity? Magnitude and Direction of Acceleration Case #2: Imagine a train speeding up as it moves to the right … What can you say about velocity? STATION 1 STATION 2 velocity’s magnitude increases (from 5 m/s to 10 m/s) so Δv will be positive

acceleration will be positive Magnitude and Direction of Acceleration Case #2: Imagine a train speeding up as it moves to the right … What can you say about acceleration? STATION 1 STATION 2 acceleration will be positive because the change in velocity (Δv) is positive

Magnitude and Direction of Acceleration Case #3: Imagine a train going in the positive direction, but it is slowing down (not stopping) as it approaches the next station STATION 1 STATION 2 What can you say about displacement, velocity, change in velocity and acceleration?

displacement is positive because Δx (xf – xi) is > 0 Magnitude and Direction of Acceleration Case #3: Imagine a train going in the positive direction, but it is slowing down as it approaches the next station STATION 1 STATION 2 displacement is positive because Δx (xf – xi) is > 0

Magnitude and Direction of Acceleration velocity is still positive Case #3: Imagine a train going in the positive direction, but it is slowing down as it approaches the next station STATION 1 STATION 2 velocity is still positive (for example: vi = 10 m/s and vf = 2 m/s) but Δv is negative (vf – vi) is < 0

Magnitude and Direction of Acceleration acceleration is also negative Case #3: Imagine a train going in the positive direction, but it is slowing down as it approaches the next station STATION 1 STATION 2 Because Δv is negative acceleration is also negative

Magnitude and Direction of Acceleration Case #4: Imagine a train going in the negative direction, and it is speeding up as it approaches the next station STATION -1 STATION 1 velocity is negative (for example: vi = - 2 m/s and vf = - 10 m/s) and Δv is negative (vf – vi) is < 0

Because Δv is negative acceleration is also negative Magnitude and Direction of Acceleration Case #4: Imagine a train going in the negative direction, and it is speeding up as it approaches the next station STATION 1 STATION 2 Because Δv is negative acceleration is also negative Notice: acceleration is negative and the object is speeding up!

Magnitude and Direction of Acceleration Case #5: Imagine a train going in the negative direction, and it is slowing down as it approaches the next station STATION -1 STATION 1 velocity is negative (for example: vi = - 10 m/s and vf = - 2 m/s) and Δv is positive (vf – vi) is > 0 (!!)

Because Δv is positive acceleration is also positive Magnitude and Direction of Acceleration Case #5: Imagine a train going in the negative direction, and it is slowing down as it approaches the next station STATION 1 STATION 2 Because Δv is positive acceleration is also positive Notice: acceleration is positive and the object is slowing down!

SUMMARY Speeding up Slowing down

If Scott is moving in the NEGATIVE direction Think About it… When Scott is out for a bike ride, he slows down on his bike as he approaches a group of hikers on a trail. Explain how his accelaration can be positive even though his speed is decreasing If Scott is moving in the NEGATIVE direction vi = -10 m/s vf = -5 m/s Slowing down a = vf − vi Δt = (−5) − (−10) Δt = −5 + 10 Δt = 5 Δt = POSITIVE

Slope of velocity (speed) vs time graph = acceleration The slope of the graph of velocity (speed) vs. time is the average acceleration Slope of velocity (speed) vs time graph = acceleration The equation for the line (y = mx +b) will have form: v = a·t + (y – intercept) v = 69t + 7.0

The slope at that point is positive = acceleration is POSITIVE The slope of the graph of velocity (speed) vs. time is the average acceleration Point A The slope at that point is positive = acceleration is POSITIVE The velocity in the positive direction is increasing (E.g. train just left the station and its speed is increasing)

The slope at that point is zero = acceleration is ZERO The slope of the graph of velocity vs. time is the average acceleration The slope at that point is zero = acceleration is ZERO Point B The velocity is constant (E.g. train is moving with out any change in velocity

The slope at that point is negative = acceleration is NEGATIVE The slope of the graph of velocity vs. time is the average acceleration Point C The slope at that point is negative = acceleration is NEGATIVE The velocity is decreasing over time (E.g. train is approaching a station, and it is slowing down

2.2 Review Questions PAGE: 52 - 53 PROBLEMS: 1, 3, 5 HOMEWORK 2.2 Review Questions PAGE: 52 - 53 PROBLEMS: 1, 3, 5

Acceleration vs. Time graph Velocity vs. Time graph Acceleration vs. Time graph Area Under the Curve = DISTANCE travelled Area Under the Curve = VELOCITY

Graphing Practice Handouts

QUIZ Thursday?