Linear Kinematics 5x6k1o

Linear Motion

Angular Motion

TRANSLATION ONLY!

ROTATION ONLY! Object rotates about a fixed point (axis)

Object maintains angular orientation (q)

but this point does NOT have to lie within the object

measured in meters - SI unit other units - inches, feet, miles, centimeters, millimeters Rectilinear motion - if path of one point on object is a straight line Curvilinear motion – if path of one point on object is curved

q

measured in radians – SI Unit other units – degrees, revolutions

q

Types of Motion • General – combination of linear and angular motion • translation and rotation

B B

A A

Kinematics is the study of motion without regard for the forces causing the motion or … the description of motion • there are three basic kinematic variables – position, velocity and acceleration

• the position of an object is simply its location in space – changes in position can be described by distance or displacement

• the velocity of an object is how fast it is changing its position • the acceleration of an object is how fast the velocity is changing

Acquisition of Position Data Frame 1 (x1,y1)

(x2,y2)

Y (x4,y4) (x5,y5)

(0,0)

(x3,y3)

X

Position data is often acquired by digitizing the x and y coordinates from film or video. Velocities and accelerations are calculated from the position data.

Position and Displacement (d) • Position (s) is the location of an object in space • units: m, cm, km, in, ft, mi • Displacement (Ds= sf - si) is the change in position of an object

s2 displacement = d

s1

d

d = s1 – s2

• Problem: • how do you describe s1 and s2? • If you put the arrow on graph paper you describe position with x- & y-coordinates Y

s2 = (x2,y2) d

d = s1 – s2 s1 = (x1,y1) X

d = s 1 – s2 d = (x1,y2) – (x2,y2)

How do you do this?

•Realize that displacement is a vector so you must determine either the Cartesian or polar coordinates Y

s2 = (x2,y2) d

d = s1 – s2 s1 = (x1,y1) X

•Two choices to describe vector •Cartesian Coordinates (dx,dy) •dx = x2 – x2 = distance in the x-direction •dy = y2 – y1 = distance in the y-direction •Polar Coordinates (d,q) •“How far and in which direction” d  ( x2  x1 ) 2  ( y2  y1 ) 2 q = measured directly from graph

Second problem: Since this movement occurs over time, displacement (as a vector) does NOT represent changes in the direction of movement well. For example – what if s1 represents you at Building A and s2 represents you at Building B 10 minutes later.

Y s2 d

d = s1 – s2 s1

X

Assuming this city is like most cities you have to walk up and down city blocks and not through buildings.

Y s2 d

s1

X

So your actual route is around the buildings, traveling up and down city blocks.

Y s2 d

s1

dy dx

X

Thus the actual distance you covered is more than displacement represents distance = the length of your travel in the x-direction (dx) plus the length of your travel in the y-direction (dy) BUT since we are only concerned with the length of travel we don’t distinguish between directions

Y distance = dx + dy d

s1

s2 dy

dx

X

Distance ( ) • distance is the length of the path traveled • it is a scalar - “How far” • units: same as displacement dx + dy = distance =

d dy

dx

Note: use “ ” for length

Example - Distance vs. Displacement

N

leg 3 = 2 miles

leg 2 = 3 miles

leg 1 = 2 miles

Total DISTANCE Traveled = 2 miles + 3 miles + 2 miles = 7 miles

Describing Displacement

N

Describing Displacement First Method (Cartesian) 3 miles East 4 miles North (3, 4) miles put ‘horizontal’ coordinate 1st put ‘vertical’ coordinate 2nd

Displacement Magnitude

Second Method (Polar) 1st - calculate length of displacement vector

N

q 3 miles

 d  32  4 2  d  25  d  5 miles

Displacement Direction

2nd – Calculate the angle using trigonometric relationships

N

q





d  tan1 ver 

  d   hor 





q  tan1 4miles   53.1  3miles 

q 3 miles

Displacement Vector (Polar Notation)

Describe the displacement vector by its length and direction

N

 d  5 miles @ 531 .

q 3 miles

Average Speed • speed is a scalar quantity • it is the rate of change of distance wrt time • units: same as velocity Speed  distance time

What is the average speed of the basketball? (80,40)

(60,10) (0,0)

l=

20  30  36 feet 2

2

l 36 speed    72 ft/s t

0.5

Average Velocity (v) • rate of change of displacement wrt time • velocity is a vector quantity – “How fast and in which direction”

• units: m/s, km/hr, mi/hr, ft/s

  Dd velocity v  Dt

NOTE: displacement (d) is a vector so must obey rules of vector algebra when computing velocity.

•When two velocities act on an object you find the net or resultant effect by adding the velocities. •Because velocity is a vector you can’t simply add the numbers. •Instead – you must use vector algebra to add the velocities. In this example the boat is propelled to the right by its motor while the river’s current carries it towards the top of the picture. This describes 2 velocities

Other examples of velocities that can be added together include the wind direction when flying.

Adding Velocities Use the laws of vector algebra. Example - the path of the swimmer is determined by the vector sum of the swimmer’s velocity and the river current’s velocity.

Example:

vswimmer = 2 m/s vriver = 0.5 m/s What is the swimmer’s resultant velocity?

Example - Solution vR = (2 m/s)2 + (0.5 m/s)2 vR = 2.06 m/s

q = 14 50 m

0.5 m/s

2 m/s

vR

Average Speed and Velocity • average speed has a greater magnitude than average velocity unless there are no direction changes associated with travel • in sports – average speed is often more important than average velocity

1996 Olympic Marathon Men 2:12:36 Josia Thugwane - RSA Women 2:26:05 Fatuma Roba - ETH Distance 26 miles + 385 yards 26 miles * 1.61 km/mile = 41.86 km 385 yards * 0.915 m/yd = 352 m Total = 41.86 km + .35 km

= 42.21 km

Average Speed & the Marathon • marathon example (cont.) t = 2:12:36 t=2 hrs (3600s/1 hr) + 12 min (60 s/ 1min) + 36 s

= 7,956 s t = 2:26:05 = 8,765 s

Average Speed and the Marathon • average speed = distance/time speed = 42,210m/7956 s

= 5.3 m/s speed = 42,210/8765 s

= 4.8 m/s average velocity???

Average vs. Instantaneous • average velocity is not very meaningful in athletic events where many changes in direction occur • e.g. marathon  – start and end in same place so

 v  0 ???

d 0

Instantaneous Values • instantaneous velocity (v) is very important – specifies how fast and in what direction one is moving at one particular point in time – magnitude of instantaneous velocity is exactly the same as instantaneous speed

Average vs. Instantaneous Speed 14

1991 World Championships - Tokyo

12

speed (m/s)

10 8 Lewis Burrell Mitchell Lewis Avg Burrell Avg Mitchell Avg

6 4 2 0 0

2

4

6 time (s)

8

10

Average Acceleration (a) • rate of change of velocity with respect to time – “How fast the velocity is changing”

• acceleration is a vector quantity • units: m/s/s or m/s2 , ms2, ft/s/s

Dv v  v acceleration  a   Dt t  t f

f

i

i

Average Acceleration

velocity (m/s)

v0.0 = 0 m/s v2.5 = 5 m/s v5.0 = 0 m/s 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0

1

2

time (s)

3

4

5

velocity (m/s)

5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0

1

2

time (s)

3

4

5

1st interval a0.0  2.5 

v v 2.5

0.0 

2.50

m 5m  0 s s

 2.0 m2 2.5s s

Note: velocity is positive and acceleration is positive.

velocity (m/s)

5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0

1

2

time (s)

3

4

5

2nd interval v v

m 5 m 0 s  2.0 m a2.5  5.0  5.0 2.5  s 2 5.0 2.5 2.5 m s s Note: velocity is positive but acceleration is negative.

velocity (m/s)

5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0

1

2

time (s)

3

4

whole interval m m v5.0 v0.0 0m  0 s 0 a0.0  5.0   s t5.0 t0.0 5s 0s s2

5

Six Cases of Acceleration 1 - speed up in positive direction = positive accel.

+ direction

t = 3 seconds

a vi = 5 m/s

Calculate average acceleration!

vf = 8 m/s

Six Cases of Acceleration 1 - speed up in positive direction = positive accel. 2 - slow down in positive direction = negative accel. + direction

t = 3 seconds

a vi = 8 m/s

Calculate average acceleration!

vf = 5 m/s

Six Cases of Acceleration 1 - speed up in positive direction = positive accel. 2 - slow down in positive direction = negative accel. 3 - speed up in negative direction = negative accel. t = 3 seconds

+ direction a

vf = -8 m/s

Calculate average acceleration! What is happening to speed?, velocity?

vi = -5 m/s

Six Cases of Acceleration 1 - speed up in positive direction = positive accel. 2 - slow down in positive direction = negative accel. 3 - speed up in negative direction = negative accel. 4 - slow down in negative direction = positive accel. + direction

vf = -5 m/s

t = 3 seconds

vi = -8 m/s

Calculate average acceleration! What is happening to speed?, velocity?

Six Cases of Acceleration 1 - speed up in positive direction = positive accel. 2 - slow down in positive direction = negative accel. 3 - speed up in negative direction = negative accel. 4 - slow down in negative direction = positive accel. 5 - reverse directions from pos to neg = negative accel. t = 3 seconds

+ direction a

vi = +1 m/s

Calculate average acceleration!

vf = -1 m/s

Six Cases of Acceleration 1 - speed up in positive direction = positive accel. 2 - slow down in positive direction = negative accel. 3 - speed up in negative direction = negative accel. 4 - slow down in negative direction = positive accel. 5 - reverse directions from pos to neg = negative accel. 6 - reverse directions from neg to pos = positive accel. + direction

t = 3 seconds a

vi = +1 m/s

Calculate average acceleration!

vf = -1 m/s

In certain activities people experience + & - accelerations. By standardizing these accelerations to the normal acceleration on earth (-9.8 m/s/s) you get an idea of how much force they are experiencing

Human Response to Sustained g’s •

6-9 Gs: "Increased chest pain and pressure; breathing difficult, with shallow respiration from position of nearly full inspiration; further reduction in peripheral vision, increased blurring, occasional tunneling, great concentration to maintain focus; occasional lacrimation; body, legs, and arms cannot be lifted at 8 G; head cannot be lifted at 9 G."

•

9-12 Gs: "Breathing difficulty severe; increased chest pain; marked fatigue; loss of peripheral vision, diminution of central acuity, lacrimation."

•

15 Gs: "Extreme difficulty in breathing and speaking; severe vise-like chest pain; loss of tactile sensation; recurrent complete loss of vision. Data primarily from: Bioastronautics Data Book, second edition, 1973, NASA)

Relationships Between s, v, & a • v is the rate of change of s wrt time • a is the rate of change of v wrt time • consider a graph of s vs. time – s on vertical axis – time on horizontal axis – rate of change is interpreted as the slope

Slope of a Curve • “Slope” = number which describes the steepness of a line – rise/run q

– Note: this is the definition for the tangent of q, opposite / adjacent

Changes in the slope • positive slope – up and to the right

• negative slope – down and to the right

• quick change – very steep slope

• slow change – very flat slope

60

P o s i t i o n

50 40 30 20 10 0 0

5

10

15

20

25

30

Time (s)

rise 21- 8 Dp slope = = = run 10 - 5 Dt

The slope of the position by time curve is the velocity.

60

P o s i t i o n

50 40 30 20 10 0 0

5

10

15

20

25

30

Time (s)

Note that this is the average velocity during the period from 5 seconds to 10 seconds.

60 50

change in direction slope = 0

40 30 20 10 0 0

5

10

15

20

25

30

The tangent of a curve is the instantaneous slope at a single point. This slope represents the instantaneous velocity.

s

v

a

Relationship of s, v, & a •the instantaneous velocity (v) curve is the plot of how the slope of the s vs. t curve changes •a similar relationship exists between a and v

Where is velocity greatest? Where is acceleration greatest? Where is velocity closest to zero? What is happening to the position curve at this point? Where is acceleration negative? What is happening to the velocity curve at this point?

7

1991 World Championships - Tokyo 6

speed (m/s)

5

Who had the largest acceleration at the beginning of the race?

4

Lewis Burrell Mitchell

3 2 1 0 0

0.5

1 time (s)

1.5

2

12.2

1991 World Championships - Tokyo

speed (m/s)

12 Lewis Burrell Mitchell

11.8 11.6 11.4 11.2

Describe the accelerations at the end of the race. 11 8

8.5

9 time (s)

9.5

10

Steps to determining v vs. t curve from s vs. t curve (1)

(2) (3) (4)

draw a set of axes (v vs t) directly under the s vs. t curve locate all minimums, maximums, asymptotes, and inflection points plot zero value points for each corresponding min, max or asym plot mins or maxes for each inflection point

negative slope but flattening out

start negative but get closer to zero

minimum = zero slope must cross time axis (i.e. v=0)

positive slope but becoming steeper

Start at zero and increase

positive slope start out flat but becoming steeper

slope stays + just not as steep slope stops becoming steeper and begins to flatten out This is known as an inflection point and corresponds to a relative maximum on velocity vs. time curve

positive slope start out steep but becoming flatter

slope flattens out as much as corresponds to a relative minimum it is going to another inflection point then slope becomes steeper

positive slope start out steep continues to become steeper

Region 1 – negative slope so negative velocity

Region 2 – positive slope so positive velocity but inflection point where slope maxes out Region 3 – positive slope so positive velocity but inflection point where slope is minimized Region 4 – positive slope so positive velocity, no special points so velocity continues to rise

The above diagram shows a typical zero-g maneuver. However, the maneuver can be modified to provide any level of g-force less than one g. Some typical glevels used on different tests and the corresponding time for each maneuver are as follows: •Negative-g (-0.1 g): Approximately 15 seconds •Zero-g: Approximately 25 seconds •Lunar-g (one-sixth g): Approximately 40 seconds •Martian-g (one-third g): Approximately 30 seconds

NASA’s KC135A “The Vomit Comet”

s

v

a

inf

max

inf

Quantitative determination of v and a from s or How to calculate v and a from s Frame Time Pos. (m) Vel. (m/s) Acc. (m/s/s) 1

0.00

0.00

2

0.10

0.59

5.90

-23.00

3

0.20

0.95

3.60

-31.00

4

0.30

1.00

0.50

-10.00

5

0.40

0.95

-0.50

-31.00

6

0.50

0.59

-3.60

Ds Dv v , a Dt Dt

Dt = 0.10 s

Calculating v and a from s Frame Time Pos. (m) Vel. (m/s) Acc. (m/s/s) 1

0.00

0.00

2

0.10

0.59

5.90

-23.00

3

0.20

0.95

3.60

-31.00

4

0.30

1.00

0.50

-10.00

5

0.40

0.95

-0.50

-31.00

6

0.50

0.59

-3.60

Note: the velocity values do not occur at the same time as the position and acceleration values and you lose data at the starting and ending frames.

First Central Difference Method x i 1  x i 1 v xi  2dt

xi is the ith frame of horizontal position data vxi is the ith frame of horizontal velocity data dt is the time interval between frames

First frame of data

Last (nth) frame of data

x 2  x1 v x1  dt

v xn

x n  x n 1  dt

First Central Difference Method Frame Time Pos. (m) Vel. (m/s) Acc. (m/s/s) 1

0.00

0.00

5.90

-11.50

2

0.10

0.59

4.75

-19.25

3

0.20

0.95

2.05

-23.75

4

0.30

1.00

0.00

-20.50

5

0.40

0.95

-2.05

-18.00

6

0.50

0.59

-3.60

-15.50

What influences the shape of the path that an object follows when it is airborne?

Gravity makes it return to earth (i.e., fall). Any initial horizontal velocity will make it move either forward or backward.

When both of these influences are present the object always follows a parabolic path.

Airborne Motion • say a person jumps up into the air • motion is influenced only by gravity while the person is in the air • the CM will follow a parabolic path

on the way up ... initially vertical velocity is high when the body leaves the ground vertical velocity then decreases due to gravity

initial velocity (positive) velocity decreases v (m/s)

top of the jump ... the body changes direction so velocity is zero

initial velocity (positive) velocity decreases v (m/s)

velocity =0

on the way down ... the jumper’s velocity decreases, it becomes negative but the magnitude gets larger - speed increases

initial velocity (positive) velocity decreases v (m/s)

velocity =0 velocity decreases

final velocity (negative)

Airborne motion is UNIFORMLY ACCELERATED MOTION

v (m/s)

the change in velocity over time is linear so we say the change in velocity is constant This constant acceleration is = -9.8

m/s2

This is the rate at which any airborne object will accelerate.

Projectile Motion: A special case of uniformly accelerated motion If air resistance is negligible then only gravity affects the path (or trajectory) of a projectile.

This path is a parabola.

Horizontal and vertical components of velocity are independent. Vertical velocity decreases at a constant rate due to the influence of gravity.

Vertical velocity = 0

Positive velocity gets smaller

Negative velocity gets larger

Horizontal velocity will remain constant.

3 Primary Factors Affecting Trajectory • Projection angle aka release angle or take-off angle

• Projection height aka relative height = release height - landing height)

• Projection velocity aka release velocity or take-off velocity

Projection Angle • The optimal angle of projection is dependent on the goal of the activity. • For maximal height the optimal angle is 90o. • For maximal distance the optimal angle is 45o. • Optimal angle changes if projection height is not equal to 0.

Projection angle = 10 degrees 10 degrees

Projection angle = 30 degrees 10 degrees 30 degrees

Projection angle = 40 degrees 10 degrees 30 degrees 40 degrees

Projection angle = 45 degrees 10 degrees 30 degrees 40 degrees 45 degrees

Projection angle = 60 degrees 10 degrees 30 degrees 40 degrees 45 degrees 60 degrees

Projection angle = 75 degrees 10 degrees 30 degrees 40 degrees 45 degrees 60 degrees 75 degrees

So angle that maximizes Range (qoptimal) = 45 degrees (or so it appears)

Projection Height • Projection height = release height - landing height

Effect of Projection Height on Range (when qrelease = 45 degrees) h1 < h2 < h3 R1 < R2 < R3 hrelease = hlanding

R1 hrelease > hlanding hrelease >> hlanding

R2 R3

Projection Height and Projection Angle interact to affect Range R45 > R40 > R30

hprojection = 0

hlanding

hrelease R30 R40 R45

When hlanding < hrelease hprojection > 0

hrelease

R45 > R40 > R30 BUT difference b/w R’s is smaller hlanding

When hlanding << hrelease hprojection > 0

R40 > R30 > R45

hrelease So … as hprojection increases the optimal qrelease decreases hlanding

It’s possible to have a negative projection height (hrelease < hlanding) In this case the optimal qrelease is greater than 45 degrees

The effect of Projection Velocity on the Range of a projectile 40

Range ~ 10 m

10 m/s @ 45 degrees

30 20 10

0 0

10

20

30

40

50

60

70

80

90

100

The effect of Projection Velocity on the Range of a projectile 40 30

10 m/s @ 45 degrees

Range ~ 10 m

20 m/s @ 45 degrees

Range ~ 40 m

20 10

0 0

10

20

30

40

50

60

70

80

90

100

The effect of Projection Velocity on the Range of a projectile 40 30

10 m/s @ 45 degrees

Range ~ 10 m

20 m/s @ 45 degrees

Range ~ 40 m

30 m/s @ 45 degrees

Range ~ 90 m

20

So R a v2

10 0 0

10

20

30

40

50

60

70

80

90

•Because R a v2, it has the greatest influence on the Range of the projectile

100

Long Jump • What is the optimum angle of takeoff for long jumpers?

Projection height > 0 (take-off height > landing height) Optimum Angle should be slightly less than 45 degrees research shows that it should be 42-43 degrees

The Best of the Best Athlete Mike Powell (USA) Bob Beamon (USA) Carl Lewis (USA) Ralph Boston (USA) Igor Ter-Ovanesian (USSR) Jesse Owens (USA)

Distance of Jump Analyzed (m) 8.95 8.90 8.79 8.28 8.19 8.13

Speed of Takeoff (m/s) 9.8 9.6 10.0 9.5 9.3 9.2

Optimum Angle of Takeoff for Given Speed (deg) 43.3 43.3 43.4 43.2 43.2 43.1

Elena Belevskaya (USSR) Heike Dreschler (GDR) Jackie Joyner-Kersee (USA) Anisoara Stanciu (Rom) Vali Ionescu (Rom) Sue Hearnshaw (GB)

7.14 7.13 7.12 6.96 6.81 6.75

8.9 9.4 8.5 8.6 8.9 8.6

43.0 43.2 42.8 42.9 43.0 42.9

Actual Angle of Takeoff (deg) 23.2 24.0 18.7 19.8 21.2 22.0 19.6 15.6 22.1 20.6 18.9 18.9

Actual Angle of Takeoff ~ 17-23 degrees

Long Jump • when a jumper is moving at 10 m/s – the foot is not on the ground long enough to generate a large takeoff angle – so jumpers maintain speed and live with a low takeoff angle

• v is the most important factor in projectile motion

VALUES FOR HYPOTHETICAL JUMPS UNDER DIFFERENT CONDITIONS

Variable Speed of Takeoff

Angle of Takeoff

Values for Actual Jump (1)

Speed of Takeoff Increased 5% (2)

Angle of Takeoff Increased 5% (3)

Relative Height of Takeoff Increased 5% (4)

8.90 m/s

9.35 m/s

8.90 m/s

8.90 m/s

20

20

21

20

Relative Ht of Takeoff

0.45 m

0.45 m

0.45 m

0.47 m

Horizontal Range

6.23 m

6.77 m

6.39 m

6.27 m

0.54 m

0.16 m

0.04 m

7.54 m

7.16 m

7.04 m

Change in Horiz Range Distance of Jump

--

7.00 m

Suppose a zookeeper must shoot the banana from the banana cannon to the monkey who hangs from the limb of a tree. This particular monkey has a habit of dropping from the tree the moment that the banana leaves the muzzle of the cannon. If the monkey lets go of the tree the moment that the banana is fired, will the banana hit the monkey?

Banana’s gravity-free path

If there is no gravity then the monkey floats AND you throw directly at the monkey, then the path of the banana will be a straight line (the “gravity-free path”). Since this path will cross the point where the monkey floats the monkey can catch and eat the banana!

When you take gravity into consideration you STILL aim at the monkey! Banana’s Gravity free path Monkey’s Gravity free path is “floating” at height of limb

Fall thru same height

YEAH! It works! Since both banana and monkey experience the same acceleration each will fall equal amounts below their gravity-free path. Thus, the banana hits the monkey.

What happens when you throw the banana slower?

Monkey’s Gravity free path is “floating” at height of limb

Banana’s Gravity free path Fall thru same height

As long as you aim at the monkey he will still catch it. The only difference is that the monkey will fall farther before he catches it because it takes longer to travel the necessary horizontal distance.

Eqns of Constant Acceleration Motion

ECAM’s Eqn 1

Eqn 3

Eqn 2

Eqn 4

v f  vi  at

d  (vi  v f )t 1

d  vi t  21 at 2

v 2f  vi2  2ad

2

d = displacement (d = sf – si) vi = initial velocity vf = final velocity a = acceleration t = time

: when there is no change in direction then displacement and distance are the same thing so …

Often times it is useful to consider these equations being applied separately for x- and y-directions

Eqns of Constant Acceleration Motion

ECAM’s Eqn 1

Eqn 3

Eqn 2

Eqn 4

v f  vi  at

d  vi t  21 at 2

d  (vi  v f )t

v 2f  vi2  2ad

1

2

Eqn

d

vi

vf

a

t

1 2 3 4

   

   

   

   

   

that d = sf - si

ECAM Examples • Example problem – a cyclist es the midpoint of a race moving at a speed of 10 m/s – she accelerates at an average rate of 3 m/s/s for 3 s – how fast is she moving at the end of this period?

ECAM Examples An object falls 10 meters from the top of a tower. What is the velocity and how much time does it take to reach the ground? Steps: 1. Draw a picture. 2. List values for any parameters that are given. 3. Find equations in which all of the variables are known except the one that you are trying to find. 4. Substitute values for variable and solve.

Example A runner starts from rest, uniformly accelerates at 3 m/s2 for 3 seconds, then runs at a constant velocity for 5 seconds, then accelerates in the negative direction at -2 m/s2 for 2 seconds. How far does the runner travel during this 10 second period?

Diving Example Can the diver successfully complete a 2.5 somersault?

PROBLEM DESCRIPTION It is given that it takes a minimum of 0.95 s to perform a 2.5 somersault.

+

2m

Only consider the vertical component.

0.85 m 1m

0.85 m ventry = ?

1st: find time to reach peak of dive (tup)

+ 2m 0.85 m

si = sf = d= ay = vf = vi = t=

1m

Which equation should you use?

tup = tup’

Step 1 (cont.)

+ 2m

0.85 m

si = sf = d= ay = vf = vi = t=

1m

Which equation should you use?

2nd: Find time from peak of flight to time of impact with water

2.0 m

1.0 m

si = sf = d= ay = vf = vi = t=

0.85 m

Which equation should you use?

How many somersaults can the diver complete off of other boards if it take .38 s per somersault?

• 3-m springboard (CM reaches 5M above water) • tup = 0.48 s • tdown = 0.92 s • ttotal = 1.40 s

1.4/0.38 = 3.7 3.5 somersaults

• 5-m platform (CM reaches 1.25 m above platform) • tup = 0.28 s (raise CM only 0.4 m above initial pos) • tdown = 0.95 s 1.33/0.38 = 3.5 • ttotal = 1.33 s 3.5 somersaults (if perfect)

Other Boards (cont.)

• 10-m platform (CM reaches 1.25m above platform) • tup = 0.28 s • tdown = 1.46 s • ttotal = 1.74 s

1.74/0.38 = 4.6 4.5 somersaults

• 20-m cliff (CM reaches 1.25 m above cliff) • tup = 0.28 s (raise CM only 0.4 m above initial pos) • tdown = 2.04 s 2.32/0.38 = 6.1 • ttotal = 2.32 s 6 somersaults (if you’re crazy!)

Speed of Impact Know: vi = ay = d= vf = 2.15 m

• 1 m board d = 2.15 m

vf = 6.5 m/s

(14.5 mph)

vf = 9.0 m/s

(20.1 mph)

• 3 m board d = 4.15 m

• 5 m platform d = 5.4 m

vf = 10.3 m/s (23.0 mph)

• 10 m platform d = 10.4 m

vf = 14.3 m

(32.0 mph)

• 20 m cliff d = 20.4 m

vf = 20.0 m/s (44.7 mph)