Chapter 2 Motion Along a Straight Line

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Chapter 2 Motion Along a Straight Line. Linear motion In this chapter we will consider moving objects: Along a straight line With every portion of an object moving in the same direction and at the same rate (particle-like motion). Types of physical quantities - PowerPoint PPT Presentation

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Chapter 2

Motion Along a Straight Line

Linear motion

In this chapter we will consider moving objects:

• Along a straight line

• With every portion of an object moving in the same direction and at the same rate (particle-like motion)

Types of physical quantities

• In physics, quantities can be divided into such general categories as scalars, vectors, matrices, etc.

• Scalars – physical quantities that can be described by their value (magnitude) only

• Vectors – physical quantities that can be described by their value and direction

Distance, position, and displacement

• Distance (scalar) a total length of the path traveled regardless of direction (SI unit: m)

• In each instance we choose an origin – a reference point, convenient for further calculations

• Position of an object (vector) is described by the shortest distance from the origin and direction relative to the origin

• Displacement (vector) – a change from position xi to position xf

if xxx

Velocity and speed

• Average speed (scalar) - a ratio of distance traveled (over a time interval) to that time interval (SI unit: m/s)

• Average velocity (vector) - a ratio of displacement (over a time interval) to that time interval

• Instantaneous velocity (vector) – velocity at a given instant

• Instantaneous speed (scalar) – a magnitude of an instantaneous velocity

t

xvavg

t

xv

t

0

lim

if

if

tt

xx

dt

dx

Velocity and speed

Velocity and speed

Instantaneous velocity

• The instantaneous velocity is the slope of the line tangent to the x vs. t curve

• This would be the green line

• The light blue lines show that as Δt gets smaller, they approach the green line

Acceleration

• Average acceleration (vector) - a ratio of change of velocity (over a time interval) to that time interval (SI unit = (m/s)/s = m/s2)

• Instantaneous acceleration (vector) – a rate of change of velocity at a given instant

2

2

dt

xd

t

vaavg

if

if

tt

vv

t

va

t

0

limdt

dv

dt

dx

dt

d

Acceleration

• The slope (green line) of the velocity-time graph is the acceleration

• The blue line is the average acceleration

Chapter 2Problem 15

An object moves along the x axis according to the equation x(t) = (3.00 t2 - 2.00 t + 3.00) m, where t is in seconds. Determine (a) the average speed between t = 2.00 s and t = 3.00 s, (b) the instantaneous speed at t = 2.00 sand at t = 3.00 s, (c) the average acceleration between t = 2.00 s and t = 3.00 s, and (d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.

Case of constant acceleration

• Average and instantaneous accelerations are the same

• Conventionally

• Then

0it

t

vaa avg

if

if

tt

vv

0

t

vv if

atvv if

tt f

Case of constant acceleration

• Average and instantaneous accelerations are the same

• Conventionally

• Then

0it tt f

t

xvavg

0

t

xx if tvxx avgif

2221 fi

avg

vvvvv

2

atvi

if

if

tt

xx

2

)( atvv ii

2

2attvxx iif

Case of constant acceleration

dt

dxv

dt

dva

atvv if

2

2attvxx iif

Case of constant acceleration

Case of constant acceleration

To help you solve problems

Equations Missing variables

2

2attvxx iif

atvv if

)(222ifif xxavv

2

)( tvvxx fiif

2

2attvxx fif

fv

if xx

t

a

iv

Chapter 2Problem 28

A particle moves along the x axis. Its position is given by the equation x = 2 + 3t - 4t2, with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.

Case of free-fall acceleration

• At sea level of Earth’s mid-latitudes all objects fall (in vacuum) with constant (downward) acceleration of

a = - g ≈ - 9.8 m/s2 ≈ - 32 ft/s2

• Conventionally, free fall is along a vertical (upward) y-axis

gtvv if

2

2gttvyy iif

Chapter 2Problem 38

A ball is thrown directly downward, with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does the ball strike the ground?

Alternative derivation

Using definitions and initial conditions

we obtain

dt

dva

2

2attvxx iif

dtadv dt

dxv vdtdx

Graphical representation

Graphical representation

Graphical representation

Graphical representation

Graphical representation

Graphical representation

Graphical representation

Graphical representation

Graphical representation

Graphical integration

f

i

t

t

if dttvxx )(

dt

dxv

0lim ( )

f

in

t

xn n xttn

v t v t dt

Graphical integration

f

i

t

t

if dttvxx )(

dt

dxv

f

i

t

t

if dttavv )(

dt

dva

Answers to the even-numbered problems

Chapter 2

Problem 4:

(a) 50.0 m/s (b) 41.0 m/s

Answers to the even-numbered problems

Chapter 2

Problem 6:

(a) 27.0 m(b) 27.0 m + (18.0 m/s)∆t + (3.00 m/s2)(∆t)2

(c) 18.0 m/s

Answers to the even-numbered problems

Chapter 2

Problem 12:

(b) 1.60 m/s2; 0.800 m/s2

Answers to the even-numbered problems

Chapter 2

Problem 20:

(a) 6.61 m/s(b) −0.448 m/s2

Answers to the even-numbered problems

Chapter 2

Problem 38:

1.79 s

Answers to the even-numbered problems

Chapter 2

Problem 48:

(b) 3.00 × 10−3 s (c) 450 m/s (d) 0.900 m

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