Objectives 1.Describe motion in terms of frame of reference, displacement, distance, vector, and scalar. 2.Understand relationship between displacement.

objectives1. Describe motion in terms of frame of

reference, displacement, distance, vector, and scalar.

2. Understand relationship between displacement and distance, vectors and scalars.

3. Compare distance traveled to displacement.

4. Draw diagrams showing displacement and/or distance from a point of origin.

Description of motion depends on the frame of reference

• A frame of reference can be thought of as any spot you are doing your measurement from as long as it is not accelerating. The reference point is arbitrary, but once chosen, it must be used throughout the problem.

Displacement vs. Distance

• DISTANCE – The complete length of the path traveled

by a moving object

• DISPLACEMENT – the length of the straight-line path from

a moving object’s origin to its final position

Scalar vs. Vector

• SCALAR– A measured quantity that has NO DIRECTION– Examples

• Distance, Time, Mass, Volume

• VECTOR– A measured quantity that includes

DIRECTION– SIGN SHOWS DIRECTION– Example

• Displacement

A ball rolls 5 meters north.

Distance =

Displacement =

5 m

+5 m

Sign Conventions

PositiveNORTHEAST

RIGHT

NegativeSOUTHWESTLEFT

A cat runs 8 meters west.

Distance =

Displacement =

A bird flies 5 meters north, then 7 meters south

Distance =

Displacement =

8 m

-8 m

12 m

-2 m

Example

• A man drives his car 3 miles north, then 4 miles east.

What distance did he travel?What is his displacement from his point of origin?

3 miNorth

4 miEast

Displacement5 mi

Northeast

Distance7 mi

Example• Three men leave the same house on foot. The first man

walks 30 feet north, then 40 feet west. The second man walks 90 feet south, then 88 feet north. The third man walks 10 feet east, then 50 feet west.

• Which man has traveled the greatest distance?

• Who is farthest from the house?

• Who is closest to the house?

The second man

The first man

The second man

• There could be many distances between xf and xi many be many, distance depends on the path.

• There is only one displacement between xf and xi. displacement refers to shortest distance between the xf and xi and direction from xi to xf

Distance vs. Displacement

Displacement = change in position = final position – initial position

∆x = xf - xi

∆ denotes change

xi xf

example

Distance (m) Displacement (m)AB 180 +180

BC 140 -140

CD 100 +100

TOTAL 420 +140

• Use the diagram to determine the resulting displacement and the distance traveled by the skier during these three minutes.

Example• Consider the motion depicted in the

diagram below. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.

Distance:

Displacement:

12m

0

Distance vs. Time GraphsDuring what time interval was the object NOT MOVING?

Distance vs. Time

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4 5 6 7

Time (s)

Dis

tan

ce (

m)

2 – 3 secondsThe interval on the graph wherethe distance remains constant!

Displacement vs. Time Graphs

Displacement vs. Time

-4

-3

-2

-1

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7

Time (s)

Dis

pla

cem

en

t (m

)

When the displacement isnegative, the object has a

position to the left of the origin1 – 2 and 4 – 5 seconds

Constant displacement means thatthe object doesn’t move

The object’s final position isat +1 meter (1 meter to the

right of the origin)

During what time interval(s) was the object to the left of the origin?During what time interval(s) was the object NOT MOVING?At what distance from the origin does the object stop?

Objectives • Know:

– - Definitions of velocity and speed. – - Equation for average velocity/speed.

• Understand: – - Relationship between speed, distance and time. – - Relationship between vectors and scalars.

• Be able to: – - Use velocity/speed equation to find unknown. – - Draw and interpret d/t and v/t graphs.

Homework – castle leaning – must show work on a separate sheet of paper

Velocity vs. Speed

• VELOCITY – change in DISPLACEMENT occurring over time– Includes both MAGNITUDE and DIRECTION

• VECTOR

• The direction of the velocity vector is simply the same as the direction that an object is moving.

• SPEED – change in DISTANCE occurring over time– Inclues ONLY MAGNITUDE

• SCALAR

As an object moves, it often undergoes changes in speed

Types of speed and velocity

• Initial speed

• Final speed

• Average speed

• Instantaneous speed

• Initial velocity

• Final velocity

• Average velocity

• Instantaneous velocity

Calculate Average Speed and Average Velocity

• The average speed during the course of a motion is often computed using the following formula:

• In contrast, the average velocity is often computed using this formula

Does NOT include DIRECTION!

vavg = ∆x∆t = tf - ti

xf - xi dt

=

Average Velocity

dv

t

What does this remind you of?Position vs. Time

Time

Po

sit

ion

What is happening in thisgraph?

SLOPE OF A GRAPH!

CONSTANTZERO

SLOPE

MotionlessObject

Position vs. Time

Time

Po

sit

ion

CONSTANTPOSITIVE

SLOPE

Moving withCONSTANT

positive velocity

Position vs. Time

Time

Po

sit

ion

INCREASINGSLOPE Moving with

INCREASINGvelocity

Example• Sally gets up one morning and

decides to take a three mile walk. She completes the first mile in 8.3 minutes, the second mile in 8.9 minutes, and the third mile in 9.2 minutes.

– What is her average speed during her walk?

vavg = d / tvavg = 3 mi / (8.3 min + 8.9 min + 9.2 min)

vavg = 0.11 mi / min

Example• Tom gets on his bike at 12:00 in the

afternoon and begins riding west. At 12:30 he has ridden 8 miles.

– What was his average velocity during his ride?

vavg = d / tvavg = 8 mi / 30 min

vavg = 0.27 mi / min WEST

Example – finding displacement

• During a race on level ground, Andre runs with an average velocity of 6.02 m/s to the east. What displacement does Andre cover in 137 s?

Answer: 825 m East

vavg = ∆x∆t (∆t )(∆t )

∆x = vavg (∆t ) = (6.02 m/s)(137 s) = 825 m

example• What is the coach's average speed and average

velocity?

average speed = (35 + 20 + 40) yd / 10min = 9.5 yd/min

average velocity = (-35 + 20 - 40) yd / 10 min = - 5.5 yd/min

Instantaneous Speed

• Instantaneous Speed - the speed at any given instant in time.

Average Speed versus Instantaneous Speed

Average Speed - the average of all instantaneous speeds; found simply by a distance/time ratio.

During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour. Yet, on average, you were moving with a speed of

In conclusion• Speed and velocity are kinematics quantities

that have distinctly different definitions. Speed, being a _______quantity, is the rate at which an object covers ___________. The average speed is the _____________ (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a _________quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the ______________ or position change (a vector quantity) per time ratio.

scalardistance

distance

vector

displacement

Find displacement

Using v-t GraphsVelocity vs. Time

-20

-15

-10

-5

0

5

10

15

20

25

30

35

0 1 2 3 4 5 6 7 8 9 10 11

Time (s)

Velo

cit

y (

m/s

) Find distancetraveled

What can we DO with a v-t graph?

Find averagevelocity

How do you use the v-t graph to findAVERAGE VELOCITY?

How do you use the v-t graph to find DISTANCE TRAVELED?

How do you use the v-t graph to find DISPLACEMENT?

Area under the graph

Area on TOP and BOTTOMboth considered POSITIVE

Area under the graph

Area on TOP = POSITIVEArea on BOTTOM = NEGATIVE

objective

1. Construct and interpret graphs of position vs. time and velocity vs. time.

Graphical Interpretation of Velocity: position vs. time graph

Straight line - Constant Velocity (positive)

Curved line – Changing Velocity (increasing)

• The Meaning of Shape for a p-t Graph

As the slope goes, so goes the velocity

Slow, Positive, Constant Velocity

Fast, Positive, Constant Velocity

Slow, Negative Constant Velocity

Fast, Negative, Constant Velocity

Check Your Understanding• Use the principle of slope to describe the motion

of the objects depicted by the two plots below.

Slope is positive, increasing

velocity is positive, increasing (faster)

slope is negative, increasing

velocity is negative, increasing (faster)

The slope of the line on a position-time graph is equal to the velocity of the object

Slope = constant = +10 m/sVelocity = constant = +10 m/s

Slope is changing –increasing in positive direction velocity is changing –increasing in positive direction

• Initial velocity

• Final velocity

• Average velocity

• Instantaneous velocity

Velocity information on P-t graph

• +10 m/s

• +10 m/s

• +10 m/s

• +10 m/s

•Initial velocity•Final velocity•Average velocity•Instantaneous velocity

• 0 m/s• ~+22 m/s• +10 m/s• Slope of the tangent

line

As the slope goes, so goes the velocity

Slope is negative, increasing (steeper)

velocity is negative, increasing (faster)

Slope is negative, decreasing (flatter)

velocity is negative, decreasing (slower)

example

Describe the velocity of the object between 0-5 s and between 5-10 s.

The velocity is 5 m/s between 0-5 seconds

The velocity is zero between 5-10 seconds

Practice – determine average velocity

7.3 m/s

objectives

1. Describe motion in terms of changing velocity.

2. Compare graphical representations of accelerated and nonaccelerated motions.

3. Apply kinematics equations to calculate distance, time, or velocity under conditions of constant acceleration.

• AccelerationAcceleration

Questions: 1. “If an object has a large velocity, does it necessarily have

a large acceleration? 2. If an object has a large acceleration, does it necessarily

have a large velocity?”

aavg = ∆v∆t = tf - ti

vf - vi

Acceleration = (change in velocity) / time

change in VELOCITY occuring over TIME

Vector

Measured in m/s2

• Anytime an object's velocity is changing, the object is said to be accelerating; it has an acceleration.

• What are the three ways to accelerate your car?Gas pedal, break, steering wheel

The Direction of the Acceleration Vector

• The direction of the acceleration vector depends on whether the object is speeding up or slowing down

• If an object is speeding up, then its acceleration is in the same direction of its motion.– moving in the + direction, the acceleration + direction– moving in the - direction, the acceleration - direction

• If an object is slowing down, then its acceleration is in the opposite direction of its motion.– moving in the + direction, the acceleration - direction– moving in

• The direction of velocity and acceleration do not have to be the same!!!

Positive Velocity

Positive Acceleration

Speeding up in + direction

Negative Velocity

Negative Acceleration

Speeding up in - direction

Positive Velocity

Negative Acceleration

Slowing downEventually speeds up in – direction!

Negative Velocity

Positive Acceleration

Slowing downEventually speeds up in + direction!

+v

+a

The car will…

-v

-a

The car will… The car will… The car will…

+v

-a

-v

+a

Equations

t

vaavg

atvv if

“Acceleration is a rateof change in velocity”

“The slope of a v-tgraph tells what the

ACCELERATION ISDOING!”

“An object’s velocity at any point in time canbe found by considering:

- its starting velocity

- its acceleration

- the amount of time over which it accelerates”

Example – average acceleration• What is the acceleration of an amusement

park ride that falls from rest to a speed of 28 m/s in 3.0 s?

aavg = ∆v∆t = tf - ti

vf - vi

aavg = (28 m/s – 0) / 3.0 s = 9.3 m/s/s down

Example – finding time of accelertion

• A shuttle bus slows to a stop with an average acceleration of -1.8 m/s2. How long does it take the bus to slow from 9.0 m/s to 0.0 m/s?

aavg = ∆v∆t

∆t = ∆vaavg

= (0.0 m/s – 9.0 m/s) / -1.8 m/s2 = 5.0 s

∆v

aavg ∆t

Example

• Two cars start at the same point . Car A starts with a velocity of -5 meters per second while Car B starts with a velocity of +3 meters per second. At the end of 15 seconds, Car A has a velocity of +25 meters per second.

– What is Car A’s acceleration?

– If Car B has the same acceleration, what is its speed at 15 seconds?

vf = vi + at25 m/s = -5 m/s + a(15 s)

a = +2 m/s2

vf = vi + atvf = +3 m/s + (2 m/s2)(15 s)

vf = +33 m/s

The Meaning of Constant Acceleration

The velocity is changing by a constant amount - in each second of time.

constant velocity: displacement is changing by a constant amount - in each second of time.

constant acceleration: displacement is increasing in each second of time. Velocity is changing by a constant amount - in each second of time

Constant motion vs. accelerated motion ticker tape

Describing Motion with Velocity vs. Time Graphs

Velocity: ConstantSlope = 0Acceleration = 0

Velocity: increasingSlope: +, constantAcceleration: +, constant

The slope value of any straight line on a velocity-time graph is the acceleration of the object

Speeding up or slowing down?

• Speeding up means that the magnitude (or numerical value) of the velocity is getting large.

Example: describe velocity and acceleration for each v-t diagram

Velocity is positive. increasing, speeding up

Acceleration is constant, positive

Velocity is positive decreasing, slowing down

Acceleration is constant, negative

Velocity is negative decreasing, slowing down

Acceleration is constant, positive

Velocity is negative, increasing, speeding up

Acceleration is constant, negative

Determine acceleration

From 0 s to 4 s:From 4 s to 8 s:

Example 1

0 m/s2

2 m/s2

Example 2• The velocity-time graph for a two-stage rocket is shown

below. Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals.

1. t = 0 - 1 second

2. t = 1 - 4 second

3. t = 4 - 12 second

+40 m/s/s

+20 m/s/s

-20 m/s/s

objectives• Understand Relationships for all kinematics

equations. • Understand Relationship between slope of

kinematics graphs and quantities they represent. • Use any kinematics equation to find unknown. • Draw and interpret all kinematics graphs.

Homework – castle learning

vavg = vf + vi

2vavg =

∆x∆t

= vf + vi

2

∆x∆t

∆x = ½ (∆t)(vf + vi)

The displacement equals to the area under the velocity vs. time graph

Determining the Area on a v-t Graph

For velocity versus time graphs, the area bound by the line and the axes represents the displacement.

The shaded area is representative of the displacement

area = base x height

area = ½ base x height

area = ½ base x ( height1 + height2)

example• Determine the displacement (i.e., the area) of the object

during the first 4 seconds (Practice A) and from 3 to 6 seconds (Practice B).

120 m 90 m

example

• Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B).

5 m 45 m

example• Determine the displacement of the object during the time

interval from 2 to 3 seconds (Practice A) and during the first 2 seconds (Practice B).

25 m 40 m

Recap

• Acceleration:

• Final velocity: vf = vi + a∆t

• Displacement: ∆x = ½ (∆t)(vf + vi)

a = ∆v∆t = tf - ti

vf - vi

• Displacement: ∆x = vi∆t + ½ a∆t2

• Average velocity: v = ∆x ∆t = 2

vf + vi

Another equation for final velocity

• Final velocity: vf = vi + a∆t

• Displacement: ∆x = ½ (∆t)(vf + vi)

• Final velocity: vf2= vi

2 + 2a∆x

Constant Non-Zero Acceleration

2

2

1attvd i

advv if 222

“Distance or Displacement” Equation

“Timeless or ShortcutEquation”

DOES NOT INCLUDETIME!

In one dimensional

motion:

∆x = d

• A bicyclist accelerates from 5.0 m/s to a velocity of 16 m/s in 8 s. Assuming uniform acceleration, what displacement does the bicyclist travel during this time interval?

Example

84 m

∆x = ½ (vf + vi)(∆t)

Example

• A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and breaking system, and comes to rest 5.5 s later. Find how far the car moves before stopping.

120 m

Example• A train starts from rest and leaves

Greenburg station and travels for 500. meters with an acceleration of 1.20 meters per second2.

– What is the train’s final velocity?

– How long does it take the train to reach its final velocity?

vf2 = vi

2 + 2advf

2 = 0 + 2(1.20 m/s2)(500. m)vf = 34.6 m/s

vf = vi + at34.6 m/s = 0 + (1.20 m/s2) t

t = 28.8 s

Example• A driver traveling at 85. miles per hour sees a

police car hiding in the trees 2.00 miles ahead. He applies his brakes, decelerating at -500. miles per hour2.

– If the speed limit is 55 mph, will he get a ticket?

– What would his acceleration need to be to not get a ticket?

vf2 = vi

2 + 2advf

2 = (85. mph)2 + 2(-500. mph2)(2.00 mi)vf = 72.3 mph *YES*

vf2 = vi

2 + 2ad(55 mph)2 = (85. mph)2 + 2a(2.00 mi)

a = -1050 mph2

Objective• Apply kinematics equations to calculate

distance, time, or velocity under conditions of constant acceleration.

example• A barge moving with a speed of 1.00 m/s increases

speed uniformly, so that in 30.0 s it has traveled 60.2 m. What is the magnitude of the barge’s acceleration?

a 6.71 10−2 m/s2

example• A person pushing a stroller starts from rest, uniformly

accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

vf = +2.18 m/s

example• An aircraft has a landing speed of 302 km/h. The landing

area of an aircraft carrier is 195 m long. What is the minimum uniform accelerating required for a safe landing?

a = -18 m/s2

example• A plane starting at rest at one end of a runway

undergoes a uniform acceleration of 4.8 m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off?

vf = 72 m/s

∆x = 540 m

Objectives

1. Relate the motion of a freely falling body to motion with constant acceleration.

2. Calculate displacement, velocity, and time at various points in the motion of a freely falling object.

3. Compare the motion of different objects in free fall.

Two important motion characteristics that are true of free-falling objects

• Free-falling objects do not encounter air resistance.

• All free-falling objects (on Earth) accelerate downwards at a constant rate.

“g” - The “Magic” Number

• “g” means ACCELERATIONACCELERATION DUE TO GRAVITY

• Each planet, star, moon, or other large object has its own value for “g”

Examples • “g” is 1.62 m/s2 on the Moon• “g” is 26 m/s2 on Jupiter• “g” is 9.81 m/s2 on Earth

If the feather and elephant experiment were performed on the moon, would they still fall at the same rate, just like on Earth?

• Dropped from rest

• Throwing downward

• Throwing upward

• Throwing side ways

a = -g = -9.81m/s2 for all free fall objects.

(the ‘-’ sign means “downward”)

• Since free fall motion has constant acceleration, we can apply all kinematics equations on free fall motion.

Free fall – motion with constant acceleration a = -g

To solve free fall motion problems, we can use kinematics equations with constant acceleration.

a = -g∆x = ∆y

Velocity of a dropped object• An object falls from rest. What is its velocity at the end

of one second? Two seconds? Three seconds?

t = 0 vi = 0 m/s

t = 1 s v = -9.81 m/s

t = 2 s v = -19.62 m/s

t = 3 s v = -29.43 m/s

atvv if a = -g = -9.81 m/s2 (‘-’ means “downward”)

Displacement of a dropped object• An object falls from rest. How far has it

fallen at the end of one second? Two seconds? Three seconds?

t = 0 d = 0 m

t = 1 s d = -4.905 m

t = 2 s d = -19.62 m

t = 3 s d = -44.145 m

2

2

1attvd i

+

Velocity and distance of a free falling object dropped from rest

Throwing Downward• An object is thrown downward from

the top of a 175 meter building with an initial speed of 10 m/s.

– What acceleration does it experience?

• -9.81 m/s2

– What is the object’s initial velocity?• -10 m/s

vi = -10 m/sa = -9.81 m/s2

“a” and “vi” in SAME DIRECTION

Building Example (cont.)

• What is the object’s velocity as it hits the ground?

• vi = -10 m/s

• d = -175 m• a = -9.81 m/s2

• vf = ?

• How long does it take the object to hit the ground?

• t = ?

vf2 = vi

2 + 2advf

2 = 0 + 2(-9.81 m/s2)(-175 m)vf = -58.6 m/s

vf = vi + at-58.6 m/s = -10 m/s + (-9.81 m/s2) t

t = 4.95 s

Throwing Upward

• A cannon fires a shot directly upward with an initial velocity of 50 m/s.

– What acceleration does the cannonball experience?

• -9.81 m/s2

– What is the cannonball’s initial velocity?

• +50 m/s

t = 0 svi = +50 m/s

t = 1 sv = +40.19 m/s

t = 2 sv = +30.38 m/s

“a” and “vi” in OPPOSITE DIRECTION

atvv if

Cannon (cont.)• What is the object’s velocity as it reaches the

top of its flight?

• How long does it take the cannonball to reach the top of its flight?

• What is the maximum height of the cannonball?

0 m/s (All objects momentarily STOP at the top of their flight)

vf = vi + at0 = +50 m/s + (-9.81 m/s2) t

t = 5.1 s

d = vit + ½ at2

d = (+50 m/s)(5.1 s) + ½ (-9.81 m/s2)(5.1 s)2

d = +127.4 m

Question

• If a ball thrown into the air, and caught at the same point you it was released. Is the speed of the ball equal to, more than, or less than the initial speed when it is caught?

• To accelerate at -10 m/s/s means to change the velocity by -10 m/s each second.

Time (s) Velocity (m/s)

0 20

1 10

2 0

3 -10

4 -20

Acceleration is the rate at which an object changes its velocity.

• If the velocity and time for a free-falling object being tossed upward from a position with speed of 20 m/s were tabulated, then one would note the pattern.

Question • If a ball thrown into the air, and caught at the

same point you it was released. Is the speed of the ball equal to, more than, or less than the initial speed when it is caught?

vi vf

given: vi, a = -9.81 m/s2; ∆y = 0 m;

Find: vf vf2 = vi

2 + 2a∆y

vf2 = vi

2 + 0

vf2 = vi

2

The ball has the same velocity, but in the opposite direction

Class work• A ball is thrown straight up into the air at an initial

velocity of 39.24 m/s. Fill in the table showing the ball’s position, velocity, and acceleration each second for the first 4.00 s of its motion.

t (s) Y (m) v (m/s) a (m/s2)

0 0 19.62 -9.81

1

2

3

4

Free Fall by Graphs of a dropped object

A position versus time graph for a free-falling object

A velocity versus time graph for a free-falling object

velocity increase in the negative direction – slope is neg. & increasing

velocity increases in the negative direction at constant rate – slope is neg. & constant

time

velo

city Upward: velocity is big, positive, decreasing,

slope is constant (a = -9.8 m/s/s).

Top, velocity is zero. Slope remains the same (acceleration is still -9.8 m/s/s)

Downward: velocity increases in negative direction at the same constant rate of -9.81 m/s/s (slope remains the same), reaches the same speed as it started upward.

Free Fall Graphs of upward object

time

pos

itio

n Upward: displacement increases, slope is positive, decreasing (velocity is positive, decreasing)

Top: slope = 0 (its velocity is zero)

Downward: displacement decreases, its slope increases in negative direction (velocity is negative, increasing)