MotionChapter 2: Motion

Homework: All questions on the “Multiple-Choice” and the odd-numbered questions on“Exercises” sections at the end of the chapter.

Chapter 2- Motion

Physics:The Most Fundamental Physical Science

• Physics is concerned with the basicprinciples that describe how theuniverse works.

• Physics deals with matter, motion,force, and energy.

Intro

Chapter 2- Motion

Physics – Areas of Study

• Classical mechanics

• Waves and sounds

• Thermodynamics

• Electromagnetism

• Quantum mechanics

• Atomic and nuclear physics

• Relativity

Intro

Chapter 2- Motion

Motion

• Motion is everywhere – walking, driving,flying, etc.

• This chapter focuses ondefinition/discussion of: speed, velocity,and acceleration.

• There are two basic kinds of motion:

– Straight line

– Circular

Section 2.1

Chapter 2- Motion

Defining Motion

• Position – the location of an object

– A reference point must be given in order todefine the position of an object

• Motion – an object is undergoing acontinuous change in position

• Description of Motion – the time rate ofchange of position

– A combination of length and time describesmotion

Section 2.1

Chapter 2- Motion

Speed and Velocity

• In Physical Science ‘speed’ and‘velocity’ have different (distinct)meanings.

• Speed – a scalar quantity, onlymagnitude

– A car going 80 km/h

• Velocity – vector, both magnitude &direction

– A car going 80 km/h north

Section 2.2

Chapter 2- Motion

Vectors

• Vector quantities may be representedby arrows. The length of the arrow isproportional to magnitude.

40 km/h 80 km/h

-40 km/h -80 km/h

Section 2.2

Chapter 2- Motion

Vectors

Note that vectors may be both positive and negative.

Section 2.2

Chapter 2- Motion

Speed

– ( where D means ‘change in’)

– Over the entire time interval, speed is anaverage

• Distance – the actual path length traveled

• Instantaneous Speed – the speed of anobject at an instant of time (Dt is verysmall)– Glance at a speedometer

Section 2.2

distance traveledtime to travel distance

• Average Speed =

• v = d/t or v = Dd/Dt

Chapter 2- Motion

Instantaneous speed

Section 2.2

Chapter 2- Motion

Velocity

• Velocity is similar to speed except a direction isinvolved.

• Displacement = straight line distancebetween the initial and final position w/direction toward the final position

• Instantaneous velocity – similar toinstantaneous speed except it has direction

Section 2.2

displacementtotal travel time

• Average velocity =

Chapter 2- Motion

Displacement and Distance

Displacement is a vector quantity between two points.Distance is the actual path traveled.

Section 2.2

Chapter 2- Motion

Constant Velocity - Example

• Describe the speed of the car above.

• GIVEN: d = 80 m & t = 4.0 s

• SOLVE: d/t = 80 m/4.0 s = 20 m/s = averagespeed

• Velocity would be described as 20 m/s in thedirection of motion (east?)

Section 2.2

• EQUATION: v = d/t

Chapter 2- Motion

Constant Velocity - Confidence Exercise

• How far would the car above travel in 10s?

• EQUATION: v = d/t

• REARRANGE EQUATION: vt = d

• SOLVE: (20 m/s) (10 s) = 200 m

Section 2.2

Chapter 2- Motion

Example: How long does it take sunlight toreach earth?

• GIVEN:

– Speed of light = 3.00 x 108 m/s = (v)

– Distance to earth = 1.50 x 108 km = (d)

• EQUATION:

– v = d/t rearrange to solve for t t = d/v

• t = 0.500 x 103 s = 5.00 x 102 s = 500 seconds

Section 2.2

• SOLVE: t = d/v =1.50 x 1011 m3.00 x 108 m/s

Chapter 2- Motion

Sunlight

Section 2.2

Chapter 2- Motion

What is the average speed in mi/h of theearth revolving around the sun?

• GIVEN:

– t = 365 days (must convert to hours)

– Earth’s radius (r) = 9.30 x 107 miles (p. 19)

• CONVERSION:

– t = 365 days x (24h/day) = 8760 h

• MUST FIGURE DISTANCE (d):

• d = ? Recall that a circle’s circumference = 2pr

• d=2pr (and we know p and r!!)

• Therefore d = 2pr = 2 (3.14) (9.30 x 107 mi)

Section 2.2

Chapter 2- Motion

What is the average speed in mi/h of theearth revolving around the sun?

• SOLVE EQUATION:

• ADJUST DECIMAL POINT:

Section 2.2

• d/t = 2 (3.14) (9.30 x 107 mi) = 0.00667 x 107 mi/h

8760 h

• v = d/t

• v = avg. velocity = 6.67 x 104 mi/h = 66,700 mi/h

Chapter 2- Motion

Confidence Exercise:What is the average speed in mi/h of aperson at the equator as a result of the

Earth’s rotation?

• Radius of the Earth = RE = 4000 mi

• Time = 24 h

• Diameter of Earth = 2pr = 2 (3.14)(4000mi)

– Diameter of Earth = 25,120 mi

• v = d/t = (25,120 mi)/24h = 1047 mi/h

Section 2.2

Chapter 2- Motion

Acceleration

• Changes in velocity occur in three ways:– Increase in magnitude (speed up)

– Decrease in magnitude (slow down)

– Change direction of velocity vector (turn)

• When any of these changes occur, theobject is accelerating.

• Faster the change Greater theacceleration

• Acceleration – the time rate of changeof velocity

Section 2.3

Chapter 2- Motion

Acceleration

• A measure of the change in velocityduring a given time period

Section 2.3

• Avg. acceleration = change in velocitytime for change to

occur

• Units of acceleration = (m/s)/s = m/s2

• In this course we will limit ourselves tosituations with constant acceleration.

• a = =Dvt

vf – vo

t(vf = final & vo = original)

Chapter 2- Motion

Constant Acceleration of 9.8 m/s2

• As the velocityincreases, thedistancetraveled by thefalling objectincreases eachsecond.

Section 2.3

Chapter 2- Motion

Finding Acceleration -- Example

• A race car starting from rest acceleratesuniformly along a straight track,reaching a speed of 90 km/h in 7.0 s.Find acceleration.

• GIVEN: vo = 0, vf = 90 km/h, t=7.0s

• WANTED: a in m/s2

Section 2.3

• vf = 90 km/h = 25 m/s)( //278.0

hkmsm

• a = = = 3.57 m/s2vf –vo

t25 m/s – 0

7.0 s

Chapter 2- Motion

Using the equation for acceleration

• Rearrange this equation:

• at = vf – vo

• vf = vo + at (solved for final velocity)

• This equation is very useful in computingfinal velocity.

Section 2.3

vf – vo

t• Remember that a =

Chapter 2- Motion

Finding Acceleration – Confidence Exercise

• If the car in the preceding example continues toaccelerate at the same rate for three moreseconds, what will be the magnitude of itsvelocity in m/s at the end of this time (vf )?

• a = 3.57 m/s2 (found in preceding example)

• t = 10 s

• vo = 0 (started from rest)

• Use equation: vf = vo + at

• vf = 0 + (3.57 m/s2) (10 s) = 35.7 m/s

Section 2.3

Chapter 2- Motion

Acceleration is a vector quantity, sincevelocity is a vector quantity

Section 2.3

Acceleration (+) Deceleration (-)

Chapter 2- Motion

Constant Acceleration = Gravity = 9.8 m/s2

• Special caseassociated withfalling objects

• Vector towardsthe center of theearth

• Denoted by “g”• g = 9.80 m/s2

Section 2.3

Chapter 2- Motion

Graph – Acceleration due to Gravity

0

10

20

30

40

50

60

70

0 2 4 6 8

Time in seconds

Velocity

The increase in velocity is linear.

Section 2.3

Chapter 2- Motion

Frictional Effects

• If frictional effects (air resistance) areneglected, every freely falling object on earthaccelerates at the same rate, regardless ofmass. Galileo is credited with thisidea/experiment.

• Astronaut David Scottdemonstrated the principle on themoon, simultaneously dropping afeather and a hammer. Each fell atthe same acceleration, due to noatmosphere & no air resistance.

Section 2.3

Chapter 2- Motion

Figure 2.1

• Galileo is alsocredited withusing the LeaningTower of Pisa asan experimentsite.

Section 2.3

Chapter 2- Motion

What about the distance adropped object will travel?

• d = ½ gt2

• This equation will compute the distance(d) an object drops due to gravity(neglecting air resistance) in a giventime (t).

Section 2.3

Chapter 2- Motion

Solving for distance dropped - Example

• A ball is dropped from a tall building.How far does the ball drop in 1.50 s?

• GIVEN: g = 9.80 m/s2, t = 1.5 s

• SOLVE:

d = ½ gt2 = ½ (9.80 m/s2) (1.5 s)2

= ½ (9.80 m/s2) (2.25 s2) = ?? m

Section 2.3

Chapter 2- Motion

Solving for distance dropped - Example

• A ball is dropped from a tall building.How far does the ball drop in 1.50 s?

• GIVEN: g = 9.80 m/s2, t = 1.5 s

• SOLVE:

d = ½ gt2 = ½ (9.80 m/s2) (1.5 s)2

= ½ (9.80 m/s2) (2.25 s2) =

Section 2.3

11.0 m

Chapter 2- Motion

Solving for final speed –Confidence Exercise

• What is the speed of the ball in theprevious example 1.50 s after it isdropped?

• GIVEN: g = a = 9.80 m/s2, t = 1.5 s

• SOLVE:

• vf = at = (9.80 m/s2)(1.5 s)

• Speed of ball after 1.5 seconds = 14.7m/s

Section 2.3

Chapter 2- Motion

Constant Acceleration = Gravity 9.8 m/s2

• Distance isproportional to t2

– (d = ½ gt2)

• Velocity isproportional to t– (vf = at )

Section 2.3

Chapter 2- Motion

v proportional to time (t)d proportional to time squared (t2)

0

50

100

150

200

0 2 4 6 8

Velocity

Distance

Time in seconds

Section 2.3

Chapter 2- Motion

Up and Down – Gravity slows the ball,then speeds it up

• Acceleration dueto gravity occursin BOTHdirections.– Going up (-)

– Coming down (+)

• The ball returnsto its startingpoint with thesame speed ithad initially. vo =vf

Section 2.3

Chapter 2- Motion

Acceleration In Uniform Circular Motion

• Although an object in uniform circular motionhas a constant speed, it is constantlychanging directions and therefore its velocityis constantly changing directions.

– Since there is a change in direction there is achange in acceleration.

• What is the direction of this acceleration?

• It is at right angles to the velocity, andgenerally points toward the center of thecircle.

Section 2.4

Chapter 2- Motion

Centripetal (“center-seeking”) Acceleration

• Supplied by friction of the tires of a car

• The car remains in a circular path as long asthere is enough centripetal acceleration.

Section 2.4

Chapter 2- Motion

Centripetal Acceleration

• This equation holds true for an object movingin a circle with radius (r) and constant speed(v).

• From the equation we see that centripetalacceleration increases as the square of thespeed.

• We also can see that as the radiusdecreases, the centripetal accelerationincreases.

Section 2.4

v2

r• ac =

Chapter 2- Motion

Finding Centripetal AccelerationExample

• Determine the magnitude of thecentripetal acceleration of a car going12 m/s on a circular cloverleaf with aradius of 50 m.

Section 2.4

Chapter 2- Motion

• Determine the magnitude of the centripetalacceleration of a car going 12 m/s on a circularcloverleaf with a radius of 50 m.

• GIVEN: v = 12 m/s, r = 50 m

• SOLVE:

Finding Centripetal AccelerationExample

Section 2.4

• ac = = = 2.9 m/s2v2

r(12 m/s)2

50m

Chapter 2- Motion

Finding Centripetal AccelerationConfidence Exercise

• Compute the centripetal acceleration inm/s2 of the Earth in its nearly circularorbit about the Sun.

• GIVEN: r = 1.5 x 1011 m, v = 3.0 x 104

m/s

• SOLVE:

Section 2.4

• ac = = =v2

r(3.0 x 104 m/s)2

1.5 x 1011 m9.0 x 108 m2/s2

1.5 x 1011 m

• ac = 6 x 10-3 m/s2

Chapter 2- Motion

Figure 2.14

• An object thrownhorizontally willfall at the samerate as an objectthat is dropped.

Multiflash photograph of two ballsSection 2.5

Chapter 2- Motion

The velocity in the horizontal direction doesnot affect the velocity and acceleration in

the vertical direction.

Section 2.5Section 2.5

Chapter 2- Motion

Projectile Motion

• An object thrown horizontally combinesboth straight-line and vertical motioneach of which act independently.

• Neglecting air resistance, a horizontallyprojected object travels in a horizontaldirection with a constant velocity whilefalling vertically due to gravity.

Section 2.5

Chapter 2- Motion

Projected at an angle (not horizontal)

Combined Horz/Vert.Components

VerticalComponent

HorizontalComponent

+=

Section 2.5

Chapter 2- Motion

In throwing a football the horizontal velocityremains constant but the vertical velocity

changes like that of an object thrownupward.

Section 2.5

Chapter 2- Motion

If air resistance is neglected, projectileshave symmetric paths and the maximum

range is attained at 45o .

Section 2.5

Chapter 2- Motion

Under real-world conditions, air resistancecauses the paths to be non-symmetric. Airresistance reduces the horizontal velocity.

Section 2.5

Chapter 2- Motion

Projectiles - Athletic considerations

• Angle of release

• Spin on the ball

• Size/shape of ball/projectile

• Speed of wind

• Direction of wind

• Weather conditions (humidity, rain, snow, etc)

• Field altitude (how much air is present)

• Initial horizontal velocity (in order to make itout of the park before gravity brings it down, itmust leave the bat at a high velocity)

Section 2.5

Chapter 2- Motion

Important Equations – Chapter 2

• d = ½at2 (distance traveled, starting from rest)

• d = ½gt2 (distance traveled, dropped object)

• g = 9.80 m/s2 = 32 ft/s2 (acceleration, gravity)

• vf = vo + at (final velocity with constant a)

• ac = v2/r (centripetal acceleration)

Review

Dvt

vf –vo

t• a = = (constant acceleration)

• v = d/t (average speed)