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Two Dimensional Motion

Chapters 3 & 4

VECTORS Geometric/Graphical Representation

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Vectors

• For 1-D vectors size is designated by a number and direction is designated by a sign. – E.g. +23m/s, -9.8m/s2

• For 2-D vectors size is designated by a number

and direction is designated by an angle. – E.g. 23 m/s @ 65°, 9.8 m/s @ -90°

• Alternatively, 2-D vectors can be designated by a

pair of signed numbers (components).

Graphical Representation

Vectors can be represented graphically by directed rays (i.e. arrows). The length of the arrow is proportional to the magnitude of the vector and the direction the arrow points is the direction the vector points.

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Adding Vectors Geometrically

If you move from point A to point B, and then from point B to point C, you have moved two displacements: AB and BC. Your net displacement is given by the vector sum, or resultant, AC. We can represent the relation among the three vectors by the following vector equation: bas

Adding Vectors Geometrically

With a ruler and protractor:

1. Sketch vector a to some convenient scale at the proper angle

2. Sketch vector b to the same scale, with its tail at the head vector a, again at the proper angle

3. The vector sum s is the vector that extends from the tail of a to the head of b.

4

Vector Addition: Commutative Law

The order of addition of two or more vectors does not matter. Adding a to b gives the same result as adding b to a.

abba

Vector Addition: Associative Law

When there are more than two vectors, we can group them in any order as we add them. We can add a and b and then add c to the resultant, or we can add b and c and then and a. We get the same result either way.

)()( cbacba

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Negative Vectors

The vector –b is a vector with the same magnitude as b, but the opposite direction. Adding the two vectors would yield zero.

0)( bb

Vector Subtraction

Adding –b has the same effect as subtraction b We use this property to define the difference between two vectors.

)( babad

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Checkpoint 1 The magnitudes of displacements a and b are 3 m and 4 m respectively, and c =a + b. Considering various orientations of a and b, what is: a) the maximum possible magnitude c? b) the minimum possible magnitude?

Sample 1 In orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three straight-line moves. You may use the following displacements in any order: a = 2.0 km due east b = 2.0 km 30° north of east c = 1.0 km due west Alternatively, you may substitute either –b for b or –c or c. What is the greatest distance you can be from base cap at the end of the third displacement?

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VECTORS Analytic/Numerical Representation

Vector Components

Vector components are projections of a vector onto a perpendicular axes. The components, ax and ay , of vector a are shown.

The process of finding the components of a vector is called resolving the vector.

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Vector Components

When the components of a vector are moved to form a head-to-toe chain, we see that a right triangle is formed whose hypotenuse is the magnitude of the vector.

Vector Components

Trigonometry defines three basic relationships for right triangles:

hypotenuse

q

opposite

adjacent

cos(q) = a/h sin(q) = o/h tan(q) = o/a

SOH CAH TOA

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Checkpoint 2 In the figure, which of the indicated methods of combining the x and y components of vector a are proper to determine that vector?

Sample 2 A small airplane leaves an airport on an overcast day and is later sighted 215 km away, in a direction making an angle of 22° east of due north. How far east and north is the airplane from the airport when sighted?

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Sample 3 A velocity vector is 25 m/s at 53° south of east. What are the sizes and signs of the x- and y-components of this velocity?

Component Notation

The components of a vector can be used to describe a vector just as accurately as its magnitude and angle. This is known as component notation.

Magnitude-angle notation: a = 100 m @ 30°

Component notation: a = <86.7m, 50m>

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Combining Components

Because the components of a vector form a right triangle with the vector itself, we can use the following geometric functions to combine the components of a vector, or change its components into a magnitude and an angle.

x

y

yx

a

a

aaa

qtan

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Sample 4 For two decades, spelunking teams sought a connection between the Flint Ridge cave system and the Mammoth Cave, which are in Kentucky. When the connection was finally discovered, the combined system was declared the world’s longest cave (more than 200 km long). The team that found the connection had to crawl, climb, and squirm through countless passages, traveling a net 2.6 km westward, 3.9 km southward, and 25 m upward. What was their displacement from start to finish?

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Angles – Degrees and Radians

Angles that are measured relative to the positive direction of the x axis are positive in the counterclockwise direction and negative if measured clockwise. For example 210° and -150° are the same angle. Angles may be measured in degrees or radians (rad). To relate the two measures, recall that a full circle is 360° and 2π rad. Ex: 40° = 0.70 rad

70.0

360

240

Inverse Trig Functions When the inverse trig functions are taken on a calculator, you must consider the reasonableness of the answer you get, because there is usually another possible answer that the calculator does not give. The range of operation for a calculator in taking each inverse trig function is indicated on the graphs. The correct answer is the one that seems more reasonable for the given situation.

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Unit Vectors

• A unit vector is a vector that has a magnitude of exactly 1 and points in a particular direction. It lacks both dimension and unit. Its sole purpose is to point – that is, to specify a direction.

• The unit vectors in the positive directions of the x, y, and z axes are labeled and .

• The arrangement of axis shown below is known as the right-handed coordinate system and we will be using it exclusively.

,ˆ,ˆ ji k̂

Unit-Vector Notation

Unit vectors can be used to express other vectors via the scalar components of those vectors. This is known as unit-vector notation.

Magnitude-angle notation: a = 100 m @ 30°

Component notation: a = <86.7 m, 50 m>

Unit-vector notation: a = (86.7 m)î + (50 m)ĵ

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Adding (and Subtracting) Vectors by Components

Vectors can be added and subtracted analytically using their components.

1. Resolve the vectors into their scalar components

2. Combine the scalar components, axis by axis, to get the components of the sum

3. Combine the components and express in magnitude-angle notation, component notation, or unit-vector notation.

jAiAA yxˆˆ

jBiBB yxˆˆ

jCiCC yxˆˆ

jCBAiCBACBA

jBAiBABA

yyyxxx

yyxx

ˆ)(ˆ)(

ˆ)(ˆ)(

Checkpoint 3 (a) In the figure, what are the signs of the x components of d1 and d2? (b) What are the signs of the y components d1 and d2? (c) What are the signs of the x and y components of d1 + d2?

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Sample 5 The shows the following three vectors: What is their vector sum r?

jmc

jmimb

jmima

ˆ)7.3(

ˆ)9.2(ˆ)6.1(

ˆ)5.1(ˆ)2.4(

Sample 6 A fellow camper is to walk away from you in a straight line (vector A), turn, walk in a second straight line (vector B) and then stop. How far must you walk in a straight line (vector C) to reach her? The three vectors are related by C = A + B. A has a magnitude of 22.0 m and is directed at an angle of -47° from the positive direction of an x axis. B has a magnitude of 17.0 m and is directed counterclockwise from the positive direction of the x axis by angle φ. C is in the positive direction of the x axis. What is the magnitude of C?

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Sample 7 The desert ant Cataglyphis fortis lives in the plains of the Sahara desert. When one of the ants forages for food, it travels from its home nest along a haphazard search path. The ant may travel more than 500 m along such a complicated path over flat, featureless sand that contains no landmarks. Yet, when the ant decides to return home, it turns and then runs directly home. How does the ant know the way home with no guiding clues on the desert plain? According to experiments, the desert ant keeps track of its movements along a mental coordinate system. When it wants to return to its home nest, it effectively sums its displacements along the axes of the system to calculate a vector that points directly home. As an example of the calculation, let’s consider an ant making five runs of 6.0 cm each on an xy coordinate system, in the direction shown in the figure on the next slide, starting from home. At the end of the fifth run, what are the magnitude and angle of the ant’s net displacement vector dnet? And what are those of the homeward vector dhome

that expends from the ant’s final position back to home.

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MULTIPLYING VECTORS

Multiplying a Vector by a Scalar

Magnitude:

Direction:

Same as original vector if scalar is positive

Opposite original vector is scalar is negative

baba

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Multiplying a Vector by a Vector

Two ways:

One results in a scalar (Scalar Product)

One results in a new vector (Vector Product)

Scalar Product (Dot Product)

A dot product can be regarded as the product of two quantities: (1) the magnitude of one of the vectors and (2) the scalar component of the second vector along the direction of the first vector.

cosabba

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Scalar Product (Dot Product)

• If the angle φ between the two vectors is 0°, the component of one vector along they other is maximum, and so also is the dot product of the vectors.

• If φ is 90°, the component of one vector along the other is zero, and so is the dot product.

)ˆˆˆ()ˆˆˆ( kbjbibkajaiaba zyxzyx

zzyyxx babababa

Checkpoint 4 Vectors C and D have magnitudes of 3 units and 4 units, respectively. What is the angle between the directions of C and D if C ∙ D equals (a) zero (b) 12 units (c) -12 units

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Sample 8 What is the angle φ between and ? jia ˆ0.4ˆ0.3

kib ˆ0.3ˆ0.2

Vector Product (Cross Product)

• The vector product produces a new vector that is perpendicular to both of the original vectors, i.e. perpendicular to the plan that contains both of the original vectors.

• If a and b are parallel or antiparallel, a x b = 0. • The magnitude of a x b, is maximum when a and b

are perpendicular to each other.

qsinabba

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Right-hand Rule The direction of the resultant vector of a cross product can be found using the right-hand rule. 1. Place the two vectors you are

“crossing” tail to tail without altering their orientations

2. Place your right hand at the origin of the two vectors such that you fingers would sweep from the first vector into the second through the smallest angle between them.

3. Your outstretched thumb points in the direction of the vector product.

Checkpoint 5 Vectors C and D have magnitudes of 3 units and 4 units, respectively. What is the angle between the directions of C and D if the magnitude of the vector product C x D is (a) zero (b) 12 units (c) -12 units

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Sample 9 Vector a lies in the xy plane, has a magnitude of 18 units and points in a direction 250° from the +x direction. Also, vector b has a magnitude of 12 units and points in the +z direction. What is the vector product c = a x b?

Sample 10 If and , what is ? kib ˆ3ˆ2

jia ˆ4ˆ3

bac

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Common Errors with Cross Products

1. Failure to arrange vectors tail to tail is tempting when an illustration presents them head to tail. Redraw one vector to the proper arrangement.

2. Failing to us the right hand in applying the right-hand rule is easy when the right hand is occupied with a calculator or pencil.

3. Failure to sweep the first vector of the product into the second vector can occur when the orientations of the vectors require an awkward twisting of your hand to apply the right-hand rule. Sometimes that happens when you try to make the sweep mentally rather than actually using your hand.

4. Failure to work with the right-handed coordinate system results when you forget how to draw such a system. Practice drawing other prespectives.

MOTION QUANTITIES IN VECTOR FORM

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Position

The general way of locating a particle is with a position vector r, which is a vector that extends from a reference point (or origin) to the particle.

kzjyixr ˆˆˆ

Displacement

If a particles position vector changes during a certain time interval, then the particles’ displacement vector Δr is given by:

kzjyixr

kzzjyyixxr

rrr

ˆˆˆ

ˆ)(ˆ)(ˆ)( 121212

12

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Sample 11 The position vector for a particle initially is And then later is . What is the particle’s displacement Δr from r1 to r2?

kmjmimr ˆ)0.5(ˆ)0.2(ˆ)0.3(1

kmjmimr ˆ)0.8(ˆ)0.2(ˆ)0.9(2

Sample 12 A rabbit runs across a parking lot on which a set of coordinate axes has, strangely enough, been drawn. The coordinates of the rabbit’s position as functions of time are given by: (a) At t = 15 s, what is the rabbit’s position vector r in unit-vector

notation and in magnitude angle notation? (b) Graph the rabbits path for t = 0 to t = 25s.

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Average Velocity

Average velocity = displacement/time interval

kt

zj

t

yi

t

xv

t

rv

avg

avg

ˆˆˆ

Instantaneous Velocity

The direction of the instantaneous velocity v of a particle is always tangent to the particle’s path at the particle’s position

kvjvivv

kdt

dzj

dt

dyi

dt

dxv

dt

rdv

zyxˆˆˆ

ˆˆˆ

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Checkpoint 7 The figure shows a circular path taken by a particle. If the instantaneous velocity of the particle is , through which quadrant is the particle moving at that instant if it is traveling (a) clockwise and (b) counterclockwise around the circle? For both cases, draw v on the figure.

jsmismv ˆ)/2(ˆ)/2(

Sample 13 For the rabbit in Sample 12, find the velocity v at time t= 15s, in unit-vector notation and in magnitude-angle notation.

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Average Acceleration

Average acceleration = change in velocity/time interval

t

v

t

vvaavg

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Instantaneous Acceleration

kajaiaa

kdt

dvj

dt

dvi

dt

dva

dt

vda

zyx

zyx

ˆˆˆ

ˆˆˆ

Caution: When an acceleration vector is drawn it does not extend from one position to another. Rather, it shows the direction of acceleration for a particle located at its tail, and its length (representing the acceleration magnitude) can be drawn to any scale.

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Checkpoint 8 Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane: (1) and (2) and (3) (4) Are the x and y acceleration components constant? Is acceleration a constant?

jittr

jtitr

ttx

ttx

ˆ3ˆ)24(

ˆ)34(ˆ2

43

243

3

2

3

2

65

46

2

2

ty

tty

Sample 14 For the rabbit in Example 12 & 13, find the acceleration a at time t = 15 s, in unit-vector notation and in magnitude-angle notation.

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Sample 15 A particle with velocity (in meters per second) at t = 0 undergoes a constant acceleration a of magnitude a = 3.0 m/s2 at an angle θ = 130° from the positive direction of the x axis. What is the particle’s velocity v at t = 5.0 s?

jiv ˆ0.4ˆ0.20

PROJECTILE MOTION

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Projectile Motion

A projectile is a particle that moves in a vertical plane with some initial velocity v0 but its acceleration is always the free-fall acceleration g = 9.8 m/s2.

For the time being, we will ignore any effects due to air resistance.

Checkpoint 9 Which of the following objects undergo projectile motion? (a) Tennis ball (b) Airplane (c) Diver (d) Frog (e) Bird (f) Rocket

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Projectile Motion

A projectile is launched with an initial velocity v

The components can then be found if we know the angle θ0 between v0 and the positive x direction.

jvivv yxˆˆ

000

000

000

sin

cos

q

q

vv

vv

y

x

Axes Independence

During its two-dimensional motion, the projectile’s position vector r and velocity vector v change continuously, but its acceleration vector a is constant and always directed vertically downward. There is NO horizontal acceleration. In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. Projectile motion can be broken into two separate one-dimensional problems, one for the horizontal motion (constant motion problem) and one for the vertical motion (freefall/constant accelerated motion problem)

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Checkpoint 10 At a certain instant, a fly ball has velocity (the x axis is horizontal, the y axis is upward, and v is in meters per second). Has the ball passed its highest point?

jiv ˆ9.4ˆ25

Projectile Motion – Horizontal

Because there is no acceleration in the horizontal direction, the horizontal component vx of the projectile’s velocity remains unchanged from its initial value v0x throughout the motion.

tvxx

tvxx x

)cos( 000

00

q

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Projectile Motion – Vertical

The vertical motion is identical to freefall, and has a constant acceleration.

)(2)sin(

sin

2

1)sin(

2

1

0

2

00

2

00

2

000

2

00

yygvv

gtvv

gttvyy

gttvyy

y

y

y

q

q

q

Trajectory Equation

2

00

2

0)cos(2

)(tanq

qv

gxxy

0

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Horizontal Range The horizontal range R of a projectile is the horizontal distance the projectile has traveled when it returns to its initial (launch) height. The horizontal range R is maximum for a launch angle of 45°, for these conditions. Caution: This equation does not give the horizontal distance traveled by a projectile when the final height is not the launch height

0

2

0 2sin qg

vR

Checkpoint 11 A fly ball is hit to the outfield. During its flight (ignore the effects of the air), what happens to its (a) horizontal and (b) vertical components of velocity? What are the (c) horizontal and (d) vertical components of its acceleration during ascent, during descent and at the topmost point of its flight?

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Sample 16 A pirate ship is 560 m from a fort defending a harbor entrance. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/s. (a) At what angle θ0 from the horizontal must a ball be fired to hit the

ship? (b) What is the maximum range of the cannonballs?

Sample 17 A rescue plane flies at 198 km/h and constant height h = 500 m toward a point directly over a victim, where a rescue capsule is to land. (a) What should be the angle φ of the pilot’s line of sight to the victim

when the capsule release is made? (b) As the capsule reaches the water, what is its velocity v in unit-vector

notation and in magnitude-angle notation?

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Sample 18 The figure illustrates the ramps for the current world-record motorcycle jump, set by Jason Renie in 2002. The ramps were H = 3.00 m high, angled at θR = 12.0°, and separated by distance D = 77.0 m. Assuming that he landed halfway down the landing ramp and that the slowing effects of the air were negligible, calculate the speed at which he left the launch ramp.

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UNIFORM CIRCULAR MOTION

Uniform Circular Motion

• A particle is in uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed. Although the speed does not vary, the particle is accelerating because the velocity changes in direction.

• This acceleration is always directed radially inward, while the velocity vector at any instant is tangential to the circle.

• Due to this fact, the acceleration is known as centripetal “center seeking” acceleration.

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Uniform Circular Motion Equations

v

rT

r

vac

2

2

Centripetal acceleration

Period

The period of revolution is the time for a particle to go around a closed path exactly once.

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Checkpoint 12 An object moves at constant speed along a circular path in a horizontal xy plane, with the center at its origin. When the object is at x = -2 m, its velocity is –(4 m/s)ĵ. Give the object’s (a) velocity and (b) acceleration at y = 2m.

Sample 19 “Top gun” pilots have long worried about taking a turn too tightly. As a pilot’s body undergoes centripetal acceleration, with head toward the center of the curvature, the blood pressure decreases, leading to loss of brain function. There are several warning signs to signal a pilot to ease up. When the centripetal acceleration is 2g or 3g, the pilot feels heavy. At about 4g, the pilot’s vision switches to back and white and narrows to “tunnel vision.” If that acceleration is sustained or increased, vision ceases and, soon after, the pilot is unconscious – a condition known as g-LOC for “g-induced loss of consciousness.” What is the centripetal acceleration, in g units, of a pilot flying an F-22 at speed v = 2500 km/h through a circular arc having a radius of curvature r = 5.80 km?

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RELATIVE MOTION

Relative Motion

Suppose you are riding on the school bus. You are watching the trees pass outside of your window. You “know” that you are moving and the trees are stationary, but it “looks” like the trees are moving backward relative to you. Science says that EITHER of these interpretations, known as reference frames, are valid and the laws of physics apply to them both. Most of the time, we choose a reference frame to be the ground because it is “not” moving.

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Relative Motion in One Dimension

Suppose that a police officer, Alex, is parked by the side of a highway. We will place the origin of a stationary reference frame (Frame A) at this point. A second police officer, Barbara, is driving along the highway at a constant speed. We add a second reference frame (Frame B). Frame B is moving with the same constant speed as Barbara so that she is always at the origin. Suddenly, a Porsche (P) speeds by and both officers measure the position of the car at the same instant.

Relative Motion in One Dimension By interpreting the situation we can see that: The coordinate of the Porsche as measured by Alex is equal to the coordinate of the Porsche as measured by Barbara plus the coordinate of Barbara as measured by Alex. Or, if we take the derivative of the above equation: The velocity of the Porsche as measured by Alex is equal to the velocity of the Porches as measured by Barbara plus the velocity of Barbara as measured by Alex. If we take another derivative we find that both Alex and Barbara measure the same acceleration of the Porsche. Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving particle.

BAPBPA xxx

BAPBPA vvv

PBPA aa

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Sample 20 Suppose that Barbara’s velocity relative to Alex is a constant vBA = 52 km/h and the Porsche (P) is moving in the negative direction of the x axis. (a) If Alex measures a constant vPA = -78 km/h for P, what velocity vPB will

Barbara measure? (b) If P brakes to a stop relative to Alex in time t = 10 s at constant

acceleration, what is its acceleration aPA relative to Alex? (c) What is the acceleration aPB of P relative to Barbara during the

braking?

Checkpoint 13 The table gives velocities (km/h) for Barbara and the Porsche for three situations. For each, what is the missing value and how is the distance between Barbara and the Porsche changing?

Situation vBA vPA vPB

(a) +50 +50

(b) +30 +40

(c) +60 -20

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Relative Motion in Two Dimensions

We can analyze relative motion in two dimensions in the same way we did in one dimensions except that we must now use position and velocity vectors instead of just 1D coordinates.

PBPA

BAPBPA

BAPBPA

aa

vvv

rrr

The same rule applies: Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving particle.

Sample 21 A plane moves due east while the pilot points the plane somewhat south of east, toward a steady wind that blows to the northeast. The plane has velocity vPW relative to the wind, with an airspeed (speed relative to the wind) of 215 km/h, directed at angle θ south of east. The wind has velocity vWG relative to the ground with speed 65.0 km/h, directed 20.0° east of north. What is the magnitude of the velocity vPG of the plane relative to the ground, what is θ?

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Checkpoint 14 In Sample 19, suppose the pilot turns the plane to point it due east without changing the airspeed. Do the following magnitudes increase, decrease, or remain the same: (a) vPG,y, (b) vPG,x, (c) vPG? (You can answer without computation.)