Do Now:

Do Now:

• Joe and Jane are running towards each other. Joe is running at a rate of 15 m/s

[North] and Jane is running at a rate of 10 m/s [South]. If they start at 50 m apart, at what time would they meet?

Quick Review/Recap

Physics Data Sheet http://www.wpwagner.com/

• We use significant digits to describe the precision of measurements.

Which digits are not significant?

- leading zeros

Which digits are significant?

- non-zero digits - trailing zeros

e.g. 0.00072100 m

This measurement has 5 significant digits.

What’s the difference between 3 cm and 3.0 cm?

3 cm

(range: 2.5 to 3.4) 3.0 cm

(range: 2.95 to 3.04)

Data with the lowest number of sigdig’s limits the number of sigdig’s in a calculation.

Example:

Distance = (8.3 m/s)(17.57 s) = 1.5 x 10² m

or = 0.15 km

145.831

Uniform motion: when an object’s speed is not changing. This includes object which are

stopped or moving at a constant speed.

Some Important Terms:

Accelerated motion: when an object’s speed is changing. It can be speeding up or slowing down.

Speed is the distance traveled per unit time.

Average Speed: calculated over a time interval.

Instantaneous Speed: Speed during one moment in time.

t v  d

A line of best fit is an averaging technique.

GRAPHING MOTION:

Distance versus Time for a Slow Pedestrian

Graph Checklist:

 Title

 Labeled Axes

 Data Points

 Line of Best Fit

Side Bar: It is possible to have a curve of best fit.

Average Total Length as a Function of Age for Sturgeon

How old is this fish?

Using Slope to Calculate Speed

Example: A baby sea turtle craws toward the sea. A biologist attaches him to a ticker tape in order to measure his speed. She gathers the following data. Plot a d-t graph and a v-t graph for this motion.

Time

(s) Distance (cm)

0.00 0.0

0.10 1.2

0.20 2.2

0.30 3.8

0.40 4.8

0.50 5.7

• The slope is a

measure of the rate of change in distance

with respect to time.

• Therefore, the slope of a d-t graph

represents the turtle’s speed.

RUN

RISE

Fast object  Large slope Slow object  small slope

d

t

d

t

Stopped object  Slope = 0

d

t

Relative to a reference point (in this case a point in space)

The find the speed of the turtle, calculate the slope of the d-t graph.

Step 1: Select two new points on the line of best fit.

Step 2: Read the

values for each of the new points.

Step 3: Apply the slope formula:

+

+

0.48 s 0.12 s

1.4 cm 5.6 cm

y₁ y₂

x₁ x₂ (INCLUDE UNITS!)

1 2

1 2

x x

y slope y



 

s cm

s s

cm cm

/ 12

12 . 0 48

. 0

4 . 1 6

. 5





 

• Recall the turtle example.

• Divide the motion into 0.10 s intervals and determine the

average speed at each interval.

Time (s) Distance (cm)

0.00 0.0

0.10 1.2

0.20 2.2

0.30 3.8

0.40 4.8

0.50 5.7

Distance traveled during each interval (cm)

Midpoint Time (average

time) (s)

Speed (cm/s)

1.2 - 0.0 = 1.2 0.05 1.2cm / 0.10s

= 12 cm/s

2.2 - 1.2 = 1.0 0.15 1.0 cm / 0.10 s

= 10 cm/s

3.8 - 2.2 = 1.6 0.25 1.6 cm / 0.10 s

= 16 cm/s

4.8 - 3.8 = 1.0 0.35 1.0 cm / 0.10 s

= 10 cm/s

5.7 - 4.8 = 0.9 0.45 0.9 / 0.10 s = 9.0 cm/s

total ave

total

f i

ave

v d

t

d d

v t



 



t (s)

V (cm/s)

0.000 0.100 5.0

*

• The average speed = the instantaneous speed at each intervals “midpoint time”.

• Constant

speed (uniform motion)

slope is

approximately zero in this

case

Labs: v-t and d-t graphs for uniform motion (Handouts)

Summary of in d-t graphs:

Uniform motion

constant (-) velocity (moving toward ref. pt.) constant (0)

velocity (stopped) constant (+) velocity

(moving away from ref. pt.)

d

t

d

t

d

t

“fast”

“slow”

• The slope of a d-t graph is velocity.

SPEED,

ACCELERATION AND VECTORS

• Acceleration describes how much an object’s velocity changes per unit time.

An Example of Accelerated Motion:

A penguin is dropped. His motion during freefall is measured using a ticker-tape timer. The following data is gathered. Plot a d-t graph and a v-t graph for this motion. Determine the

acceleration of the falling penguin using the v-t graph.

Time

(s) Distance (cm)

0.00 0.0

0.10 5.6

0.20 22.7 0.30 50.6 0.40 88.9 0.50 136.8

• The Penguin’s d-t graph:

• Recall that the slope of the d-t graph is speed, but the slope is changing!

• The slope of the tangent line from any point on the curve gives the instantaneous

speed at that point in time.

• As it falls, the

penguin is speeding up (accelerating)

slow

medium fast

FYI: A curve made of tangent lines:

Penguin’s v-t data:

Time

(s) Distance (cm)

0.00 0.0

0.10 5.6

0.20 22.7

0.30 50.6

0.40 88.9

0.50 136.8

Midpoint

Time (s) Speed (cm/s)

0.05 56

0.15 171

0.25 279

0.35 383

0.45 479

The falling penguin’s v-t graph

+

+

0.40 s 0.12 s

430 cm/s

y₂

x₁

160 cm/s

y₁

x₂

1 2

1 2

x x

y slope y



 

s s cm

s s

s cm s

cm

/ / 1071

12 . 0 40

. 0

/ 160

/ 430





 

• Convert our experimental value to m/s/s:

1071 cm/s/s = 10.7 m/s/s

• The slope of a v-t graph is a measure of the rate of change in speed. Therefore, the slope of a v-t graph represents

acceleration.

• The most common units for acceleration are m/s²

2

1

s m s

s m s

s m

 



 







 



 



 





The universally accepted average rate of vertical

acceleration due to gravity near the earth’s surface is 9.81 m/s². All objects in free fall experience this acceleration if the effects of air resistance can be ignored.

• How does our experimental value of 10.7 m/s² compare to the accepted value of 9.81 m/s²?

• Why the difference?

2 2

2

( )

100%

(10.7 / 9.81 / ) 9.81 / 100%

9.1%

Experimental Value Accepted Value

Percent Difference x

Accepted Value

m s m s

m s x difference

 

 



Accelerated Motion

(-) acceleration (+) acceleration

d

t

“small accel.”

“big accel.”

Speeding up in the positive direction

Slowing down in the negative direction

Speeding up in the negative direction

Slowing down in the positive direction

Summary of Vectors in d-t graphs con’t:

d

t

d

t

d

t

d (m)

t (s)

Found that last slide tough? You can use tangent lines to analyze

d-t graphs for accelerated motion.

Remember the slope of a d-t graph is velocity!

Arbitrary sign convention

450 m 370 m

North e) Sketch a v-t graph of the motion

d (m)

450

80

144

- +

v (m/s)

t (s) +

-

0

Summary of v-t graphs

Uniform motion

constant (+) velocity (moving away from ref. pt.) constant (0)

velocity (stopped) constant (-) velocity

(moving toward ref. pt.)

v

t

v

t

v

t

“fast”

“slow”

Accelerated Motion

(-) acceleration (+) acceleration

v t

“small accel.”

“big accel.”

Speeding up while traveling

in a positive direction

Slowing down while traveling in

a negative direction

Speeding up while traveling

in a negative direction

Slowing down while traveling

in a positive direction

Summary of Vectors in v-t graphs con’t:

v t v t v t

Practice Problems

• Page 30 #1 - 4

Do Now

• Draw a d-t graph for the following situations:

– An object speeding up in a positive direction – An object slowing down in a negative direction

• Draw v-t graph for the above situations

Lab Activity

• Measuring acceleration due to gravity of different objects.

Penguin’s v-t data:

Time

(s) Distance (cm)

0.00 0.0

0.10 5.6

0.20 22.7

0.30 50.6

0.40 88.9

0.50 136.8

Midpoint

Time (s) Speed (cm/s)

0.05 56

0.15 171

0.25 279

0.35 383

0.45 479

The falling penguin’s v-t graph

+

+

0.40 s 0.12 s

430 cm/s

y₂

x₁

160 cm/s

y₁

x₂

1 2

1 2

x x

y slope y



 

s s cm

s s

s cm s

cm

/ / 1071

12 . 0 40

. 0

/ 160

/ 430





 

Lab: Motion Sensors

Part A: Match the Graph

Procedure: Follow the instructions on the laminated card.

Each group member must get a chance to match both a d-t and v-t graph.

Question 1: Using graph paper, sketch a d-t and a v-t graph for the following story: (include numbers on the time scale for only for your d-t graph, and numbers on both scales for your v-t graph.)

Starting at his shed, Jim-Bob accelerates east away from his shed for 2.0 s until he reaches a velocity of 4.5 m/s, east. He maintains this velocity for 3.0 s. He then slows to a stop in 2.5 s. After remaining stationary for 2.0 s, Jim-Bob accelerates west toward his shed for 1.5 s, reaching a velocity of 6.0 m/s, west which he maintains

for 3.2 s, at which time, he smashes into the shed.

Due to the shed’s flimsy construction, Jim-Bobs speed is unaltered and 2.8 s later he is still traveling with a velocity of 6.0 m/s, west.

Part B: Rolling Ball

Procedure: Set up your motion sensor and eave as demonstrated by teacher. Roll the ball. Be sure that you’ve performed an adequate trial.

Analysis:

• Using the data lists in your calculator, plot d-t and v-t graphs describing the motion of the ball. (Plot them with the same scale for time – like in the notes.)

• Label each graph with the following:

- Positive velocity - Negative velocity - Speeding up - Slowing down

- Positive acceleration - Negative acceleration

Question 2: Sketch d-t and v-t graphs for an object which is thrown 2.0 m into the air and caught at a height 0.5 m below the height from which it was thrown.

Solving Simple Motion Problems – speed, velocity, and acceleration

Let’s review the equations you should know so far:

v_ave= average velocity (m/s) d = total displacement (m) t = time (s)

a = acceleration (m/s²) v_f = final velocity (m/s) v_i = initial velocity (m/s)

Example Problems:

1) How long does it take a grizzly to run 150 m if she is

running at a speed of 32 km/h?

2) A taunting baboon approaches a male lion at a velocity of 4.0 km/h, east for 4.5 s, and then runs at a velocity of 20 km/h, west for 3.8 s.

a) Calculate the baboon’s average speed.

b) Calculate the baboon’s average velocity.

Converting “Combo” Units

e.g. you need to convert 20 km/h to m/s

1000 m= 1 km, therefore, and equal 1.

3600 s = 1 h, therefore, and also equal 1.

20 km/h = x x = 5.555… m/s

km m 1

1000

m km 1000

1

s h 3600

1 h

s 1 3600

h km 1 20

km m 1

1000

s h 3600

1

2) Sparky drops off the end of a dock. If he falls for 0.80 s. How fast is he moving when he hits the water?

3) A cat jumps upward, leaving the ground with a speed of

9.50 m/s.

a) How long does she remain in the air if she lands at the

same height as she jumped?

b) Sketch a v-t graph of the cat’s motion.

Practice:

• P.45 Q 1,2

• P.47 Q 6, 7, 8

• P.66 Q 1-4

• P.69 Q 9-12

More Powerful Methods for Analyzing Motion

More Powerful Methods for Analyzing Motion: Formulas for almost any occasion!

• The area under a v-t graph is equal to the displacement of the object

V (m/s)

t (s)

1.5

Right

Left

A cat is carried at a constant velocity of 1.5 m/s, right for 3.0 s

d = vt = (1.5 m/s)(3.0s) = 4.5m, right

3.0

Area = (height)(base) = (1.5 m/s) (3.0s) = 4.5 m, right

height

base

A cat jumps straight up and then falls to the ground again (see graph below).

Calculate the height of the cat’s jump.

(+) d

The area under the graph represents displacement.

+

-

(-) d

m

s m s

v t

height base

d

00 . 5

) / 00

. 10 )(

00 . 1 (

) )(

(

) )(

(

2 1 2 1 2 1













 



INSERT MORE COMPLEX AREA EXAMPLE HERE!!!!

Practice:

P. 61 Q 24, 25

Answers:

25 a) 75 m b) 150 m c)125 d) 500 m

26 a) b) 6.0 m c) 24 m d) 102 m v

t

Do Now

• Answer the following question:

• If a car can accelerate from 0 – 100 km/hr in 3 seconds, how far will the car have

travelled if it constantly accelerates for 4.2 seconds?

We can use the fact that the area under a v-t graph is displacement to derive more powerful formulas for accelerated motion.

v_i v_f

v

t

The area under the graph is equal to displacement.

Area A Area B Formula #1

Data:

 d

 v_i

 v_f

 a

 t

 

2 2

1 2 1 2 1

) (

t a t

v d

t at

t v

t v

v t

v

Area Area

d

i i

i f

i

B A



 





















t a v

t v v

a v^f ⁱ _f _i 

    



Example Problem 1:

When you fall off of a cliff, your friends decide to time your fall in order to determine the height of the cliff. If you remain in the air for 2.2 s, determine the height of the cliff.

24 m

Example Problem 2:

An cannonball is launched straight up. Calculate how high it goes if it stays in the air for 5.9 s.

43 m

Example Problem 3:

A train traveling 100 m/s hits the accelerator. As it does so, it travels 400 m in 2.9 s. Calculate the train’s acceleration.

Compare your answer to the acceleration due to gravity.

26 m/s²: That’s 2.7g

v_i v_f

v

t

… a slight variation on the previous formula:

Area_total Area_above

The area under the graph = displacement Formula #2

Data:

 d

 v_i

v_f

a

 t

 

2 2

1 2 1 2 1

) (

t a t

v d

t at

t v

t v

v t

v

Area Area

d

f f

i f

f

above total



 





















v_i v_f

v

t

and…

therefore…

Formula #3

Data:

 d

 v_i

v_f

a

 t

t v

d

t v d

ave ave

 

 





2

i f

ave

v v v



  



v t d v ^{f i} 



 



 

 2



 

Example Problem 4:

Calculate the final speed of a kangaroo that begins at 3.0 m/s and uniformly accelerates over a distance of 25 m in a time of 5.50 s.

6.1 m/s

Do Now

• Try and answer the following problem:

• Joe starts at a velocity of 1.0 m/s right. He accelerates at a constant rate of 3.0 m/s². Find his velocity after 5.0 m of

displacement.

Formula #4 Let’s start with Formula #3

… and

Substitute for t

Data:

 d

 v_i

 v_f

 a

 t

v t d v^{f i} 



 



 

 2



 

t v a v^{f i}



  

 a

v t v^f  ⁱ

  



d a v

v

a v v

a

v v

v v

v v

a

v v

v v

a v v

v d v

i f

i f

i f

i f

i f

i f

i f

i f

i f

 









































 

2 2

2 2

) )(

(

2

2 2

2 2

2 2





 





 



 



 



 



 



 



Example Problem 5

During a high speed chase a police car accelerates to a final speed of 180 km/h at a rate of acceleration of 8.50 m/s², over a distance of 100 m. Determine the initial

speed of the police car.

28.3 m/s or 102 km/h

Practice:

P. 72 Q. 13-16 P. 74 Q. 17-20 P. 75 Q. 21-24 P. 77 Q. 25-26

P. 79, 80 Q. 27-31

Example Problem 6:

A clown is launched straight up at a speed 20 m/s out of a helicopter hovering 14m

above the ground. While the clown is airborne, the

helicopter flies away.

Calculate how long the clown remains in the air.

4.7 s

Do Now

• A clown is launched straight up at a speed 20 m/s out of a helicopter hovering 14m

above the ground. While the clown is airborne, the helicopter flies away.

Calculate how long the clown remains in the air.

4.7 s