PHYSICAL SCIENCE PSCI 1CHAPTER 2: MOTION

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Intro

Physics: The Most Fundamental Physical Science

Physics is concerned with the basic principles that describe how nature works.Physics deals with matter, motion, force, and energy.

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Intro

Physics – Areas of StudyClassical mechanics Waves and soundsThermodynamicsElectromagnetismOpticsRelativityQuantum mechanicsAtomic and nuclear physics

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Intro

MotionThis chapter focuses on definition/discussion of: speed, velocity, and acceleration.There are two basic kinds of motion:

Straight line Circular (more generally, curvilinear)

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Section 2.1

Defining MotionPosition – the location of an object

A reference point and a measurement scale must be given in order to define the position of an objectMotion – an object is undergoing a continuous change in positionDescription of Motion – the time rate of change of position

A combination of length and time describes motion

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Speed and VelocityIn Physical Science ’speed’ and ’velocity’ have different (distinct) meanings.Speed is a scalar quantity, with only magnitude

A car going 80 km/hVelocity is a vector, and has both magnitude & direction

A car going 80 km/h north

�7Section 2.2

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40 km/h 80 km/h

Section 2.2

Vectors

Vector quantities may be represented by arrows. The length of the arrow is proportional to magnitude.

Vectors

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Note that for motion along one axis only, vectors may be expressed as positive or negative numbers to represent direction.

Section 2.2

Speed

Distance: the actual path length traveledInstantaneous Speed: v, the speed of an object at an instant of time (Δt is very small)

Glance at a speedometer�10

Section 2.2

v = dt

Average Speed v = ΔdΔt

= d − d0t − t0

Where:d = distance, t = time, v = speedGreek ∆ means “change in”∆d means change in distance∆t means change in time

Instantaneous speed

�11Section 2.2

Velocity

Velocity is similar to speed except a direction is involved.

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Displacement – straight line distance between the initial and final position with direction toward the final position !Instantaneous velocity – similar to instantaneous speed except it has direction

Section 2.2

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

Section 2.2

Displacement and Distance

Constant Velocity - Example

Describe the speed of the car above.GIVEN: d = 80 m & t = 4.0 s

�14Section 2.2

v = dt= 80 m

4.0 s= 40 m

s= 40 m/s

Constant Velocity – Confidence Exercise

�15Section 2.2

How far would the car above travel in 10s?Equation: v = d/tRearrange equation: vt = dSolve: (20 m/s) (10 s) = 200 m

Example: How long does it take sunlight to reach 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

�16Section 2.2

t = dv= 1.5 ×1011m3.00 ×108m/s

t = 0.5 ×103s = 5.00 ×102s = 500s

Sunlight

�17Section 2.2

What is the average speed in mi/h of the Earth revolving around the Sun?

�18Section 2.2

GIVEN: t = 365 days (must convert to hours) Earth’s radius, r = 9.30 x 107 milesCONVERSION: t = 365 days x (24h/day) = 8760 hMUST CALCULATE DISTANCE (d):

Recall that a circle’s circumference = 2πr d = 2πr (and we know π and r ) Therefore: d = 2πr = 2 (3.14) (9.30 x 107 mi)

What is the average speed in mi/h of the Earth revolving around the Sun?

SOLVE EQUATION:

�19Section 2.2

v = dt

v =2 3.14( ) 9.3×107 mi( )

8,760 h= 6.67 ×104 mi/h

Or: v = 66,700 mi/h

Confidence Exercise: What is the average speed in mi/h of a person at the equator as a result of the Earth’s rotation?

Radius of the Earth = 4000 miTime = 24 h (actually, 23 h, 56 min, 4 s)Diameter of Earth = 2πr = 2 (3.14) (4000mi)

Diameter of Earth = 25,120 miv = d/t = (25,120 mi)/24h = 1,047 mi/h

�20Section 2.2

AccelerationChanges in velocity occur in three ways: Change in speed Change in directionWhen either or both of these changes occur, the object is accelerating.Faster the change -> Greater the accelerationAcceleration: the time rate of change of velocity

�21Section 2.3

AccelerationA measure of the change in velocity during a given time period

�22Section 2.3

a = ΔvΔt

= v − v0t − t0

The units of acceleration are:

velocitytime

= m/ss

= m/s2

Constant Acceleration of 9.8 m/s2

As the velocity increases, the distance traveled by the falling object increases each second.

�23Section 2.3

Finding Acceleration - ExampleA race car starting from rest accelerates uniformly along a straight track, reaching a speed of 90 km/h in 7.0 s. Find its average acceleration.GIVEN: v0 = 0, vf = 90 km/h, t0 = 0, tf = 7.0sWANTED: average acceleration, a, in m/s2

�24Section 2.3

a = ΔvΔt

=vf − v0

t f − t0= 90 km/h

7.0 s

a = 90 km7.0 s h

103 m1 km

⎛⎝⎜

⎞⎠⎟

1 h3600 s

⎛⎝⎜

⎞⎠⎟ = 3.6 m/s2

Using the equation for acceleration

This last equation is very useful in computing final velocity.

�25Section 2.3

a =vf − v0Δt

a Δt( ) = vf − v0vf = a Δt( )+ v0

vf = at + v0You could also write

if it’s clear to you what t represents.

Finding Acceleration – Confidence Exercise

If the car in the preceding example continues to accelerate at the same rate for three more seconds, what will be the magnitude of its velocity in m/s at the end of this time (vf )?a = 3.57 m/s2 (found in preceding example)t = 10 s v0 = 0 (started from rest)Use equation: vf = v0 + atvf = 0 + (3.57 m/s2) (10 s) = 35.7 m/s

�26Section 2.3

Acceleration is a vector quantity, since velocity is a vector quantity

�27Section 2.3

Acceleration (+) Deceleration (-)

Constant Acceleration = Gravity = 9.8 m/s2

Special case associated with falling objectsVector towards the center of the earthDenoted by “g”g = 9.80 m/s2

�28Section 2.3

Graph – Acceleration due to Gravity

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The increase in velocity is linear because the acceleration is constant.

Section 2.3

Acceleration due to GravityVe

loci

ty, m

/s

10

20

30

40

Time, s0 1 2 3 4

Frictional (Dissipative) EffectsIf frictional effects (air resistance) are neglected, every freely falling object on earth accelerates at the same rate, regardless of mass. Galileo is credited with this idea/experiment.

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Astronaut David Scott demonstrated the principle on the moon, simultaneously

dropping a feather and a hammer. Each fell at the same acceleration, due to no

atmosphere & no air resistance.Section 2.3

Figure 2Galileo is also credited with using the Leaning Tower of Pisa as an experiment site.

�31Section 2.3

What about the distance a dropped object will travel in a given time?

�32Section 2.3

vf = at + v0vf = gt

v =vf + v0( )2

=vf2= 12gt

Recall:

For v0 = 0, a = g:

Average velocity:

d = vt

d = 12gt × t = 1

2gt 2

Distance fallenin time t.

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

What about the distance a dropped object will travel in a given time?

d = 12gt 2

m = ms2

⎛⎝⎜

⎞⎠⎟ s2( )Units check:

Solving for distance dropped - Example

An object is dropped from a tall building. How far does the object drop in 1.50 s?GIVEN: g = 9.80 m/s2, t = 1.5 sSOLVE:

d = ½ gt2 = ½ (9.80 m/s2) (1.5 s)2 = ½ (9.80 m/s2) (2.25 s2) = ?? m

�34Section 2.3

Solving for distance dropped - Example

An object is dropped from a tall building. How far does the object drop in 1.50 s?GIVEN: g = 9.80 m/s2, t = 1.5 sSOLVE:

d = ½ gt2 = ½ (9.80 m/s2) (1.5 s)2 = ½ (9.80 m/s2) (2.25 s2) =

�35Section 2.3

11.0 m

Solving for final speed – Confidence Exercise

What is the speed of the object in the previous example 1.50 s after it is dropped?GIVEN: g = a = 9.80 m/s2, t = 1.5 sSOLVE:vf = v0 + at = 0 + (9.80 m/s2)(1.5 s)Object’s speed after 1.5 seconds = 14.7 m/s

�36Section 2.3

Constant Acceleration = Gravity 9.8 m/s2

Distance is proportional to t2 Remember that (d = ½ gt2)Velocity is proportional to t Remember that (vf = gt )

�37Section 2.3

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

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Velo

city,

m/s

; Dis

tanc

e, m

0

20

40

60

80

Time, s0 1 2 3 4

velocity distance

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

Acceleration due to gravity occurs in BOTH directions.

Going up (-)Coming down (+)

The ball returns to its starting point with the same speed it had initially. v0 = vf

�39Section 2.3

Acceleration In Uniform Circular Motion

Although an object in uniform circular motion has a constant speed, it is constantly changing directions and therefore its velocity is constantly changing.

Since there is a change in direction there is an acceleration. What is the direction of this acceleration?It is at right angles to the velocity, and points toward the center of the circle.

�40Section 2.4

Centripetal (“center-seeking”) Acceleration

Supplied by friction of the tires of a carThe car remains in a circular path as long as there is enough centripetal acceleration.

�41Section 2.4

Centripetal Acceleration

This equation holds true for an object moving in a circle with radius (r) and constant speed (v).From the equation we see that centripetal acceleration increases as the square of the speed.We also can see that as the radius decreases, the centripetal acceleration increases.

�42Section 2.4

ac =v2

r

Finding Centripetal Acceleration Example

Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular track with a radius of 50 m.GIVEN: v = 12 m/s, r = 50 mSOLVE:

�43Section 2.4

ac =v2

r=

12 m/s( )2

50 m= 2.9 m/s2

Finding Centripetal Acceleration Confidence Exercise

Compute the centripetal acceleration in m/s2 of the Earth in its nearly circular orbit about the Sun.GIVEN: r = 1.5 x 1011 m, v = 3.0 x 104 m/sSOLVE:

�44Section 2.4

ac =3.0 ×104 m/s( )2

1.5 ×1011 m= 9.0 ×108 m2 /s2

1.5 ×1011

ac = 6.0 ×10−3 m/s2

Projectile MotionAn object thrown horizontally combines both straight-line and vertical motion each of which act independently.

(Think of space as providing 3 independent degrees of freedom.)

Neglecting air resistance, a horizontally projected object travels in a horizontal direction with a constant velocity while falling vertically due to gravity.

�45Section 2.5

Figure 2.14(b)

An object thrown horizontally will fall at the same rate as an object that is dropped.

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Multi-flash photograph of two objectsSection 2.5

The velocity in the horizontal direction does not affect the velocity and acceleration in the vertical direction.

�47Section 2.5

Projected at an angle (not horizontal)

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Combined Horz./Vert. Components

Vertical Component

Horizontal Component+=

Section 2.5

In throwing a football the horizontal velocity remains constant but the vertical velocity changes like that of an object thrown upward.

�49Section 2.5

If air resistance is neglected, projectiles have symmetric paths and the maximum range is attained at 45 degrees.

�50Section 2.5

Under real-world conditions, air resistance causes the paths to be non-symmetric. Air resistance reduces the horizontal velocity.

�51Section 2.5

Projectiles – Athletic considerations

Angle of releaseSpin on the ballSize/shape of ball/projectileSpeed of windDirection of windWeather conditions (humidity, rain, snow, etc)Field altitude (how much air is present)Initial horizontal velocity

�52Section 2.5

Important Equations – Chapter 2

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

�53Review

v = dt

v = ΔdΔt

= d − d0t − t0

a = ΔvΔt

= v − v0t − t0

vf = at + v0

d = 12gt 2

ac =v2

r

Average speed:

Acceleration:

Final velocity:

Distance fallen:

Centripetal acceleration:

a = g fof the case of gravity

Defin

ition

sDe

rived