21/Fev/2018 Aula 2. 19/Fev/2018 Aula 1

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19/Fev/2018 – Aula 1 21/Fev/2018 – Aula 2 2.1 Queda livre 2.2 Movimento 2 e 3-D 2.2.1 Vetor deslocamento 2.2.2 Vetor velocidade 2.2.3 Vetor aceleração 2.3 Lançamento de projétil

2.3.1 Independência dos movimentos 2.3.2 Forma vetorial 2.3.3 ângulo de lançamento 2.3.4 Alcance 1.1 Conceitos gerais 1.1.1 Introdução 1.1.2 Unidades 1.1.3 Dimensões 1.1.4 Estimativas 1.1.5 Resolução de problemas - método 1.1.6 Escalares e vetores 1.2 Descrição do movimento 1.2.1 Distância e deslocamento 1.2.2 Rapidez e velocidade 1.2.3 Velocidade instantânea 1.2.4 Aceleração 1.2.5 Posição, velocidade e aceleração

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Aula anterior

1.1.3 Dimensões

Todas as equações físicas tem de ser dimensionalmente consistentes: os dois lados da equação têm de ter as mesmas dimensões.

A partir desta Tabela:

Distância = velocidade × tempo Velocidade= aceleração × tempo Energia = massa × (velocidade)2

Evaluating the dimensions of an expression will tell you only if the dimensions are correct, not whether the entire expression is correct. While expressing the area of a circle, for

instance, dimensional analysis will not tell you the correct expression is or

. (The correct expression is pr2.) 2pr2

pr2

!

8

|

C H A P T E R 1 Measurement and Vectors

quantities makes sense only if the quantities have the same dimensions. For example, we cannot add an area to a speed to obtain a meaningful sum. For the equation

the quantities A, B, and C must all have the same dimensions. The addition of B and C also requires that these quantities be in the same units. For example, if B is an area of and C is we must either convert B into square feet or C into square inches in order to find the sum of the two areas.

You can often find mistakes in a calculation by checking the dimensions or units of the quantities in your result. Suppose, for example, that you mistakenly use the formula for the area of a circle. You can see immediately that this cannot be correct because has dimensions of length whereas area must have dimen-sions of length squared.

Example 1-2

Dimensions of Pressure

The pressure P in a fluid in motion depends on its density and its speed . Find a simple combination of density and speed that gives the correct dimensions of pressure.

PICTURE Using Table 1-2, we can see that pressure has the dimensions density is and speed is In addition, both the dimensions of pressure and density have mass in the numerator, whereas the dimensions of speed do not contain mass. Therefore, the ex-pression must involve multiplying or dividing dimensions of density and dimensions of speed to obtain the unit of mass in the dimensions of pressure. To find out the exact rela-tionship, we can start by dividing the dimensions of pressure by those of density, and then evaluate the result with respect to the dimensions for speed.

SOLVE

1. Divide the dimensions of pressure by those of density to obtain an expression with no M in it:

L>T. M>L3_, M>1LT2_2, v r 2pr A ! 2pr 4 ft2_, 500 in.2 A ! B " C 3P4 3r4 ! M>LT2 M>L3 ! L2 T2

2. By inspection, we note that the result has dimensions of The dimensions of pressure are thus the same as the dimensions of density multiplied by speed squared:

v2_.

CHECK Divide the dimensions of pressure by the dimensions of speed squared and the re-sult is the dimensions of density

1-5

SIGNIFICANT FIGURES

AND ORDER OF MAGNITUDE

Many of the numbers in science are the result of measurement and are therefore known only to within a degree of experimental uncertainty. The magnitude of the uncertainty, which depends on both the skill of the experimenter and the apparatus used, often can only be estimated. A rough indication of the uncer-tainty in a measurement is inferred by the number of digits used. For example, if a tag on a table in a furniture store states that a table is 2.50 m long, it is say-ing that its length is close to, but not exactly, 2.50 m. The rightmost digit, the 0, is uncertain. If we use a tape measure with millimeter markings and measured the table length carefully, we might estimate that we could measure the length to mm of its true length. We would indicate this precision when giving the length by using four digits, such as 2.503 m. A reliably known digit (other than a zero used to locate the decimal point) is called a significant figure. The

#0.6

3P4>3v2_{4 ! 1M>LT}2_2>1L2_>T2_{2 ! M>L}3 _! _3r4.

Table 1-2

Dimensions of

Physical Quantities

Quantity Symbol Dimension

Area A L2 Volume V L3 Speed L T Acceleration a L T2 Force F ML T2 Pressure (F A) p M LT2 Density (M V) r M L3 Energy E ML2 _T2 Power (E T)_> P ML2_>_T3 > > > > > > > > v M LT2 ! M L3 $ L2 T2 ! 3P4 ! 3r43v2_{4 !} M L3 $ a L T b 2 2 As unidades da maior parte das

grandezas físicas podem ser

expressas como combinações das unidades da massa, do

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Aula anterior

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Aula anterior

1.2.1 Distância e deslocamento

Distância é o percurso total percorrido.

Exemplo: a distância BCA é igual a 4 km + 4 km + 2 km = 10 km.

2 km

4 km

A _B

C

Deslocamento é a variação da posição, durante um certo intervalo de tempo.

Exemplo: o deslocamento BCA é igual a -2 km, mas a distância BCA é igual a 12 km.

Δx = x

_f

− x

_i

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1.2.2 Rapidez e velocidade

Aula anterior

A velocidade é o quociente entre o deslocamento efetuado e o intervalo de tempo decorrido:

v_média = Δx Δt = 0 48,0 s − 0 = 0,0 m/s rapidez = s Δt = 50,0 m + 50,0 m 48,0 s − 0 = 2,08 m/s

v

_x

=

Δx

Δt

=

x

_{f −}

x

_i

t

_{f −}

t

_i v_walk = Δx Δt = 0 − 50,0m 48,0 s −8,0 s = −1,25 m/s 8 s 48 s

A rapidez é o quociente entre a distância

percorrida e o intervalo de tempo decorrido:

rapidez =

distância

tempo

=

s

Δt

v_sprint = Δx Δt = 50,0 m − 0 8,0 s − 0 = 6,25 m/s

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1.2.3 Velocidade instantânea

Aula anterior

Se se conseguir determinar a velocidade em intervalos de tempo cada vez mais pequenos, até se tornarem infinitamente pequenos, conseguir-se-á obter a velocidade instantânea:

v

_x

₌

dx

dt

v

_x

=

Δx

Δt

=

x

_{f −}

x

_i

t

_{f −}

t

_i

v

x

= lim

_Δt→0

Δx

Δt

Interpretação gráfica das velocidades média e instantânea:

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2.1 Queda livre

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Para cair em “queda livre”, só pode estar apenas sob a influência da gravidade. Num dado local, a aceleração devido à gravidade é uma constante (g).

Equation 2-16 is applicable only for time intervals during which the acceleration remains constant.

!

Motion with Constant Acceleration S E C T I O N 2 - 3 | 39

to the displacement !x_i during the interval !t_i. The sum of the rectangular areas is therefore approximately the sum of the displacements during the time intervals and is approximately equal to the total displacement from time t₁ to t₂. We can make the approximation as accurate as we wish by putting enough rectangles under the curve, each rectangle having a sufficiently small value for !t. For the limit of smaller and smaller time intervals (and more and more rectangles), the resulting sum approaches the area under the curve, which in turn equals the displacement. The displacement !x is thus the area under the

v_x-versus-t curve.

For motion with constant acceleration (Figure 2-13a), !x is equal to the area of the shaded region. This region is divided into a rectangle and a triangle of areas

v_1x !t and a_x(!t)2_{, respectively, where !t " t}

2 # t1. It follows that

2-13 If we set t₁ " 0 and t₂" t, then Equation 2-13 becomes

2-14

CONSTANT ACCELERATION: x (t)

where x₀ and v_0x are the position and velocity at time t " 0, and x " x(t) is the position at time t. The first term on the right, v_0xt, is the displacement that would occur if a_xwere zero, and the second term, , is the additional displacement due to the constant acceleration.

We next use Equations 2-12 and 2-14 to obtain two additional kinematic equa-tions for constant acceleration. Solving Equation 2-12 for t, and substituting for t, in Equation 2-14 gives

Multiplying both sides by 2a_x we obtain

Simplifying and rearranging terms gives

2-15

CONSTANT ACCELERATION: v_x(x )

The definition of average velocity (Equation 2-3) is: !x " v_{av x}!t

where v_{av x}!tis the area under the horizontal line at height v_{av x}in Figure 2-13b and !x is the area under the v_x versus t curve in Figure 2-13a. We can see that if , the area under the line at height v_avin Figure 2-13a and the area under the v_x versus t curve in Figure 2-13b will be equal. Thus,

2-16

CONSTANT ACCELERATION: v_{av x} AND v_x

For motion with constant acceleration, the average velocity is the mean of the ini-tial and final velocities.

For an example of an instance where Equation 2-16 is not applicable, consider the motion of a runner during a 10.0-km run that takes 40.0 min to complete. The

v_{av x}" 1₂(v_1x $ v_2x) v_{av x} "1₂(v_1x $ v_2x) v2 x" v20x$ 2ax¢x 2a_x¢x " 2v_0x(v_x # v_0x) $ (v_x # v_0x)2 ¢x " v0x v_x # v_0x a_x $ 1 2axa v_x # v_0x a_x b 2 1 2axt2 x # x₀ " v_0xt $12axt2 ¢x " v1x¢t $ 12ax(¢t)2 1 2 (a) t₁ t₂ v1x v_1x v_2x v_x ∆t ∆v_x= a_x∆t t (b) t1 t2 v1x v_{av x} v_2x v_x ∆t t

F I G U R E 2 - 1 3 Motion with constant

acceleration.

”It goes from zero to 60 in about 3 seconds.”

(© Sydney Harris.)

g

= 9,80 m/s2

2.1 Queda livre

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Velocidade em função do tempo:

a

_x

=

dv

x

dt

⇒ v

x

=

∫

a

x

dt

= a

x

∫

dt

= a

x

t + v

0x Posição em função do tempo:

v

_x

₌

dx

dt

⇒ x =

∫

(

a

x

t + v

0x

)

dt

= a

x

∫

t dt

+ v

0x

∫

dt

=

1

2 a

x

t

2

+ v

_0x

t + x

₀ Integration S E C T I O N 2 - 4

|

49

The antiderivative of a function is also called the indefinite integral of the function

and is written without limits on the integral sign, as in

Finding the function x from its derivative v

_x

(that is, finding the antiderivative) is

also called integration. For example, if v

_x

!

v

_0x

, a constant, then

where x

₀

is the arbitrary constant of integration. We can find a general rule for the

integration of a power of t from Equation 2-6, which gives the general rule for the

derivative of a power. The result is

2-19

where C is an arbitrary constant. This equation can be checked by

differen-tiating the right side using the rule of Equation 2-6. (For the special case n ! –1,

, where ln t is the natural logarithm of t.)

Because a

_x

!

dv

_x

/dt

, the change in velocity for some time interval can similarly

be interpreted as the area under the a

_x

-versus-t curve for that interval. This change

is written

2-20

We can now derive the constant-acceleration equations by computing the

indefi-nite integrals of the acceleration and velocity. If a

_x

is constant, we have

2-21

where we have expressed the product of a

_x

and the constant of integration as v

_0x

.

Integrating again, and writing x

₀

for the constant of integration, gives

2-22

It is instructive to derive Equations 2-21 and 2-22 using definite integrals instead

of indefinite ones. For constant acceleration, Equation 2-20, with t

₁

!

0, gives

where the time t

₂

is arbitrary. Because it is arbitrary, we can set t

₂

!

t

to obtain

where v

_x

!

v

_x

(t) and v

_0x

!

v

_x

(0). To derive Equation 2-22, we substitute v

_0x

"

a

_x

t

for v

_x

in Equation 2-17 with t

₁

!

0. This gives

This integral is equal to the area under the v

_x

-versus-t curve (Figure 2-26).

Evaluating the integral and solving for x gives

where t

₂

is arbitrary. Setting t

₂

!

t,

we obtain

where x ! x(t) and x

!

x(0).

x ! x

₀

"

v

_0x

t "

1₂

a

_x

t

2

x(t

₂

) # x(0) !

!

t₂ 0

(v

0x

"

a

_x

t)dt ! v

_0x

t "

1₂

a

_x

t

2

`

t₂ 0

!

v

0x

t

2

"

1 2

a

_x

t

2₂

x(t

₂

) # x(0) !

!

t₂ 0

(v

_0x

"

a

_x

t)dt

v

_x

!

v

_0x

"

a

_x

t

v

_x

(t

₂

) # v

_x

(0) ! a

_x

!

t₂ 0

dt ! a

x

(t

2

#

0)

x !

!

(v

_0x

"

a

_x

t)dt ! x

₀

"

v

_0x

t "

1₂

a

_x

t

2

v

_x

!

a

_x

dt ! a

_x

!

dt ! v

_0x

"

a

_x

t

¢v

_x

!

lim

¢tS0

a a

_i

a

ix

¢t

i

b !

!

t₂ t₁

a

_x

dt

"

t

#1

_{dt !}

ln t " C

!

t

n

_{dt !}

t

n"1

n "

1 "

C, n $ #1

x !

!

v

_0x

dt ! v

_0x

t " x

₀

x !

!

v

_x

dt

v_x = v_0x+ at Area 0₀ v_x(0) v_x(t₂) v_x(t) t t₂

F I G U R E 2 - 2 6 The area under the

v_x-versus-t curve equals the displacement .

¢x ! x(t ) # x(0)

9

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A figura mostra dois objetos numa câmara de vácuo, com massas e formas diferentes, a caírem apenas sob a ação da gravidade.

Se se considerar que o sentido +y aponta para baixo, então a = +g; se se considerar que esse é o sentido –y, então a = -g.

Note-se que o valor de g é sempre positivo, mas o de a pode ser positivo ou negativo.

Queda livre

filme

2.1 Queda livre

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Queda livre a partir do repouso:

v

_x

_{= v}

_0x

_{+ a}

_x

_{t = 0 + gt}

x =

1

2 a

x

t

2

+ v

_0x

t + x

₀

=

1

2 gt

2

+ 0 + 0

Método de resolução:

1)  Determinar o que é pedido (tempo, distância, velocidade, aceleração,…)

2)  Desenhar o objeto nas posições inicial e final. Definir o sistema de eixos.

3)  Selecionar as equações relevantes e resolvê-las. Só no fim, substituir os valores dados e calcular o resultado. 4)  Verificar se o resultado tem as

dimensões certas e o valor esperado.

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12

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Exemplo

Um geólogo mede o tempo de voo de um pedaço de lava, expelido por um vulcão. Admita que o tempo total é 4,75 s. Determine:

a)  a velocidade inicial; b) a altura máxima atingida. a)

_{x = x}

₀

_{+ v}

_0x

_{t +}

1

2 a

x

t

2 _v

x

= v

0x

+ a

x

t

Δx = 0 = t (v

₀

−

1

2 g t)

t = 0

ou

v

0 −

1

2 g t = 0

v

₀

=

1 2

g t =

1 2

(9,81 m/s

2

_{)(4,75 s) = 23,3 m/s}

b) A altura máxima ocorre para

_{t =}

t

voo

2 x − x

₀

= v

_0x

t

voo

2 +

1

2 a

x

t

_voo

2 ⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

2

= 23,3 m/s

(

)

×

(

4,75 s

)

2 +

1

2 × −9,81 m/s

2

(

)

×

4,75 s

2 ⎛

⎝

⎜

⎞

⎠

⎟

2

= 27,7 m

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2.2 Movimento 2 e 3-D

Equações do movimento:

v

_x

= v

_0x

+ a

_x

t

x = x

₀

+ v

_0x

t +

1

2 a

x

t

2

vy = v

0 y

+ a

y

t

y

₌

_y0

_{+ v}

_{0 y}

t +

1

2 a

y

t

2 64

|

C H A P T E R 3 Motion in Two and Three Dimensions

3-1

DISPLACEMENT, VELOCITY,

AND ACCELERATION

In Chapter 2, the concepts of displacement, velocity, and acceleration were used to describe the motion of an object moving in a straight line. Now we use the concept of vectors to extend these characteristics of motion in two and three dimensions.

POSITION AND DISPLACEMENT VECTORS

The position vector of a particle is a vector drawn from the origin of a coordinate system to the location of the particle. For a particle in the x, y plane at the point with coordinates (x, y), the position vector is

3-1

DEFINITION — POSITION VECTOR

Note that x and y components of the position vector are the Cartesian coordinates (Figure 3-1) of the particle.

Figure 3-2 shows the actual path or trajectory of the particle. At time t₁, the par-ticle is at P₁, with position vector ; by time t₂, the particle has moved to P₂, with position vector . The particle’s change in position is the displacement vec-tor

3-2

DEFINITION — DISPLACEMENT VECTOR

Using unit vectors, we can rewrite this displacement as

3-3

VELOCITY VECTORS

Recall that average velocity is defined as displacement divided by the elapsed time. The result of the displacement vector divided by the elapsed time interval !t " t₂ # t₁ is the average-velocity vector:

3-4

DEFINITION — AVERAGE-VELOCITY VECTOR

The average velocity vector and the displacement vector are in the same direction. The magnitude of the displacement vector is less than the distance traveled along the curve unless the particle moves along a straight line and never reverses its direction. However, if we consider smaller and smaller time intervals (Figure 3-3), the magnitude of the displacement approaches the distance along the curve, and the angle between and the tangent to the curve at the beginning of the interval approaches zero. We define the instantaneous-velocity vector as the limit of the average-velocity vector as !t approaches zero:

3-5

DEFINITION — INSTANTANEOUS-VELOCITY VECTOR

vS " lim ¢tS0 ¢rS ¢t " drS dt ¢rS vS_av " ¢r S ¢t

¢rS " Sr₂ # Sr₁ " (x₂ # x₁)in $ (y₂ # y₁)jn " ¢xin $ ¢yjn ¢rS " Sr₂ # Sr₁ ¢rS: r S 2 r S 1 r S r S _" xin $ yjn r S

F I G U R E 3 - 1 The x and y components of

the position vector for a particle are the x and y (Cartesian) coordinates of the particle.r

S Particle x (x, y) y xiˆ r = xi ˆ+ yˆj yjˆ y x O r₂ r₁ ∆r P₁ at t₁ P₂ at t₂ y x O r₁ P”₂ P₂ P’₂ ∆r” The tangent to the

curve at P₁ is by definition the direction of v at P₁

∆r’ ∆r

P₁

F I G U R E 3 - 3 As the time interval

decreases, the angle between direction of and the tangent to the curve approaches zero.¢r

S

Do not confuse the trajectory in graphs of x-versus-y with the curve in the x-versus-t plots of Chapter 2.

!

F I G U R E 3 - 2 The displacement vector

is the difference in the position vectors, Equivalently, is the vector that, when added to the initial position vector

yields the final position vector That is,

r S 1 $ ¢rS " Sr2. r S 2. r S 1, ¢rS ¢rS " Sr₂#Sr₁. ¢rS 14

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2.2 Movimento 2 e 3-D

Equações do movimento:

Displacement, Velocity, and Acceleration

S E C T I O N 3 - 1

|

65 The instantaneous-velocity vector is the derivative of the position vector with

re-spect to time. Its magnitude is the speed and its direction is along the line tangent

to the curve in the direction of motion of the particle.

To calculate the derivative in Equation 3-5, we write the position vectors in

terms of their components (Equation 3-1):

Then

or

3-6

where v

_x

!

dx

/dt and v

_y

!

dy

/dt are the x and y components of the velocity.

The magnitude of the velocity vector is given by:

3-7

and the direction of the velocity is given by

3-8

u !

tan

"1

v

_y

v

_x

v !

2v

2 x

#

v

2y

v

S

!

dx

dt

in #

dy

dt

jn ! v

x

in # v

y

jn

v

S

!

lim

¢tS0

¢r

S

¢t

!

¢tS0

lim

¢xin # ¢yjn

¢t

!

¢tS0

lim

a

¢x

¢t

b in # lim

¢tS0

a

¢y

¢t

b jn

¢r

S

!

S

r

₂

"

S

r

₁

!

(x

₂

"

x

₁

)in # (y

₂

"

y

₁

)jn !

_{¢xin # ¢yjn}

Do not trust your calculator to

always give the correct value for

$

when using Equation 3-8. Most

calculators will return the correct value

for

$

if v

_x

is positive. If v

_x

is negative,

however, you will need to add 180°

(

%

rad) to the value returned by the

calculator.

!

See

Math Tutorial for more

information on

Trigonometry

Example 3-1

The Velocity of a Sailboat

A sailboat has coordinates (x₁, y₁) ! (130 m, 205 m) at t₁ ! 60.0 s. Two minutes later, at time t₂, it has coordinates (x₂, y₂) ! (110 m, 218 m). (a) Find the average velocity for this two-minute interval. Express in terms of its rectangular components. (b) Find

the magnitude and direction of this average velocity. (c) For t & 20.0 s, the posi-tion of a second sailboat as a funcposi-tion of time is x(t) ! b₁ # b₂t and y(t) ! c₁ # c₂ t, where b₁ ! 100 m, b₂ ! 0.500 m s, c₁ ! 200 m, and c₂ ! 360 m s. Find the instan-taneous velocity as a function of time t, for t & 20.0 s.

PICTURE The initial and final positions of the first sailboat are given. Because the motion of the boat is in two dimensions, we need to express the displace-ment, average velocity, and instantaneous velocity as vectors. Then we can use Equations 3-5 through 3-8 to obtain the requested values.

#

> > vS_av vS_av 210 220 200 110 100 120 (110, 218) (130, 205) θ 130 ∆x ∆y y, m x, m ∆r SOLVE

(a) 1. Draw a coordinate system (Figure 3-4) and draw the displacement of the sailboat. Draw the average-velocity vector (it and the

displacement vector are in the same direction):

2. The x and y components of the average velocity are calculated directly from their definitions:

vS_av where so "(0.167 m/s)in # (0.108 m/s)jn ! vS_av v_{y av} ! ¢y ¢t ! 218 m " 205 m 120 s ! 0.108 m/s v_{x av} ! ¢x ¢t ! 110 m " 130 m 120 s ! "0.167 m/s vS_av ! v_{x av}in # v_{y av}jn

(b) 1. The magnitude of vS_av is found from the Pythagorean theorem: v_av _{! 4(v}_{x av})2 _# _(v 0.199 m/s

y av)2 !

2. The ratio of to gives the tangent of the angle $ between and the #x direction (we add 180° to the value of "33.0° returned by the calculator because v_x is negative):

vS_av v_{x av} v_{y av} so 147° ! "33.0° # 180° ! u ! tan"1 v_{y av} v_{x av} ! tan "1 0.108 m/s "0.167 m/s tan u ! vy av v_{x av} F I G U R E 3 - 4

Displacement, Velocity, and Acceleration

S E C T I O N 3 - 1

|

65 The instantaneous-velocity vector is the derivative of the position vector with

re-spect to time. Its magnitude is the speed and its direction is along the line tangent

to the curve in the direction of motion of the particle.

To calculate the derivative in Equation 3-5, we write the position vectors in

terms of their components (Equation 3-1):

Then

or

3-6

where v

_x

!

dx

/dt and v

_y

!

dy

/dt are the x and y components of the velocity.

The magnitude of the velocity vector is given by:

3-7

and the direction of the velocity is given by

3-8

u !

tan

"1

v

_y

v

_x

v !

2v

2_x

#

_v

2_y

v

S

!

dx

dt

in #

dy

dt

jn ! v

x

in # v

y

jn

v

S

!

lim

¢tS0

¢r

S

¢t

!

¢tS0

lim

¢xin # ¢yjn

¢t

!

¢tS0

lim

a

¢x

¢t

b in # lim

¢tS0

a

¢y

¢t

b jn

¢r

S

!

S

r

₂

"

S

r

₁

!

(x

₂

"

x

₁

)in # (y

₂

"

y

₁

)jn !

¢xin # ¢yjn

Do not trust your calculator to

always give the correct value for $

when using Equation 3-8. Most

calculators will return the correct value

for $ if v

_x

is positive. If v

_x

is negative,

however, you will need to add 180°

(% rad) to the value returned by the

calculator.

!

See

Math Tutorial for more

information on

Trigonometry

Example 3-1

The Velocity of a Sailboat

A sailboat has coordinates (x

₁

, y

₁

) ! (130 m, 205 m) at t

₁

!

60.0 s. Two minutes later, at time t

₂

,

it has coordinates (x

₂

, y

₂

) ! (110 m, 218 m). (a) Find the average velocity

for this

two-minute interval. Express

in terms of its rectangular components. (b) Find

the magnitude and direction of this average velocity. (c) For t & 20.0 s, the

posi-tion of a second sailboat as a funcposi-tion of time is x(t) ! b

₁

#

b

₂

t

and y(t) ! c

₁

#

c

₂

t

,

where b

₁

!

100 m, b

₂

!

0.500 m s, c

₁

!

200 m, and c

₂

!

360 m s. Find the

instan-taneous velocity as a function of time t, for t & 20.0 s.

PICTURE

The initial and final positions of the first sailboat are given. Because

the motion of the boat is in two dimensions, we need to express the

displace-ment, average velocity, and instantaneous velocity as vectors. Then we can use

Equations 3-5 through 3-8 to obtain the requested values.

#

>

v

S_av

v

S_av

210

220

200

110

100

120 (110, 218)

(130, 205)

θ

130 ∆x

∆y

y,

m

x,

m

∆r

SOLVE

(a) 1. Draw a coordinate system (Figure 3-4) and draw the displacement

of the sailboat. Draw the average-velocity vector (it and the

displacement vector are in the same direction):

2. The x and y components of the average velocity

are calculated

directly from their definitions:

v

S_av

where

so

"

(0.167 m/s)in # (0.108 m/s)jn

!

v

S_av

v

_{y av}

!

¢y

¢t

!

218 m " 205 m

120 s

!

0.108 m/s

v

_{x av}

!

¢x

¢t

!

110 m " 130 m

120 s

! "

0.167 m/s

v

S_av

!

v

_{x av}

in # v

_{y av}

jn

(b) 1. The magnitude of

v

S_av

is found from the Pythagorean theorem:

v

_av

_{! 4(v}

_{x av}

)

2

_#

_(v

0.199 m/s

y av

)

2

!

2. The ratio of

to

gives the tangent of the angle

$

between

and the #x direction (we add 180° to the value of "33.0°

returned by the calculator because v

_x

is negative):

v

S_av

v

_{x av}

v

_{y av}

so

147°

! "

33.0° # 180° !

u !

tan

"1

v

_{y av}

v

_{x av}

!

tan

"1

0.108 m/s

"

0.167 m/s

tan u !

v

y av

v

_{x av} F I G U R E 3 - 4

Displacement, Velocity, and Acceleration

S E C T I O N 3 - 1

|

65 The instantaneous-velocity vector is the derivative of the position vector with

re-spect to time. Its magnitude is the speed and its direction is along the line tangent

to the curve in the direction of motion of the particle.

To calculate the derivative in Equation 3-5, we write the position vectors in

terms of their components (Equation 3-1):

Then

or

3-6

where v

_x

!

dx

/dt and v

_y

!

dy

/dt are the x and y components of the velocity.

The magnitude of the velocity vector is given by:

3-7

and the direction of the velocity is given by

3-8

u !

tan

"1

v

_y

v

_x

v !

2v

2 x

#

v

2y

v

S

!

dx

dt

in #

dy

dt

jn ! v

x

in # v

y

jn

v

S

!

lim

¢tS0

¢r

S

¢t

!

¢tS0

lim

¢xin # ¢yjn

¢t

!

¢tS0

lim

a

¢x

¢t

b in # lim

¢tS0

a

¢y

¢t

b jn

¢r

S

!

S

r

₂

"

r

S₁

!

(x

₂

"

x

₁

)in # (y

₂

"

y

₁

)jn !

¢xin # ¢yjn

Do not trust your calculator to

always give the correct value for $

when using Equation 3-8. Most

calculators will return the correct value

for $ if v

_x

is positive. If v

_x

is negative,

however, you will need to add 180°

(% rad) to the value returned by the

calculator.

!

See

Math Tutorial for more

information on

Trigonometry

Example 3-1

The Velocity of a Sailboat

#

v S av where so "(0.167 m/s)in # (0.108 m/s)jn ! vS_av v_{y av} ! ¢y ¢t ! 218 m " 205 m 120 s ! 0.108 m/s v_{x av} ! ¢x ¢t ! 110 m " 130 m 120 s ! "0.167 m/s vS_av ! v_{x av}in # v_{y av}jn

y av)2 !

2. The ratio of to gives the tangent of the angle

$

between

and the #x direction (we add 180° to the value of "33.0° returned by the calculator because v_x is negative):

vS_av v_{x av} v_{y av} so 147° ! "33.0° # 180° ! u ! tan"1 v_{y av} v_{x av} ! tan"1 0.108 m/s "0.167 m/s tan u ! vy av v_{x av} F I G U R E 3 - 4

Displacement, Velocity, and Acceleration

S E C T I O N 3 - 1

|

65 The instantaneous-velocity vector is the derivative of the position vector with

re-spect to time. Its magnitude is the speed and its direction is along the line tangent

to the curve in the direction of motion of the particle.

To calculate the derivative in Equation 3-5, we write the position vectors in

terms of their components (Equation 3-1):

Then

or

3-6

where v

_x

!

dx

/dt and v

_y

!

dy

/dt are the x and y components of the velocity.

The magnitude of the velocity vector is given by:

3-7

and the direction of the velocity is given by

3-8

u !

tan

"1

v

_y

v

_x

v !

2v

2 x

#

v

2y

v

S

!

dx

dt

in #

dy

dt

jn ! v

x

in # v

y

jn

v

S

!

lim

¢tS0

¢r

S

¢t

!

¢tS0

lim

¢xin # ¢yjn

¢t

!

¢tS0

lim

a

¢x

¢t

b in # lim

¢tS0

a

¢y

¢t

b jn

¢r

S

!

S

r

₂

"

S

r

₁

!

(x

₂

"

x

₁

)in # (y

₂

"

y

₁

)jn !

_{¢xin # ¢yjn}

Do not trust your calculator to

always give the correct value for $

when using Equation 3-8. Most

calculators will return the correct value

for $ if v

_x

is positive. If v

_x

is negative,

however, you will need to add 180°

(% rad) to the value returned by the

calculator.

!

See

Math Tutorial for more

information on

Trigonometry

Example 3-1

The Velocity of a Sailboat

#

vS_av where so "(0.167 m/s)in # (0.108 m/s)jn ! vS_av v_{y av} ! ¢y ¢t ! 218 m " 205 m 120 s ! 0.108 m/s v_{x av} ! ¢x ¢t ! 110 m " 130 m 120 s ! "0.167 m/s vS_av ! v_{x av}in # v_{y av}jn

y av)2 !

2. The ratio of to gives the tangent of the angle $ between

and the #x direction (we add 180° to the value of "33.0° returned by the calculator because v_x is negative):

v S av v_{x av} v_{y av} so 147° ! "33.0° # 180° ! u ! tan"1 v_{y av} v_{x av} ! tan"1 0.108 m/s "0.167 m/s tan u ! vy av v_{x av} F I G U R E 3 - 4

Displacement, Velocity, and Acceleration

S E C T I O N 3 - 1

|

65 The instantaneous-velocity vector is the derivative of the position vector with

re-spect to time. Its magnitude is the speed and its direction is along the line tangent

to the curve in the direction of motion of the particle.

To calculate the derivative in Equation 3-5, we write the position vectors in

terms of their components (Equation 3-1):

Then

or

3-6

where v

_x

!

dx

/dt and v

_y

!

dy

/dt are the x and y components of the velocity.

The magnitude of the velocity vector is given by:

3-7

and the direction of the velocity is given by

3-8

u !

tan

"1

v

_y

v

_x

v !

2v

2_x

#

v

2_y

v

S

!

dx

dt

in #

dy

dt

jn ! v

x

in # v

y

jn

v

S

!

lim

¢tS0

¢r

S

¢t

!

¢tS0

lim

¢xin # ¢yjn

¢t

!

¢tS0

lim

a

¢x

¢t

b in # lim

¢tS0

a

¢y

¢t

b jn

¢r

S

!

S

r

₂

"

S

r

₁

!

(x

₂

"

_x

₁

)in # (y

₂

"

_y

₁

)jn !

¢xin # ¢yjn

Do not trust your calculator to

always give the correct value for

$

when using Equation 3-8. Most

calculators will return the correct value

for

$

if v

_x

is positive. If v

_x

is negative,

however, you will need to add 180°

(

%

rad) to the value returned by the

calculator.

!

See

Math Tutorial for more

information on

Trigonometry

Example 3-1

The Velocity of a Sailboat

A sailboat has coordinates (x

₁

, y

₁

) ! (130 m, 205 m) at t

₁

!

60.0 s. Two minutes later, at time t

₂

,

it has coordinates (x

₂

, y

₂

) ! (110 m, 218 m). (a) Find the average velocity

for this

two-minute interval. Express

in terms of its rectangular components. (b) Find

the magnitude and direction of this average velocity. (c) For t & 20.0 s, the

posi-tion of a second sailboat as a funcposi-tion of time is x(t) ! b

₁

#

b

₂

t

and y(t) ! c

₁

#

c

₂

t

,

where b

₁

!

100 m, b

₂

!

0.500 m s, c

₁

!

200 m, and c

₂

!

360 m s. Find the

instan-taneous velocity as a function of time t, for t & 20.0 s.

PICTURE

The initial and final positions of the first sailboat are given. Because

the motion of the boat is in two dimensions, we need to express the

displace-ment, average velocity, and instantaneous velocity as vectors. Then we can use

Equations 3-5 through 3-8 to obtain the requested values.

#

>

v

S_av

v

S_av

210

220

200

110

100

120 (110, 218)

(130, 205)

θ

130 ∆x

∆y

y,

m

x,

m

∆r

SOLVE

(a) 1. Draw a coordinate system (Figure 3-4) and draw the displacement

of the sailboat. Draw the average-velocity vector (it and the

displacement vector are in the same direction):

2. The x and y components of the average velocity

are calculated

directly from their definitions:

v

S_av

where

so

"(0.167 m/s)in # (0.108 m/s)jn

!

v

S_av

v

_{y av}

!

¢y

¢t

!

218 m " 205 m

120 s

!

0.108 m/s

v

_{x av}

!

¢x

¢t

!

110 m " 130 m

120 s

! "

0.167 m/s

v

S_av

!

v

_{x av}

in # v

_{y av}

jn

(b) 1. The magnitude of

v

S_av

is found from the Pythagorean theorem:

v

_av

_{! 4(v}

_{x av}

)

2

_#

_(v

0.199 m/s

y av

)

2

!

2. The ratio of

to

gives the tangent of the angle

$

between

and the #x direction (we add 180° to the value of "33.0°

returned by the calculator because v

_x

is negative):

v

S_av

v

_{x av}

v

_{y av}

so

147°

! "

33.0° # 180° !

u !

tan

"1

v

_{y av}

v

_{x av}

!

tan

"1

0.108 m/s

"0.167 m/s

tan u !

v

y av

v

_{x av} F I G U R E 3 - 4 15

(16)

2.2 Movimento 2 e 3-D

Uma partícula p move-se com velocidade , relativamente a um referencial A. Por sua

vez, este move-se com velocidade ,

relativamente ao referencial B. A velocidade da partícula relativamente ao referencial B é

!

v

_pA

!

v

_AB

66 |

C H A P T E R 3

Motion in Two and Three Dimensions

CHECK

The magnitude of

is greater than the absolute value of either its x or its y

com-ponent. With t in seconds, the units for the y component of in Part (c) are m!s/s

2

_"

_m/s,

which are appropriate units for velocity.

v

S

v

S_av

(c) We find the instantaneous velocity by calculating dx/dt and dy/dt:

v

S

(0.500 m/s)in #

360 m

#

s

t

2

jn

"

v

S

"

dx

dt

in $

dy

dt

jn " b

2

in # c

2

t

#2

jn

RELATIVE VELOCITY

If you are sitting in an airplane that is moving with a

ve-locity of 500 mi/h toward the east, your veve-locity is the same

as that of the airplane. This velocity might be your velocity

relative to the surface of Earth, or it might be your velocity

relative to the air outside the airplane. (These two relative

velocities would be very different if the plane were flying in

a jet stream.) In addition, your velocity relative to the

air-plane itself is zero.

The surface of Earth, the air outside the plane, and the

plane itself are frames of reference. A frame of reference (or

reference frame) is an extended object or collection of

ob-jects whose parts are at rest relative to each other. To

spec-ify the velocity of an object requires that you specspec-ify the

frame of reference that the velocity is relative to.

We use coordinate axes that are attached to reference frames to

make position measurements. (A coordinate axis is said be attached to

a reference frame if the coordinate axis is at rest relative to the

refer-ence frame.) For a horizontal coordinate axis attached to the plane,

your position remains constant. (At least it does if you remain in your

seat.) However, for a horizontal coordinate axis attached to the surface

of Earth, and for a horizontal coordinate axis attached to the air

out-side the plane, your position keeps changing. (If you have trouble

imagining a coordinate axis attached to the air outside the plane,

in-stead imagine a coordinate axis attached to a balloon that is

sus-pended in, and drifting with, the air. The air and the balloon are at rest

relative to each other, and together they form a single reference

frame.)

If a particle p moves with velocity

relative to reference

frame A, which is in turn moving with velocity

relative to

reference frame B, the velocity

of the particle relative to

refer-ence frame B is related to

and

by

3-9

For example, if a person p is on a railroad car C that is moving with

velocity

relative to the ground G (Figure 3-5a), and the person is

walking with velocity

(Figure 3-5b) relative to the car, then the

ve-locity of the person relative to the ground is the vector sum of these

two velocities:

(Figure 3-5c).

The velocity of object A relative to object B is equal in magnitude

and opposite in direction to the velocity of object B relative to

object A. For example,

is equal to

, where

is the velocity

of the person relative to the car, and

is the velocity of the car

rel-ative to the person.

v

S_Cp

v

S_pC

#

v

S_Cp

v

S_pC

v

S_pG

"

v

S_pC

$

v

S_CG

v

S_pC

v

S_CG

v

S_pB

"

v

S_pA

$

v

S_AB

v

S_AB

v

S_pA

v

S pB

v

S_AB

v

S_pA

(a)

(b)

(c)

v

_pC

_v

CG

v

_CG

v

_pC

v

pG

v

pC

v

CG

=

+

F I G U R E 3 - 5

The velocity of the person relative to the

ground is equal to the velocity of the person relative to

the car plus the velocity of the car relative the ground.

Midair refueling. Each plane is nearly at rest relative to the other, though

both are moving with very large velocities relative to Earth.

(Novastock/Dembinsky Photo Associates.)

(17)

!

e

_x

= ˆx = ˆi = (1,0,0) = versor da direção +x

!

e

_y

= ˆy = ˆj = (0,1,0) = versor da direção + y

!

e

_z

= ˆz = ˆ

k = (0,0,1) = versor da direção +z

!

A = A

_x

e

!

_x

_{+ A}

_y

e

!

_y

_{+ A}

_z

e

!

_z

=

A

!

_x

+

A

!

_y

+

A

!

_k

= A

_x

ˆx + A

_y

ˆy + A

_z

ˆz

= (A

_x

, A

_y

, A

_z

)

2.2 Movimento 2 e 3-D

(18)

Um barco está na posição (130 m, 205 m) no instante

t

₁=0,0 s. Dois minutos depois, está em (110 m, 218 m). 18 v_{x média} ₌ Δx Δt = 110 m −130 m 120 s = −0,167 m/s

!

v

_média

= v

_{x média}

ˆx + v

_{y média}

ˆy

v_{y média} = Δy Δt = 218 m − 205 m 120 s = 0,108 m/s ! v_média = (−0,167 m/s) ˆx + (0,108 m/s) ˆy v_média = (−0,167 m/s)2+ (0,108 m/s)2 = 0,199 m/s θ = arctg 0,108 m/s −0,167 m/s = 147 o

2.2 Movimento 2 e 3-D

(19)

!

r = x ˆx + yˆy

Δ

r =

!

r

!

₂

−

r

!

₁

= (x

₂

ˆx + y

₂

ˆy) − (x

₁

ˆx + y

₁

ˆy)

= Δxˆx + Δy ˆy

2.2.1 Vetor deslocamento

(20)

2.2.2 Vetor velocidade

Δ

r =

!

r

!

₂

−

r

!

₁

= (x

₂

ˆx + y

₂

ˆy) − (x

₁

ˆx + y

₁

ˆy)

= Δx ˆx + Δy ˆy

!

v

_média

₌

Δ

!

r

Δt

!

v = lim

Δt→0

Δ

r

!

Δt

!

v = lim

Δt→0

Δ

r

!

Δt

= lim

Δt→0

Δx ˆx + Δy ˆy

Δt

= lim

Δt→0

Δx

Δt

⎛

⎝

⎜

⎞

⎠

⎟ ˆx + lim

Δt→0

Δy

Δt

⎛

⎝

⎜

⎞

⎠

⎟ ˆy

!

v = v

_x

ˆx + v

_y

ˆy

_{v = v}

x

2 + v

2 y

;

θ

= arctg

v

_y

v

_x

20

(21)

2.2.3 Vetor aceleração

!

a

_média

₌

Δ

!

v

Δt

;

!

a = lim

Δt→0

Δ

v

!

Δt

!

v = v

_x

ˆx + v

_y

ˆy + v

_z

ˆz = lim

Δt→0

Δx

Δt

ˆx + Δ

y

Δt

ˆy + Δ

z

Δt

ˆz

⎛

⎝

⎜

⎞

⎠

⎟

!

a = lim

Δt→0

Δv

_x

Δt

ˆx +

Δv

_y

Δt

ˆy +

Δv

_z

Δt

ˆz

⎛

⎝

⎜

⎞

⎠

⎟

= a

x

ˆx + a

y

ˆy + a

z

ˆz

a

_x

_{= lim}

Δt→0

Δv

_x

Δt

; a

y

= lim

Δt→0

Δv

_y

Δt

; a

z

= lim

Δt→0

Δv

_z

Δt

(22)

22 A aceleração é independente da direção da velocidade inicial.

2.3 Lançamento de projétil

•  A resistência do ar é desprezada.

•  g

= 9,80 m/s2_{, dirigida} para baixo.

x(t) = x

₀

+ v

_0x

t

y(t) = y

₀

+ v

_{0 y}

t −

1 2

gt

2

v

_x

= v

_0x

v

_y

_{= v}

_{0 y}

_{− gt}

v

_x2

_{= v}

_0x2

v

2_y

_{= v}

_{0 y}2

_{− 2gΔy}

(23)

Admita que o projétil é deixado cair com uma velocidade inicial horizontal: para um observador em repouso, descreve uma trajetória curva, que

combina movimentos horizontal e vertical.

2.3.1 Independência dos movimentos

(24)

24

2.3.2 Forma vetorial

a

_x

= 0

a

_y

_{= −g}

a

_z

= 0

⇒ a

_x

ˆx + a

_y

ˆy + a

_z

ˆz = −gˆy ou

a = −

!

g, com

!

g = 9,81 m/s

2

(ao nível do mar e à latitude de 45°)

v

_x

= v

_0x

v

_y

= v

_{0 y}

− gt

⇒

v =

!

v

!

₀

+

gt ou Δ

!

v =

!

gt, com

!

v = v

_x

ˆx + v

_y

ˆy + v

_z

ˆz

!

v

₀

_{= v}

_0x

_{ˆx + v}

_{0 y}

_{ˆy + v}

_0z

_{ˆz; !g = −gˆy}

x(t) = x

₀

+ v

_0x

t

y(t) = y

₀

+ v

_{0 y}

t −

1₂

gt

2

⇒

r =

!

r

!

₀

+

v

!

₀

t +

1 2

!

gt

2

ou Δ

r =

!

v

!

₀

_{t +}

1 2

!

gt

2

, com

!

r = xˆx + yˆy + zˆz

!

r

₀

_{= x}

₀

_{ˆx + y}

₀

_{ˆy + z}

₀

ˆz

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2.3.3 Ângulo de lançamento

Ângulo de lançamento: direção da velocidade inicial, relativamente à

horizontal.

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26

horizontal.

Exemplo: em qual destes casos a velocidade de chegada à água é maior?

v

_água

2 _{= v}

₀

2 _{+ 2gh}

v

_x2

_{= v}

_0x2

v

2_y

_{= v}

_{0 y}2

_{− 2gΔy}

v

_água

= v

_x

2 + v

2 _y

v

_x2

_{= v}

₀2

v

2_y

_{= 0 − 2g 0 − h}

₍

₎

_{= 2gh}

2.3.3 Ângulo de lançamento

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horizontal.

v

_x2

_{= v}

_0x2

v

2_y

_{= v}

_{0 y}2

_{− 2gΔy}

v

_água

= v

_x

2 + v

2 _y

v

_x2

_{= v}

₀2

cos

2

_θ

v

2_y

_{= v}

₀2

sen

2

_θ

_{+ 2gh}

v

_água

2 _{= v}

₀

2 (

cos

2 _θ

_{+ sen}

2 _θ

)

_{+ 2gh}

= v

₀

2 + 2gh

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28

horizontal.

v

_água2

= v

₀2

+ 2gh

v

_água2

= v

₀2

+ 2gh

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2.3.3.1 Ângulo zero

Se o ângulo de lançamento for igual a zero, a velocidade inicial na direção de