1 Chapter 3 Motion in Two Dimensions 2 3 3.1 Position and Displacement The position of an object is...

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Chapter 3

Motion in Two Dimensions

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3.1 Position and Displacement The position of an

object is described by its position vector,

The displacement of the object is defined as the change in its position

Fig 3.1

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Average Velocity The average velocity is

the ratio of the displacement to the time interval for the displacement

The direction of the average velocity is the direction of the displacement vector, Fig 3.2

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Average Velocity, cont The average velocity between points is

independent of the path taken This is because it is dependent on the

displacement, which is also independent of the path

If a particle starts its motion at some point and returns to this point via any path, its average velocity is zero for this trip since its displacement is zero

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Instantaneous Velocity The instantaneous velocity is the limit of

the average velocity as ∆t approaches zero

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Instantaneous Velocity, cont The direction of the instantaneous

velocity vector at any point in a particle’s path is along a line tangent to the path at that point and in the direction of motion

The magnitude of the instantaneous velocity vector is the speed

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Average Acceleration The average acceleration of a particle

as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs.

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Average Acceleration, cont

As a particle moves, can be found in different ways

The average acceleration is a vector quantity directed along

v

Fig 3.3

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Instantaneous Acceleration The instantaneous acceleration is the

limit of the average acceleration as approaches zero

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3.2 Producing An Acceleration Various changes in a particle’s motion

may produce an acceleration The magnitude of the velocity The direction of the velocity

Even if the magnitude remains constant Both change simultaneously

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Kinematic Equations for Two-Dimensional Motion When the two-dimensional motion has a

constant acceleration, a series of equations can be developed that describe the motion

These equations will be similar to those of one-dimensional kinematics

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Kinematic Equations, 2 Position vector

Velocity

Since acceleration is constant, we can also find an expression for the velocity as a function of time:

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Kinematic Equations, 3 The velocity vector

can be represented by its components

is generally not along the direction of either or

Fig 3.4(a)

fi

xf xi x

yf yi y

t

v v a t

v v a t

v v a

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Kinematic Equations, 4 The position vector can also be

expressed as a function of time:

This indicates that the position vector is the sum of three other vectors:

The initial position vector The displacement resulting from The displacement resulting from

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Kinematic Equations, 5 is generally not in

the same direction as or as

and are generally not in the same direction

Fig 3.4(b)

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212

212

fi i

fi xi x

fi yi y

t t

x x v t a t

y y v t a t

r r v a

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Active Figure3.4

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SolutionConceptualize

Establish the mental representation and thinking about what the particle is doing.

Categorize Consider that because the acceleration is only in the x direction, the moving particle can be modeled as one under constant acceleration in the x direction and one under constant velocity in the y direction.

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AnalyzeIdentify vxi = 20 m/s and ax = 4.0 m/s2. The equations of kinematics for the x direction,

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Solution At t = 5.0 s, the velocity expression from part A gives

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拋射 3.3 Projectile Motion An object may move in both the x and y

directions simultaneously The form of two-dimensional motion we

will deal with is called projectile motion

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假設Assumptions of Projectile Motion

The free-fall acceleration is constant over the range of motion and is directed downward

The effect of air friction is negligible

With these assumptions, an object in projectile motion will follow a parabolic path

This path is called the trajectory This is the simplification model that will be used

throughout this chapter

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Verifying the Parabolic Trajectory Reference frame chosen

y is vertical with upward positive Acceleration components

ay = -g and ax = 0

Initial velocity components vxi = vi cos and vyi = vi sin

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Projectile Motion – Velocity Equations The velocity components for the

projectile at any time t are: vxf = vxi = vi cosi = constant

vyf = vyi – gt = vi sini – gt

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Projectile Motion – Position Displacements

Combining the equations gives:

This is in the form of y = ax – bx2 which is the standard form of a parabola

2 21 12 2

( cos )

( sin )

fi xi i i

fi yi i i

x x v t v t

y y v t gt v t gt

22 2(tan )

2 cosfi ffi i

gy x x

v

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Projectile Motion Vectors

Fig 3.6

212fi it t r r v g

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Projectile Motion Diagram

Fig 3.5

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Active Figure3.5

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Projectile Motion – Implications The y-component of the velocity is zero

at the maximum height of the trajectory The accleration stays the same

throughout the trajectory

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Horizontal Range and Maximum Height of a Projectile

When analyzing projectile motion, two characteristics are of special interest

The range, R, is the horizontal distance of the projectile

The maximum height the projectile reaches is h

Fig 3.7

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Height of a Projectile, equation The maximum height of the projectile

can be found in terms of the initial velocity vector:

2 2sin2

i ivh

g

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Range of a Projectile, equation The range of a projectile can be

expressed in terms of the initial velocity vector:

This is valid only for symmetric trajectory

2 sin2i ivR

g

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More About the Range of a Projectile

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Range of a Projectile, final

The maximum range occurs at i = 45o

Complementary angles will produce the same range The maximum height will be different for

the two angles The times of the flight will be different for

the two angles

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Active Figure3.8

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Projectile Motion – Problem Solving Hints Conceptualize

Establish the mental representation of the projectile moving along its trajectory

Categorize Confirm that the problem involves the

particle in free fall (in the y-direction) and air resistance can be neglected

Select a coordinate system

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Problem Solving Hints, cont. Analyze

Resolve the initial velocity into x and y components Remember to treat the horizontal motion independently

from the vertical motion Treat the horizontal motion using constant velocity

techniques Analyze the vertical motion using constant

acceleration techniques Remember that both directions share the same time

Finalize Check your results

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Non-Symmetric Projectile Motion Follow the general rules

for projectile motion Break the y-direction

into parts up and down or symmetrical back to

initial height and then the rest of the height

May be non-symmetric in other ways

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A stone is thrown from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, as in Figure 3.9.

Fig 3.9

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To find t, we use the vertical motion, in which we model the stone as a particle under constant acceleration. We use Equation 3.13 with yf = 45.0 m and vyi = 10.0 m/s (we have chosen the top of the building as the origin):

t = 4.22 s

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Solution The y component of the velocity just before the stone strikes the ground can be obtained using Equation 3.11, with t = 4.22 s:

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An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the pictorial representation in Figure 3.10. If the plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground, where does the package strike the ground relative to the point at which it is released?

Fig 3.10

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Solution Ignore air resistanceModel this problem as a particle in two-dimensional free-fall,a particle under constant velocity in the x direction anda particle under constant acceleration in the y direction

Define the initial position xi = 0 right under the plane at the instant the package is released.

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If we know t, the time at which the package strikes the ground, we can determine xf. To find t, we turn to the equations for the vertical motion of the package. At the instant the package hits the ground, its coordinate is 100 m.

The initial component of velocity vyi of the package is zero.

From Equation 3.13,

xf = (40.0 m/s)(4.52 s) = 181 m

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An athlete throws a javelin a distance of 80.0 m at the Olympics held at the equator, where g = 9.78 m/s2. Four years later the Olympics are held at the North Pole, where g = 9.83 m/s2. Assuming that the thrower provides the javelin with exactly the same initial velocity as she did at the equator, how far does the javelin travel at the North Pole?

50= 79.6 m

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3.4 Uniform Circular Motion Uniform circular motion occurs when an

object moves in a circular path with a constant speed

An acceleration exists since the direction of the motion is changing This change in velocity is related to an

acceleration The velocity vector is always tangent to the

path of the object

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Fig 3.11

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Centripetal Acceleration The acceleration is always

perpendicular to the path of the motion The acceleration always points toward

the center of the circle of motion This acceleration is called the

centripetal acceleration Centripetal means center-seeking

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Changing Velocity in Uniform Circular Motion The change in the

velocity vector is due to the change in direction

The vector diagram shows

Fig 3.11

fiv v v

fir r r

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avg

fi

fit t t

v v va

v rrv

avg

vt r t

rva

2

c

va

r

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Period The period, T, is the time interval

required for one complete revolution The speed of the particle would be the

circumference of the circle of motion divided by the period

Therefore, the period is

2 r

Tv

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What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?

Solution We model the Earth as a particle and approximate the Earth’s orbit as circular Although we don’t know the orbital speed of the Earth, with the help of Equation 3.18 we can recast Equation 3.17 in terms of the period of the Earth’s orbit, which we know is one year:

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3.5 Tangential Acceleration The magnitude of the velocity could also

be changing As well as the direction

In this case, there would be a tangential acceleration

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Total Acceleration

The tangential acceleration causes the change in the speed of the particle

The radial acceleration comes from a change in the direction of the velocity vector

Fig 3.12

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Total Acceleration, equations The tangential acceleration:

The radial acceleration:

The total acceleration: Magnitude The direction of the acceleration is the

same or opposite that of the velocity

2

r c

va a

r

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Active Figure3.12

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3.6 Relative Velocity Two observers moving relative to each other generally

do not agree on the outcome of an experiment For example, the observer on the side of the road

observes a different speed for the red car than does the observer in the blue car

Fig 3.13

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Relative Velocity, generalized

Reference frame S is stationary

Reference frame S’ is moving

Define time t = 0 as that time when the origins coincide

Fig 3.14

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Relative Velocity, equations The positions as seen from the two reference

frames are related through the velocity

The derivative of the position equation will give the velocity equation

This can also be expressed in terms of the observer O’

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A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The river has a current such that the water moves with uniform speed of 5.00 km/h due east relative to the ground.

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bwˆ10.0 km/hjv

wEˆ5.00 km/hiv

bE bw wE

ˆ ˆ(5.00 10.0 ) km/hi j

v v v

2 2bE bw wE

2 2 10.0 5.00 11.2 km/h

v v v

1 1wE

bw

5.00tan tan

10.0

26.6

vv

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1 1wE

bw

5.00sin sin

10.0

30.0

vv

2 2bE bw wE

2 2 10.0 5.00 8.66 km/h

v v v

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3.7 Acceleration of Autos The lateral acceleration

橫向加速度 is the maximum possible centripetal acceleration the car can exhibit without rolling over in a turn

The lateral acceleration depends on the height of the center of mass of the vehicle and the side-to-side distance between the wheels

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Beamon 於 1968 在 Mexico 創下 8.90 m 的跳遠紀錄，由照片得知，他起跳時重心高度約為 1.0 m ，在最高點處重心高度為 1.90 m ，落地時重心高度 0.15 m 。由這些數據，求 (a) 他在空中停留時間。 (b) 起跳時速度之水平分量及垂直分量。 (c) 起跳角度。

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An enemy ship is on the east side of a mountain island, as shown in Figure P3.61. The enemy ship has maneuvered to within 2500 m of the 1800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?

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Problem1, 3, 10, 13, 15, 19, 20, 27, 31, 36, 38, 40, 41, 48, 51, 55, 60

1 Chapter 3 Motion in Two Dimensions 2 3 3.1 Position and Displacement The position of an object is...

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