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Martha Casquete

Would you risk your life driving drunk?

Intro

Assignments: For next class: Finish reading Ch. 2, read Chapter 3

(Vectors) HW3 Set due next Wednesday, 9/11 HW3 will be in weebly.

Question/Observation Mondays

Research Q/O Wednesday with HW (due date

Wednesday)

Motion

Speed and Velocity

Acceleration

Acceleration in Uniform Circular Motion

Projectile Motion

What today’s lesson has to do with drunk drivers?

Units: Length, Mass, and Time

Order of magnitude calculations

Dimensional Analysis

Conversion of Units

Significant digits (on your own)

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

This chapter focuses on definition/discussion of: speed, velocity, and acceleration.

There are two basic kinds of motion: ◦ Straight line

◦ Circular

Section 2.1

Position – the location of an object ◦ A reference point must be given in order to define

the position of an object

Motion – an object is undergoing a continuous change in position

Description of Motion – the time rate of change of position ◦ A combination of length and time describes motion

Section 2.1

Any motion involves three concepts ◦ Displacement/distance ◦ Velocity/speed ◦ Acceleration

Displacement is a change of position in time.

It is a vector quantity.

It has both magnitude and direction: + or - sign

It has units of [length]: meters.

x1 (t1) = + 2.5 m x2 (t2) = - 2.0 m Δx = -2.0 m - 2.5 m = -4.5 m

x1 (t1) = - 3.0 m x2 (t2) = + 1.0 m Δx = +1.0 m + 3.0 m = +4.0 m

Sometimes displacement is confused with distance. It is important that you understand the difference between distance and displacement. Distance – the actual path length traveled

If I start walking when x1 = - 3.0m and then I return to the same place, then x2 = - 3.0m. a) What is my displacement? b) What is my distance?

a) What is my displacement?

b) What is my distance?

x1 = - 3.0 m x2 = - 3.0 m Δx = -3.0 m – (- 3.0 m) = 0 m

x1 = 3.0 m x2 = 3.0 m

x1 + x2 = 3.0 m + 3.0 m = 6.0m

Every day, I travel a distance of 108 miles from Brownsville/Rancho Viejo to Edinburg and back.

Every day, at 8:20pm I noticed that I have displaced myself zero miles. Where am I?

What is the distance between one side of the wall and the other in the classroom?

If I pace it, what will be my displacement?

Displacement is not Distance.

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

Speed – a scalar quantity, only magnitude (numerical value and unit measurement)/scalar ◦ A car going 80 km/h

Velocity – vector, both magnitude & direction ◦ A car going 80 km/h north

Section 2.2

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

40 km/h 80 km/h

-40 km/h -80 km/h

Section 2.2

Note that vectors may be both positive and negative.

Section 2.2

◦ ( where D means ‘change in’) ◦ Over the entire time interval, speed is an average

Instantaneous Speed – the speed of an object at an instant of time (Dt is very small) ◦ Glance at a speedometer

Section 2.2

distance traveled

time to travel distance • Average Speed =

• v = d/t or v = Dd/Dt

Section 2.2

Velocity is similar to speed except a direction is involved.

Instantaneous velocity – similar to instantaneous speed except it has direction

Section 2.2

displacement

total travel time • Average velocity =

What is the speed of a car if it has traveled 100m in 4.0 s.

GIVEN: d = 1000 m & t = 4.0 s

SOLVE: d/t = 100 m/4.0 s = 25 m/s = average speed

Velocity would be described as 25 m/s in the direction of motion (east?)

Section 2.2

• EQUATION: v = d/t

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

Section 2.2

(it is on your book – do the math)

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

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

Radius of the Earth = RE = 4000 mi

Time = 24 h

Circumference of Earth = 2pr = 2 (3.14) (4000mi) ◦ Circumference of Earth = 25,120 mi

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

Section 2.2

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, the object is accelerating.

Faster the change Greater the acceleration Acceleration – the time rate of change of

velocity

Section 2.3

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

Section 2.3

• Avg. acceleration = change in velocity time for change to

occur

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

• In this course we will limit ourselves to

situations with constant acceleration.

• a = = Dv

t

vf – vo

t (vf = final & vo = original)

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

Section 2.3

2-28

Rearrange this equation:

at = vf – vo

vf = vo + at (solved for final velocity)

This equation is very useful in computing final velocity.

Section 2.3

vf – vo

t • Remember that a =

If a car with an acceleration of 3.57 m/s2

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

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

Section 2.3

Acceleration (+) Deceleration (-)

Special case associated with falling objects

Vector towards the center of the earth

Denoted by “g” g = 9.80 m/s2

Section 2.3

Copyright © Houghton Mifflin Company. All rights reserved. 2-32

If 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.

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

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

Section 2.3

2-34

d = ½ gt2

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

Section 2.3

A ball is dropped from the top of an 80 m building. Does it reach the ground in 4.0s?

GIVEN: g = 9.80 m/s2, t = 4.0 s

SOLVE:

d = ½ gt2 = ½ (9.80 m/s2) (4.0 s)2

= ½ (9.80 m/s2) (16.00 s2) = ?? m

Section 2.3

d = 78 m

What is the speed of the ball in the previous example 4 s after it is dropped?

Use equation: vf = vo + at vo = 0

GIVEN: g = a = 9.80 m/s2, t = 4 s

SOLVE:

vf = at = (9.80 m/s2)(4 s)

Speed of ball after 4 seconds = 39.2 m/s

Section 2.3

Distance is proportional to t2 ◦ (d = ½ gt2)

Velocity is proportional to t ◦ (vf = at )

Section 2.3

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. vo = vf

Section 2.3

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 generally points toward the center of the circle.

Section 2.4

Supplied by friction of the tires of a car

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

Section 2.4

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.

Section 2.4

v2

r • ac =

Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular cloverleaf with a radius of 50 m.

Section 2.4

Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular cloverleaf with a radius of 50 m.

GIVEN: v = 12 m/s, r = 50 m

SOLVE:

Section 2.4

• ac = = = 2.9 m/s2 v2

r

(12 m/s)2

50m

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/s

SOLVE:

Section 2.4

• ac = = = v2

r

(3.0 x 104 m/s)2

1.5 x 1011 m 9.0 x 108 m2/s2

1.5 x 1011 m

• ac = 6 x 10-3 m/s2

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

Multiflash photograph of two balls Section 2.5

2-46 Section 2.5 Section 2.5

2-47

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

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

Section 2.5

Combined Horz/Vert.

Components

Vertical

Component

Horizontal

Component + =

Section 2.5

2-49 Section 2.5

Copyright © Houghton Mifflin Company. All rights reserved. 2-50

Section 2.5

2-51 Section 2.5

2-52

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

it out of the park before gravity brings it down, it must leave the bat at a high velocity)

Section 2.5

2-53

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

Dv

t

vf –vo

t • a = = (constant acceleration)

• v = d/t (average speed)