1 honors physics chapter 3 acceleration. 2 honors physics turn in chapter 2 homework, worksheet,...

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Honors Physics Chapter 3

Acceleration

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Honors Physics

Turn in Chapter 2 Homework, Worksheet, & Lab

Lecture Q&A

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Chapter 2 Review

Quantity Symbol Unit

Position

Displacement

Distance

Time

Velocity

Speed

d m

d m

m

s

v

tD

s

m

sm

s

4

Acceleration

• How fast is your position changing? Velocity

• How fast is your velocity changing? Acceleration

• Both velocity and acceleration are vector quantities with both direction and magnitude.

• But velocity and acceleration are two different concepts.

5

Average Acceleration

Unit

va

t

v

at

2 1

2 1

f i

f i

v vv v

t t t t

v

t /m s

s

m

s s

2

m

s

6

Example (64-6)

A race car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0-s time interval. What is its average acceleration?

1 24.0 , 36 , 4.0 , ?m m

v v t s as s

va

t

2 1v v

t

2

36 4.0 32.8.0

4.0 4.0

m m mms s s

s s s

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Practice (64-9)

A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in 3.0s.

a) What is the average acceleration of the bus while braking?

25 , 3.0 ,i f

mv t s v

s

va

t

What does the negative acceleration mean?

What direction has been defined as the positive direction?

Opposite to + direction.

Direction of motion.

0

?a

2

0 258.3

3.0f i

mv v mst s s

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Instantaneous Acceleration

• Instantaneous acceleration is average acceleration when the time interval becomes very, very small.

• On a velocity-time graph, the instantaneous acceleration at any time is given by the slope of the line tangent to the curve at that time.

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12

tt

vv

t

v

run

riseslopea

– Draw line tangent to curve at given time.– Locate two points on tangent line and find coordinates. (t1, v1)

and (t2, v2)

– Find slope using equation

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Instantaneous acceleration and slope

Two points are _______ and ___________.

v (m/s)

riseslope

run

at=3s= 8.3 m/s2t (s)0 4321 65

20

10

50

40

30

60

•

•

(1.8s, 0)(6.0s, 35m/s)

35 / 0

6.0 1.8

m s

s s

2

35 /8.3

4.2

m s m

s s

a3s= 8.3 m/s2

•

Draw tangent line at 3 s.

What is the acceleration at 3 s?

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Example

v (m/s)

t (s)1

-10

0

2

10

a) 0.0 , 2.0 , 10 , 10i f i f

m mt s t s v v

s s

a

2) 10 /c a m s

•

•

) ?b a

a. Draw a velocity-time graph for an object whose velocity is constantly decreasing from 10 m/s at t = 0.0s to –10 m/s at t = 2.0s. Assume it has constant acceleration.

b. What is its average acceleration between 0.0s and 2.0s?

c. What is its acceleration when its velocity is 0 m/s?

f i

f i

v v

t t

2

10 10

2.0 0.0

2010

2.0

m m

s ss sm

mss s

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Negative acceleration

Acceleration is a vector quantity. Negative sign of acceleration indicates

direction only. Negative acceleration does not necessarily

mean slowing down Think of acceleration as a push in a direction

(though not exactly correct.)

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Speeding Up or Slowing down

v and a are in the same direction (or have the same sign) ___________

v: speeding up

v: speeding up

v:

v: slowing down

slowing down

a: a:

a: a:

v and a are in the opposite direction (or have the opposite signs) ____________

speeding up

slowing down

+

+

-

-

+

-

-

+

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Constant acceleration motion

a = constant. Also, let ti = 0:

2

2 2

2

2

1

2

f i f

f i i f f

f i

i f

v v at

d d v t at

v v a

v v

d

v

21

2i f fd v t at

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Example

An airplane starts from rest and accelerates at a constant +3.00 m/s2 for 30.0 s before leaving the ground. What is its displacement during this time?

20, 30.0 , 3.00 , ?i f

mv t s a d

s

21

2i f fd v t at

2

2

13.00 30.0

2

ms

s

1350m

21

2 fat0

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Practice

22 , 0, 2.0

) ?

i f f

mv v t s

sa a

What does the negative mean?

f i fv v at

?) db

fat

a

Define the direction of motion to be the + direction

A driver brings a car traveling at +22 m/s to a full stop in 2.0 s. Assume its acceleration is constant.

a. What is the car’s acceleration?

b. How far does it travel before stopping?

f iv v

2

0 2211.

2.0f i

f

mv v mst s s

21

2i f fd v t at

2

2

122 2 11 2 22

2

m ms s m

s s

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Another approach to b)

2 2 2f iv v a d

d

2

2

0 2222

2 11

m

sm

m

s

2 2

2f iv v

a

17

221 , 3.0 , 535 , ?i f

m mv a d m v

s s

2 2 2f iv v a d

fv

2fv

2

22 3.0 535 21 60.

m m mm

s s s

Practice 4: An airplane accelerates from a velocity of 21 m/s at the constant rate of 3.0 m/s2 over +535 m. What is its final velocity?

2 2iv a d

22 ia d v

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Free-Fall Motion

• Assume no air resistance. (Valid when speed is not too fast.)

• a = g, downward (g = 9.8 m/s2) Acceleration can be positive or negative, depending on what we

define as the positive direction. g is always a positive number, equivalent to 9.81 m/s2. Does not matter if the object is on its way up, on its way down,

or at the very top.

g is acceleration due to gravity (It is not gravity.) g does not depend on mass of object.

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Signs of v and a

v: a:

v

v: a:

a:

Define: up = +

v:

Define: down = +

a:

v a:

v: a:

+ -

= 0 -

- -

- +

= 0 +

+ +

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Free-Fall motion equations

These equation are valid only when downward is defined as the positive direction.

Not valid when upward is defined as the positive direction. (Must replace every g with –g.)

No need to remember these equations.

2 2

2

2

1

2

f i f

f i

i f f

v v gt

v v g d

d v t gt

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Terms to remember

Drop, release initial velocity is zero with respect to hand. (Initial means at the moment right after it leaves hand.)

Throw initial velocity is not zero with respect to hand.

When it hits the ground right before it hits the ground.

Rest v = 0 At top of ascent v = 0

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Example 5

Define downward as the positive direction, then a = g

?)

0,0.1,

ta

vmdga i

21

2id v t at

) ?fb v

21

2at

2 dt

a

at2

9.8 0.45 4.4m m

ss s

A man falls 1.0 m to the floor.

a. How long does the fall take?

b. How fast is he going when he hits the floor?

0

f iv v at 0

2

2 1.00.45

9.8

ms

m

s

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Practice

Define upward as the positive direction, then a = -g

27 ,

) 0, ?

i

f

mv a g

sa v t

?) db

f iv v at

21

2id v t at

t

2

2

127 2.8 9.8 2.8 37

2

m ms s m

s s

A pitcher throws a baseball straight up with an initial speed of 27 m/s.

a. How long does it take the ball to reach its highest point?

b. How high does the ball rise above its release point?

2

0 272.8

9.8

f i

mv v s s

mas

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Practice

Define upward as the positive direction, then a = -g, let d = 0 at hand.

0, 2.2 , 0,

) ?

i f

f

d t s v a g

a d

) ?ib v

21

2f i id d v t at

f iv v at

You throw a beanbag in the air and someone catches it 2.2 s later at its highest point.

a. How high did it go?b. What was its initial velocity?

2

2

121.6 2.2 9.8 2.2 23.8

2

m ms s m

s s

iv 20 9.8 2.2 21.6f

m mv at s

s s

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Position, Velocity, and acceleration Graphs

Position (vs. Time) graph

v = slope

a = slope

d = Area under curve

v = Area under curve

Velocity (vs. Time) graph

Acceleration (vs. Time) graph