Chapter 2 Motion Along a Line. MFMcGraw- PHY 1410Ch_02b-Revised 5/31/20102 Motion Along a Line Position & Displacement Speed & Velocity Acceleration Describing.

Chapter 2

Motion Along a Line

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Motion Along a Line

• Position & Displacement

• Speed & Velocity

• Acceleration

• Describing motion in 1D

• Free Fall

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Introduction

• Kinematics - Concepts needed to describe motion - displacement, velocity & acceleration.

• Dynamics - Deals with the effect of forces on motion.

• Mechanics - Kinematics + Dynamics

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Goals of Chapter 2

Develop an understanding of kinematics that comprehends the interrelationships among

• physical intuition

• equations

• graphical representations

When we finish this chapter you should be able to move easily among these different aspects.

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Kinematic Quantities Overview

The words speed and velocity are used interchangably in everyday conversation but they have distinct meanings in the physics world.

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The position (x) of an object describes its location relative to some origin or other reference point.

Position & Displacement

0 x2

0 x1

The position of the red ball differs in the two shown coordinate systems.

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0 x (cm)

212 1

The position of the ball is cm 2x

The + indicates the direction to the right of the origin.

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0 x (cm)

212 1

The position of the ball is

cm 2xrx

The indicates the direction to the left of the origin.

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The displacement is the change in an object’s position. It depends only on the beginning and ending positions.

if xxx

All Δ quantities will have the final value 1st and the inital value last.

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0 x (cm)212 1

cm 4

cm 2 cm 2

if xxx

Example: A ball is initially at x = +2 cm and is moved to x = -2 cm. What is the displacement of the ball?

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Example:

At 3 PM a car is located 20 km south of its starting point. One hour later its is 96 km farther south. After two more hours it is 12 km south of the original starting point.(a) What is the displacement of the car between 3 PM

and 6 PM?

xi = –20 km and xf = –12 km

km 8km 20 km 12

if xxx

Use a coordinate system where north is positive.

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(b) What is the displacement of the car from the starting point to the location at 4 pm?

(c) What is the displacement of the car from 4 PM to 6 PM?

Example continued

xi = 0 km and xf = –96 km

km 96km 0 km 96

if xxx

xi = –96 km and xf = –12 km

km 84km 96 km 12

if xxx

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Velocity is a vector that measures how fast and in what direction something moves.

Speed is the magnitude of the velocity. It is a scalar.

Velocity: Rate of Change of Position

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t

xv x

,av In 1-dimension the average velocity is

vav is the constant speed and direction that results in the same displacement in a given time interval.

tripof time

traveleddistancespeed Average

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On a graph of position versus time, the average velocity is represented by the slope of a chord.

x (m)

t (sec)

t1 t2

x1

x2

12

12,av velocityAverage

tt

xxv x

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t

xv

tx

lim

0

velocityousInstantane

This is represented by the slope of a line tangent to the curve on the graph of an object’s position versus time.

x (m)

t (sec)

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The area under a velocity versus time graph (between the curve and the time axis) gives the displacement in a given interval of time.

vx (m/s)

t (sec)

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Example (text problem 2.11): Speedometer readings are obtained and graphed as a car comes to a stop along a straight-line path. How far does the car move between t = 0 and t = 16

seconds?

Since there is not a reversal of direction, the area between the curve and the time axis will represent the distance

traveled.

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Example continued:

The rectangular portion has an area of

Lw = (20 m/s)(4 s) = 80 m.

The triangular portion has an area of

½bh = ½(8 s) (20 m/s) = 80 m.

Thus, the total area is 160 m. This is the distance traveled by the car.

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The Most Important Graph- V vs T

Area under the curve gives DISTANCE.

The slope of the curve gives the ACCELERATION.

The values of the curve gives the instantaneous VELOCITY.

Negative areas are possible.

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Acceleration: Rate of Change of Velocity

t

va x

x

av,onaccelerati Average

t

va x

tx

lim

0

onaccelerati ousInstantane

These have interpretations similar to vav and v.

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Example (text problem 2.29): The graph shows speedometer readings as a car comes to a stop. What is the magnitude of the

acceleration at t = 7.0 s?

The slope of the graph at t = 7.0 sec is

2

12

12av m/s 5.2

s 412

m/s 200

tt

vv

t

va x

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Motion Along a Line With Constant Acceleration

xavv

tavvv

tatvxxx

xixfx

xixfxx

xixif

2

2

1

22

2

For constant acceleration the kinematic equations are:

2,av

,av

fxixx

x

vvv

tvx

Also:

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A Modified Set of Equations

2f i ix x

fx ix x

2 2fx ix x

1x = x + v Δt + a Δt

2v = v + a Δt

v = v + 2a Δx

For constant acceleration the kinematic equations are:

f i av,x

ix fxav,x

x = x + v Δt

v + vv =

2

Also:

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Visualizing Motion Along a Line with Constant Acceleration

Motion diagrams for three carts:

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Graphs of x, vx, ax for each of the three carts

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A stone is dropped from the edge of a cliff; if air resistance can be ignored, we say the stone is in free fall. The magnitude of the acceleration of the stone is afree fall = g = 9.80 m/s2, this acceleration is always directed toward the Earth.

The velocity of the stone changes by 9.8 m/s every sec.

Free Fall

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Free FallAssumption: acceleration due to gravity is g

g = 9.8 m/s2 ≈ 10 m/s2

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Example: You throw a ball into the air with speed 15.0 m/s; how high does the ball rise?

Given: viy = +15.0 m/s; ay = 9.8 m/s2

2

2

1tatvy yiy

x

y viy

ay

tavv yiyfy

To calculate the final height, we need to know the time of flight.

Time of flight from:

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sec 53.1m/s 9.8

m/s 0.15

0

2

y

iy

yiyfy

a

vt

tavvThe ball rises until vfy = 0.

2iy y

2

1Δy = v Δt + a Δt

21

= 15.0 1.53 + -9.8 1.53 2

= 11.5 m

The height:

Example continued:

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Example (text problem 2.45): A penny is dropped from the observation deck of the Empire State Building 369 m above the ground. With what velocity does it strike the ground? Ignore air

resistance.

369 m

x

y

Given: viy = 0 m/s, ay = 9.8 m/s2,

y = 369 m

Unknown: vfy

Use:

2 2fy iy y

y

fy y

v = v + 2a Δy

= 2a Δy

v = 2a Δy

ay

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fy yv = 2a Δy = 2 -9.8 -369 = 85.0 m/s

How long does it take for the penny to strike the ground?

sec 7.82

2

1

2

1 22

y

yyiy

a

yt

tatatvy

(downward)

Example continued:

Given: viy = 0 m/s, ay = 9.8 m/s2, y = 369 m

Unknown: t

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Summary• Position

• Displacement Versus Distance

• Velocity Versus Speed

• Acceleration

• Instantaneous Values Versus Average Values

• The Kinematic Equations

• Graphical Representations of Motion

• Free Fall