DO NOW The position-time graph above represents the motion of a basketball coach during the last sixteen seconds of overtime. (a) Determine the total distance.

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DO NOWThe position-time graph above represents the motion of a basketball coach during the last sixteen seconds of overtime.

(a) Determine the total distance walked by the coach during these 16 seconds.

(b) Determine the resulting displacement of the coach during these 16 seconds.

(c) What was the fastest speed which the coach walked? During which time interval was he walking at this speed?

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AIM: How are velocity vs. time graphs analyzed?SWBAT: • construct and interpret graphs of position, velocity, or

acceleration versus time (S4.5.1.i)• determine and interpret slopes and areas of motion

graphs (S4.5.1.ii)

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Graphs of Motion• Position (displacement) vs. Time

• Slope of position-time graph = velocity over that interval of time

time time time

posi

tion

posi

tion

posi

tion

time

posi

tion

Slope is zero ∴velocity is zero(object at rest)

Slope is positive ∴velocity is constant, positive

Slope is negative ∴velocity is constant, negative

Slope is curve ∴velocity is not constant(object accelerating)

* the steeper the slope, the higher the speed (value of velocity)

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Graphs of Motion• Velocity vs. Time

• Slope of v-t graph = acceleration over that interval of time

time time time

velo

city

velo

city

velo

city

time

velo

city

Slope is zero ∴acceleration is zero(at rest or with constant velocity)

Slope is positive ∴acceleration is constant, positive

Slope is negative ∴acceleration is constant, negative

Slope is curve ∴acceleration is not constant

* the steeper the slope, the higher the value of acceleration

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• BEWARE: Just because the slope of the v-t graph is negative (acceleration negative), does not always mean the object in motion is decelerating (slowing down)!

• Nor does the slope being positive mean the object is accelerating (speeding up)

velo

city

timeSpeeding up

Speeding up

Slowing down

Slowing down

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Graphs of Motion• Velocity vs. Time

• Area under v-t graph = displacement and/or distance over that interval of time

time time time

velo

city

velo

city

velo

city

Line along x-axis No area

∴ no displacement

Line above x-axisPositive area

Line below x-axisNegative area

* Sum of areas with signs: displacement* Sum of areas without signs: distance

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A

C

B

D

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0

1

2

3

4

5

6

0 2 4 6 8 10 12

Time (s)

Vel

oci

ty (

m/s

)

What is the acceleration from: • t = 0s to t = 4s?• t = 4s to t = 8s?• t = 8s to t = 10s?

What is the distance traveled? Is it the same as the object’s displacement?

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0

1

2

3

4

5

6

0 2 4 6 8 10 12

Time (s)

Velo

cit

y (

m/s

)

position-time acceleration-time

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0 5 10 15 20 250

5

10

15

20

25

30

35

Time (s)

Vel

oci

ty (

m/s

)

Picture from Problem 3.82/3.87

Draw the p-t graph and a-t graph for this object’s motion.

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0 5 10 15 20 250

5

10

15

20

25

30

35

Time (s)

Vel

oci

ty (

m/s

)

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0 5 10 15 20 250

5

10

15

20

25

30

35

Time (s)

Vel

oci

ty (

m/s

)

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Two Equations from Monday• Position with constant acceleration:

• Velocity with constant acceleration:

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Equations of Motion for Uniform (Constant) Acceleration

Equation Variables Initial Conditions

t, vf, a vi

t, df, a di, vi

df, vf, a di, vi

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Example: 3.27• A race car travels on a racetrack at 44 m/s and slows at a

constant rate to a velocity of 22 m/s over 11 s. How far does it move during this time?

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Problem 3.28

• A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m/ How long does it take to achieve this speed?

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Problem 3.29

• A bike rider pedals with constant acceleration to reach a velocity of 7.5 m/s over a time of 4.5 s. During the period of acceleration, the bike’s displacement is 19 m. What was the initial velocity of the bike?

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Problem 3.30

• A man runs at a velocity of 4.5 m/s for 15 minutes. When going up an increasingly steep hill, he slows down at a constant rate of 0.05 m/s2 for 90.0 seconds and comes to a stop. How far did he run?