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•  CAPA set 2 due at 11:59pm Tuesday.

•  Remember the 4 digit CAPA ID pin number changes weekly.

•  You can review your old CAPA homework to see the correct answer. Solutions are available on D2L.

•  Remember, help room (G2B90) is staffed 9-5 M-F.

Web page: http://www.colorado.edu/physics/phys1110/

Announcements: 1D Kinematics & Vectors

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What velocity versus time graph goes with the acceleration versus time graph on the right? It is known that the object is initially moving right and eventually moving left.

Clicker question 1 Set frequency to BA

A. B. C.

E. More than one could be correct

( )2m/s xa

(s) t

(s) t (s) t (s) t (s) t

( )m/s xv( )m/s xv ( )m/s xv ( )m/s xv

D.

A,B,C could all work except we are told the initial velocity is positive and the final velocity is negative so it must be B

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Example problem 2 Usain Bolt accelerates at 4 m/s2 to his top speed of 12 m/s which he maintains until the finish. How long does he take to run the 100 m dash? Step 1: Draw picture.

1 Start

.

2 Reach top speed 3 Finish

Include coordinate system & label points

. .

x (m) 100 0

Step 2: Write down known information and what is needed.

m 01 =xm/s 01 =xv

21 m/s 4=xa

?2 =x

22 m/s 0&4=xa

m 1003 =xm/s 123 =xv

23 m/s 0=xa

01 =t ?2 =t ?3 =t

m/s 122 =xv

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Example problem 2 Step 3: Motion diagram

1 Start 2 Reach top speed 3 Finish x (m) 100 0

0=aStep 4: Draw approximate graphs of , , and vs time xaxvx

100 12

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Step 5: Solve the problem.

This problem has two parts. Point 2 is the final for part one but the initial for part two.

( )2m/s xa

(s) t

(m/s) xv

(s) t

(m) x

(s) t

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Any other equation can be used to get x2.

Solving problem 2 1 Start 2 Reach top speed 3 Finish

x (m) m 01 =x m/s 01 =xv2

1 m/s 4=xa

?2 =x2

2 m/s 0&4=xam/s 122 =xv

m 1003 =xm/s 123 =xv

23 m/s 0=xa

01 =t ?2 =t ?3 =t

Between 1 & 2 have constant acceleration. Can find t2: vx = v0 x + axΔt

s 3m/s 4m/s 012

21

1212 =−=

−+=

x

xxavvtt

v2x = v1x + a1x t2 − t1( )becomes

Solving for t2:

x = x0 +12 v0 x + vx( )Δt

x2 = x1 +12 v1x + v2x( ) t2 − t1( ) = 0+ 1

2 0+12 m/s( ) 3 s− 0( ) =18 mbecomes

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Solving problem 2 1 Start 2 Reach top speed 3 Finish

x (m) m 01 =x m/s 01 =xv2

1 m/s 4=xa

m 182 =x2

2 m/s 0&4=xam/s 122 =xv

m 1003 =xm/s 123 =xv

23 m/s 0=xa

01 =t s 32 =t ?3 =t

Between 2 & 3 have constant velocity. Can find t3:

x = x0 + vxΔt ( )23223 ttvxx x −+=

s 8.9m/s 12m 18m 100 s 0.3

2

2323 =−+=

−+=

xvxxtt

becomes

Does the answer make sense? Yes, his world record is 9.58 seconds so this was just a slow race for him.

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Gravity On the surface of the Earth gravity causes a downward acceleration of magnitude 9.8 m/s2 which is labeled “g”: Note that g is positive (it is the magnitude of the free fall acceleration). However, we usually define up as positive so the acceleration is 2

fall free m/s 8.9−=−== gaayGalileo figured out that all objects are subject to the same acceleration

Free fall is just a particular type of constant acceleration

Note: objects moving up can still be in free fall! Free fall means acceleration is down; it says nothing about velocity!

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A.  more than 9.8 m/s2 B.  9.8 m/s2 C.  less than 9.8 m/s2 D.  depends on the mass of the ball

Clicker question 2 Q. If you drop an object in the absence of air resistance, it accelerates downward at 9.8 m/s2. If you throw it downward, the magnitude of its downward acceleration after release is

Set frequency to BA

After release, the only acceleration is due to gravity which is 9.8 m/s2

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Constant acceleration formulas vx = v0 x + axΔtx = x0 + v0 xΔt +

12 ax Δt( )2

vx2 = v0 x

2 + 2ax Δxx = x0 +

12 v0 x + vx( ) Δt

You can replace x with y in these formulas when your problem is in the vertical direction.

If the object is in free fall, 2

fall free m/s 8.9−=−== gaay

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Some problem solving hints Draw pictures

Split into parts if necessary

Take account of everything you can figure out about the problem. Some things may not be obvious

If two objects collide then you know they have the same position at that time

If an object is in free fall you know the acceleration

An object with constant velocity has no acceleration

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2D & 3D Kinematics •  Displacement, velocity, & acceleration are vectors and

so have magnitude and direction (which is why we needed to remember signs in 1D)

•  We will mostly work in 2D but everything generalizes to 3D

Trigonometry refresher: SOHCAHTOA

θAdjacent

Opposite Hypotenuse

ypotenuse

pposite

HOsin =θ

ypotenuse

djacent

HAcos =θ

djacent

pposite

AOtan =θ

Need to understand trig functions (including inverse) on calculator and understand degrees and radians

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Vector representations •  Can represent as an arrow •  Can represent as ordered list of

numbers in a known coordinate system (giving the components)

•  Can represent as a sum of components times unit vectors – Unit vectors have magnitude 1

and point along an axis: axis- along ˆ axis,- along ˆ axis,- along ̂ zkyjxi

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Vector representation y

x 4

3 A! Vector is represented by arrow A

!

scoordinateCartesian in )3,4(=A!

),( scoordinatepolar in )37,5( θrA °=! rsunit vecto using ˆ3ˆ4 jiA +=!

The vector is written with an arrow A!

The magnitude is written without the arrow: AA =!

Components can be found by trigonometry. If angle is measured from +x axis then θθ sin and cos AAAA yx ==