Motion Diagrams - Santa Rosa Junior College

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Motion Diagrams

Transcript of Motion Diagrams - Santa Rosa Junior College

Page 1: Motion Diagrams - Santa Rosa Junior College

Motion Diagrams

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Skiing through the woods

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Motion Graphs

What kind of motion is this?

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What kind of motion is this?

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xvt

∆=∆

3400 1 /400

mv m ss

−= = −2 0 /v m s=

1400 2 /200

mv m ss

= =

What is the velocity during each segment? Graph it!

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Which position-versus-time graph represents the motion shown in the motion diagram?

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Which position-versus-time graph represents the motion shown in the motion diagram?

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Motion Graphs What kind of motion is this?

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2( ) 4 2x t t t= − +

24 /xa m s=

24 /a m s=( ) 4 4v t t= − +

Deriving Graphs from Graphs

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Here is a position graph of an object: At t = 3.0 s, the object’s velocity is A. 40 m/s. B. 20 m/s. C. 10 m/s. D. –10 m/s. E. None of the above.

QuickCheck 2.7

Slide 2-50

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Here is a position graph of an object: At t = 3.0 s, the object’s velocity is A. 40 m/s. B. 20 m/s. C. 10 m/s. D. –10 m/s. E. None of the above.

QuickCheck 2.7

Slide 2-51

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Which velocity-versus-time graph or graphs goes with this acceleration-versus-time graph? The particle is initially moving to the right and eventually to the left.

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Which velocity-versus-time graph or graphs goes with this acceleration-versus-time graph? The particle is initially moving to the right and eventually to the left.

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Free Fall Unless told otherwise, ignore air resistance for

free fall problems!

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Falling from Rest

2 21 52

y at t∆ = =

2

:~ 10 /

Estimatea g m s=

10v at t= =

20 /20

v m sy m=

∆ =

10 /5

v m sy m=

∆ =

30 /45

v m sy m=

∆ =

40 /80

v m sy m=

∆ =

50 /125

v m sy m=

∆ =

+

0

20

12

fv v gt

y v t gt

= +

∆ = +

0 0v =

!v y≠ ∆How FAR is not How FAST!

Take down as +y:

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How Far: y(t) ~ t2

0fv v at= +

20

12

y v t at∆ = +

How Fast: v(t) ~ t1

+

How Fast How Fast is Changing! 29.80 /g m s=

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FIRST: Define Reference Frame In this reference frame,what is the sign of a? 29.80 /a m s= −

What is v at t = 3s?

0fv v at= +

20 9.80 (3 )m ss

= −

29.4 ms

= −

20: 0, 9.8 / , 3Knowns v a m s t s= = − =

: ?fUnknown v =

Negative because it is moving downward, in the negative direction!

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FIRST: Define Reference Frame

20: 0, 9.8 / , 3 , 29.4 /fKnowns v a m s t s v m s= = − = = −

: ?Unknown y∆ =

The displacement is negative because it is moves downward, in the negative direction but “how far” is a distance – a scalar – and is positive!

How far did the ball fall in those 3 seconds?

20

1

2y v t at∆ = +

22

1

20 ( 9.8 )(3 )m s

s= + −

44.1m= −

The ball fell 44.1m.

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Throwing up is Also Free Fall! Symmetry of G Field.

2

:~ 10 /

Estimatea g m s=

0

20

12

fv v gt

y v t gt

= +

∆ = +

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What Goes Up Must Come Down Someone standing at the edge of a cliff throws one ball straight up and one straight down at the same speed. Ignoring air resistance, which ball strikes the ground with the greatest speed?

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Free Fall Question: You throw the rock down with an initial speed of 30 m/s. The rock hits the ground in 3 seconds. With what speed will the rock hit the ground?

+y

230 9.8 (3 )m m ss s

= − −

59.4fmvs

= −

0fv v at= +

How high is the cliff?

20: 30 / , 9.8 / , 3Knowns v m s a m s t s= − = − =

: ?fUnknown v =

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Free Fall

20

1

2y v t at∆ = +

2 21

2( 30 / )(3 ) ( 9.8 / )(3 )m s s m s s= − + −

134m= −

The cliff is 134 m high.

20: 30 / , 9.8 / , 3Knowns v m s a m s t s= − = − =

: ?Unknown y∆ =+y

Question: You throw the rock down with an initial speed of 30 m/s. The rock hits the ground in 3 seconds. With what speed will the rock hit the ground? How high is the cliff?

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Free Fall: Throwing Up What is the speed at the top of the path? ZERO! What is the acceleration at the top? a = -9.80 m/s2 What is the velocity at the same height on the way down? -30 m/s

+y

With what velocity will the rock hit the ground? -59.4 m/s SAME as if you threw it straight down at 30m/s!

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How long does it take to hit the ground? First try to guess!

+y 0fv v at= +

02

59.4 / 30 /9.8 /

fv v m s m sta m s− − −

= =−

9.12t s=

20: 30 / , 9.8 / , 3 , 59.4 /fKnowns v m s a m s t s v m s= = − = = −

: ?Unknown t∆ =

How long to the top? How long back to launch point? Final v increases by 30m/s?

I guess about 9 seconds!

Free Fall: Throwing Up Problem

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A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration ay is A. Positive. B. Negative. C. Zero.

QuickCheck 2.18

Slide 2-96

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A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration ay is A. Positive. B. Negative. C. Zero.

QuickCheck 2.18

Slide 2-97