Kinematics Review. DISTANCE ( m ) TIME (s) 35 30 25 20 15 10 5 0 -5 -10 -15 -20 -25 -30 -35 -40 2 4...

Kinematics Review

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A. What time interval (if any) has negative velocities in positive positions?

B. What time interval(s) if any has a positive acceleration?

C. Does the object ever change directions?

D. What is the total distance for the 40 seconds?

E. Construct a VT graph that shows the same motion as this DT graph.

If an object travels 78.4 km/hr for 45 min, and then 50.0 km/hr for 20.0 km, what was the average velocity for the whole trip?

An object is tossed 24.0 m/s upward. How high does it go?

Sketch a VT graph for each DT graph:

A. B. C.

35

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VE

LOC

ITY

(m

/s)

TIME (s)

A.What is the total distance for the 20 s?

B.What is the displacement for the 20 s?

Two objects start at the same place and travel 300 m. Both start from rest at the same time and reach 300m at the same time. One object travels at constant speed, and the other accelerates. If the first object has an acceleration of 4 m/s2, what must the constant speed of the other object be?

What was the initial velocity of a ball thrown from a 80.0 m tall cliff, if it strikes the base of the cliff with a velocity of 62.0 m/s?

THINKER QUESTION:

A brick is dropped from rest from the top of a 40 m tall building. On the way down it is observed passing a 1.2 m tall window in .10 s. What is the height of the bottom of the window?

?

40 m

Now you will check and correct your answers

Take the time to really understand each problem

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A. What time interval (if any) has negative velocities in positive positions?

ANSWER : 32-40 s

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B. What time interval(s) if any has a positive acceleration?

+ direction

+ faster & faster

+ a

ANSWER: 4 – 12 s

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C. Does the object ever change directions?

ANSWER: Yes , at t = 32s it starts going in the

negative direction

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D. What is the total distance for the 40 seconds?

35 m

30 m

10 m

ANSWER: 75 m

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(12,-30)

(22,-15)

Find velocity (slope):

-15 – (-30) 15

22 – 12 10V = ________ = _____ = 1.5 m/s

(28,30)


30 - (-15) 45

28 - 22 6V = ________ = _____ = 7.5 m/s


(38, 25)

(40, 20)

25 - 20 5

38 - 40 -2V = ________ = _____ = - 2.5 m/s

VE

LO

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Y(

m/s

)

TIME (s)


8

7

65

4

3 2

1 0

-1

-2

-3

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-15 – (-30) 15

22 – 12 10V = ________ = _____ = 1.5 m/s


30 - (-15) 45

28 - 22 6V = ________ = _____ = 7.5 m/s


25 - 20 5

38 - 40 -2V = ________ = _____ = - 2.5 m/s

If an object travels 78.4 km/hr for 45 min, and then 50.0 km/hr for 20.0 km, what was the average velocity for the whole trip?

First Part The Rest Whole Trip

d d d

t t t v = v = v =

d 20

.75 t 78.4 = 50.0 = v =

In hr

Solve:

d = 58.8 km

Solve:

t = .4 hr

78.8

1.15

Solve:

V = 68.5 km/hr

An object is tossed 24.0 m/s upward. How high does it go?

vi = 24 So right here I set my frame of reference…UP is positive

a = -9.8 + Pos direction- Getting slower

vf = 0 Goes UP until it stops!

d = ?

vf2 = vi

2 + 2ad

0 = 242 + 2(-9.8)d

0 = 576 - 19.6d

19.6d = 576

d = 29.4 m

Sketch a VT graph for each DT graph:

A. B. C.

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VE

LOC

ITY

(m

/s)

TIME (s)

A. What is the total distance for the 20 s?

B. What is the displacement for the 20 s?

You do NOT have to split them The way that I did – as longAs you have all rectangles

And trianglesWe should get the same

answers

30

22562.5

10

1090

30 + 225 + 62.5 + 10 + 10 + 90 = 427.5 m

10 + 10 + 90 – 30 – 225 – 62.5 = - 207.5 m

Two objects start at the same place and travel 300 m. Both start from rest at the same time and reach 300 m at the same time. One object travels at constant speed, and the other accelerates. If the first object has an acceleration of 4 m/s2, what must the constant speed of the other object be?

Constant Speed Object

d

tv =

Accelerating Object

vi = 0

d = 300

a = 4

t = ?

d = vit + ½ at2

300 = ½ (4)t2

t = 12.2 s300

12.2v =

v = 24.6 m/s

What was the initial velocity of a ball thrown from a 80.0 m tall cliff, if it strikes the base of the cliff with a velocity of 62.0 m/s?

vi = ?

d = 80 so down is POS

a = 9.8

vf = 62

vf2 = vi

2 + 2ad

622 = vi2 + 2(9.8)(80)

vi = +- 47.7 m/s

THINKER QUESTION:A brick is dropped from rest from the top of a 40 m tall building. On the way down it is observed passing a 1.2 m tall window in .10 s. What is the height of the bottom of the window?

?

40 m

Passing window:

d = 1.2

a = 9.8

t = .10

vi = ?

d = vit + ½ at2

1.2 = vi(.10) + ½ (9.8)(.10)2

vi = 11.5 m/s

From top of building:

d = ?

a = 9.8

vf = 11.5

vi = 0

vf2 = vi

2 + 2ad

11.52 = 2(9.8)d

d = 6.76 m

32.0m

Kinematics Review. DISTANCE ( m ) TIME (s) 35 30 25 20 15 10 5 0 -5 -10 -15 -20 -25 -30 -35 -40 2 4...

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Transcript of Kinematics Review. DISTANCE ( m ) TIME (s) 35 30 25 20 15 10 5 0 -5 -10 -15 -20 -25 -30 -35 -40 2 4...