chapter 3: acceleration and accelerated motion unit 3 accelerated motion

Chapter 3: Acceleration and Accelerated Motion

Unit 3

Accelerated Motion

Chapter 3: Acceleration and Accelerated MotionChapter 3: Acceleration and Accelerated Motion

t (time-seconds) v (velocity-m/s) x (position-m) a (acceleration-m/s2)

0 0

1 10

2 20

3 30

4 40

5 50

Velocity/Time Graphvelocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

12

12

xx

yyslope

ss

smsmslope

05

/0/50

s

smslope

/10

The slope means acceleration!

210s

m

The slope means something!


What equation can we get from this graph?

t

v

x

yslope

t

va

The constant acceleration equation!

We can also get the “how fast” equation.

bmxy

if vtav

From Graph:From Algebra:

t

va

t

vva if

if vvta fi vtav

if vtav




0 0 10

1 10 10

2 20 10

3 30 10

4 40 10

5 50 10


acceleration (m/s2)

t (seconds)

52 4

0

5

15

1 5

10

30

15 What would the acceleration/time graph look like?

Horizontal line means constant acceleration.

Let’s look at the area under the ‘curve.’bhArea

)10)(5( 2smsArea

smArea 50

5s

10 m/s2

It is the change in velocity!



0 0 0 10

1 10 10

2 20 10

3 30 10

4 40 10

5 50 10

velocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

Area of a triangle:

bhA 21

10 m/s

1 s

)10)(1(21

smsA

How do you find displacement from a velocity/time graph?

Area under the ‘curve.’

m5

5



0 0 0 10

1 10 5 10

2 20 10

3 30 10

4 40 10

5 50 10

velocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

Area of a triangle:

bhA 21

20 m/s

2 s

)2)(20(21 sA s

m m20

20



0 0 0 10

1 10 5 10

2 20 20 10

3 30 10

4 40 10

5 50 10

velocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

Area of a triangle:

bhA 21

30 m/s

3 s

msA sm 45)3)(30(2

1

45



0 0 0 10

1 10 5 10

2 20 20 10

3 30 45 10

4 40 10

5 50 10

velocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

Area of a triangle:

bhA 21

40 m/s

4 s

msA sm 80)4)(40(2

1

80



0 0 0 10

1 10 5 10

2 20 20 10

3 30 45 10

4 40 80 10

5 50 10

velocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

Area of a triangle:

bhA 21

50 m/s

5 s

msA sm 125)5)(50(2

1 What is happening to the amount of distance increased after each second?

125


0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Vel

oci

ty (

m/s

)


0 0 0 10

1 10 5 10

2 20 20 10

3 30 45 10

4 40 80 10

5 50 125 10


0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Vel

oci

ty (

m/s

)

0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Vel

oci

ty (

m/s

)

0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Vel

oci

ty (

m/s

)

0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Vel

oci

ty (

m/s

)

0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Vel

oci

ty (

m/s

)

0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Po

siti

on

(m

)

Shape: Top opening parabola (curvy up)

What is the proportionality?2xy 2Axy

221 )( tax

This is the “How Far” equation! (Starting with zero velocity)


t (time-sec) t ^2 (time^2)-sec^2) X (position-meters)

0 0

1 5

2 20

3 45

4 80

5 125

014

9

1625

y = 5x

0

20

40

60

80

100

120

140

0 5 10 15 20 25 30

t^2 (s^2)

x (

m)

bmxy 05 2 tx2

21 atx

It checks out!!


Let’s look at another way to get the “How Far” equation.

From the velocity time graph:velocity (m/s)

t (seconds)52 4

0

10

30

50

1 5

20

0

50

3

40

Area under curve = displacement

xA xbh 2

1

xvt ))((21

xvvt if ))((21

From previous “How Fast” equation:

if vtav

tavv if xatt ))((2

1

xat 2212

21 atx flip


0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Po

siti

on

(m

)

On a position time graph what is slope equal to?

Velocity

Is the slope constant in this graph?

No

You can use a tangent line to tell you the slope at a given point in time. Let’s try.

Finding the slope at 3 seconds:

Draw a tangent line, which is a straight line that touches the curve at only the desired point.

smss

mmvelocityslope /9.28

5.16

0130

smvelocity /30This is instantaneous velocity. (The velocity at that instant.)

Find Average Velocity

t

xx

t

xv ifavg

sm

avg s

mmv 25

5

0125


0

10

20

30

40

50

60

70

80

90

100

110

120

130

0 1 2 3 4 5 6

Time (s)

Po

siti

on

(m

)

Can we make a motion map of this motion?

You Bet!

x (displacement13020 40 60 80 100 1200 30 70 11010 9050 130

0s 1s 2s 3s 4s 5sv v v v v

What happens to the distance between the dots?

What is happening to the velocity?


We need to make a change one addition to our “How Far” equation.

What if you saw a velocity/time graph that looked like this?

velocity (m/s)

t (seconds)62 4 6

60

20

40

60

1 5

10

50

3

30

00

What is different about this graph than the previous velocity/time graph?

The velocity at t = 0 is 10 m/s. In other words, the car has an initial velocity of 10 m/s.


velocity (m/s)

t (seconds)62 4 6

60

20

40

60

1 5

10

50

3

30

00

Let’s see how this affects our “how far” equation.

Again, we need to find displacement. How do we do this?

Area under ‘curve’

Let’s look at the time interval of 0 – 1 sec.

This area is a goofy, irregular shape, so we need to look at this as a rectangle and a triangle together!

Green Area

xAreaWhat equation can I make for the area (displacement)? bhbh2

1

Red Area

)10)(1()10)(1(21

sm

sm ssx

mmmx 15105


xArea bhbh21

)10)(2()1030)(2(21

sm

sm ssx

mmmx 402020

velocity (m/s)

t (seconds)62 4 6

60

20

40

60

1 5

10

50

3

30

00

Green Area Red Area

Let’s do the same thing for 0 – 2 sec.

Look for the pattern:2

21

21 atbh

?bh tvi

time initial velocity

Therefore the ‘How Far’ equation becomes:

tvatx i 221


Now use that equation to find the position of the object at each second.


0 10 0 10

1 20 15 10

2 30 40 10

3 40 10

4 50 10

5 60 10

This comes from slope, which is the same as the first v/t graph.

221 )3)(10()3)(10( ssx s

msm At 3 sec. m75

221 )4)(10()4)(10( ssx s

msm

221 )4)(10()4)(10( ssx s

msm

m120

m175

75120175

From previous pages

At 4 sec.

At 5 sec.



0 10 0 10

1 20 15 10

2 30 40 10

3 40 75 10

4 50 120 10

5 60 175 10

Let’s make a position/time graph for this motion.

0102030405060708090

100110120130140150160170180

0 1 2 3 4 5

Time (s)

po

siti

on

(m

)

Notice the shape: top opening parabola (curvy up)

How can the position time graph go through (0,0) and the velocity time graph didn’t?

The car can have an initial velocity at t=0, at the ref. point.


0s 1s 2s 3s 4sv v v v v

x (displacement

18020 40 60 80 100 120 140 160 18010 50 90 130 17030 1100 15070

Let’s make a motion map for this motion also.


0 10 0 10

1 20 15 10

2 30 40 10

3 40 75 10

4 50 120 10

5 60 175 10

5s


Let’s take the case of an object slowing down…(Negative acceleration)

t (time-seconds)

v (velocity-m/s) x (position-m) a (acceleration-m/s2)

0 50

1 40

2 30

3 20

4 10

5 0

velocity (m/s)

t (seconds)

52 4

0

10

30

50

1 5

20

0

50

3

40

Calculate the slope:

ssa s

msm

05

500

sm10

What does slope of a v/t graph mean again???

Oh yeah….. Acceleration!

What does negative acceleration mean?


What does negative acceleration mean?

It can mean slowing down, but that’s not a complete picture.

It most accurately means that the object is accelerating in the negative direction.

Ex: If your put your car in reverse at the stop sign (reference pt.) and put your foot on the gas pedal, you would be speeding up in the backwards direction. This would also be negative acceleration.

Velocity Acceleration Motion

Positive Positive

Positive Negative

Negative Negative

Negative Positive

Speeding up, forward

Slowing down forwardSpeeding up, backward

Slowing down, backward


acceleration (m/s2)

0 t (seconds)

52 4

-15-15

-5

10

0 3

15

5

1

-10

5

15


0 50 -10

1 40 -10

2 30 -10

3 20 -10

4 10 -10

5 0 -10

Chapter 3: Acceleration and Accelerated Motionvelocity (m/s)

t (seconds)52 4

0

10

30

50

1 5

20

0

50

3

40Let’s make the position/time graph:

x (position-meters)

time (t-seconds)

52 40

10

30

50

70

90

110

0 3

130

40

80

120

5

60

1

100

20

Find x at t=1

bhbhx 21

)/40)(1()/10)(1(21 smssmsx m45

221 attvx i

Let’s use the “how far” equation.

221 )2)(10()2)(/50( 2 sssmx

sm m80

221 )3)(10()3)(/50( 2 sssmx

sm m105

221 )4)(10()4)(/50( 2 sssmx

sm m120

221 )5)(10()5)(/50( 2 sssmx

sm m125

chapter 3: acceleration and accelerated motion unit 3 accelerated motion

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