ch 03: kinematics in two or three …lightcat-files.s3.amazonaws.com/packets/admin_physics-3...if...

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PHYSICS - GIANCOLI CALC 4E

CH 03: KINEMATICS IN TWO OR THREE DIMENSIONS; VECTORS

CONCEPT: REVIEW OF VECTORS AND SCALARS

● Remember: When you take measurements, you always get the size (Magnitude, how much).

- Measurements with direction → [ Vectors | Scalars ]; without direction → [ Vectors | Scalars ]

Measurement Quantity Magnitude? Direction? Vector/Scalar

“It’s 60°F outside” Temperature [ Vector | Scalar ]

“I pushed with 100N north” Force [ Vector | Scalar ]

“I walked for 10 m” Distance [ Vector | Scalar ]

“I walked 10 m east” Displacement [ Vector | Scalar ]

“I drove at 80 mph” Speed [ Vector | Scalar ]

“I drove 80mph west” Velocity [ Vector | Scalar ]



CONCEPT: INTRO TO VECTOR MATH

● Adding/subtracting scalars is easy. But vectors have direction, so math with vectors is sometimes not as straightforward.

- Because vectors have direction, they’re drawn as ________________.

EXAMPLE: For each of the following situations, draw your displacement vectors and calculate the total displacement:

COMBINING

SCALARS

3 kg 4 kg

“You combine a 3kg & 4kg box”

COMBINING PERPENDICULAR

VECTORS

Total Mass: 3kg + 4kg = _______ Total Displacement: __________

“You walk 3m right, then 4m up”

● Forms _____________.

- just TRIANGLE MATH

● Simple Addition

(a) You walk 10m to the right, and then 6m to the left (b) You walk 6m to the right, and then 8m down

COMBINING PARALLEL VECTORS

“You walk 3m right, then 4m right”

Total Displacement: __________

● Add just like normal numbers



PRACTICE: Two perpendicular forces act on a box, one pushing to the right and one pushing up. An instrument tells you

the magnitude of the total force is 13N. You measure the force pushing to the right is 12N. Calculate the force pushing up.



CONCEPT: ADDING VECTORS GRAPHICALLY

● Vectors are drawn as arrows and are added by ______________ the arrows (tip-to-tail).

● The RESULTANT vector (�⃗⃗� or �⃗⃗� ) is always the SHORTEST PATH from the start of the first vector → end of the last.

- Adding vectors does NOT depend on the order (commutative), so �⃗⃗� + �⃗⃗� = �⃗⃗� + �⃗⃗� .

EXAMPLE: Find the magnitude of the Resultant Vector �⃗⃗� = �⃗⃗� + �⃗⃗� .

⇔

ADDING PERPENDICULAR VECTORS

3m

4m

ADDING ANY VECTORS

�⃗⃗�

�⃗⃗�

�⃗⃗� + �⃗⃗� �⃗⃗� + �⃗⃗�

𝒚

𝒙

�⃗⃗�

�⃗⃗�

Resultant Vector: (Total Displacement)

____________


____________


____________



PRACTICE: A delivery truck travels 8 miles in the +x-direction, 5 miles in the +y-direction, and 4 miles again in the

+x-direction. What is the magnitude (in miles) of its final displacement from the origin?

𝒚

𝒙



EXAMPLE: Find the magnitude of the Resultant Vector �⃗⃗� = �⃗⃗� + �⃗⃗� + �⃗⃗� .

𝒚

𝒙 −𝒙

−𝒚

�⃗⃗�

�⃗⃗�

�⃗⃗�



CONCEPT: SUBTRACTING VECTORS GRAPHICALLY

● Subtracting vectors is exactly like adding vectors tip-to-tail, but one (or more) of the vectors gets _______________.

EXAMPLE: Find the magnitude of the Resultant Vector �⃗⃗� = �⃗⃗� − �⃗⃗� .

⇔

ADDING VECTORS

�⃗⃗� − �⃗⃗� �⃗⃗� − �⃗⃗�

SUBTRACTING VECTORS

�⃗⃗� + �⃗⃗� �⃗⃗� + �⃗⃗�

● When adding, order [ DOES | DOES NOT ] matter

● “Negative” vector: SAME magnitude, ____________ direction

● When subtracting, order [ DOES | DOES NOT ] matter

𝒚

𝒙

𝒚

𝒙

𝒚

𝒙

�⃗⃗�

�⃗⃗�

𝒚

𝒙

Resultant → shortest path: (Total Displacement)

_______________


_______________


_______________

�⃗⃗�

�⃗⃗�

𝒚

𝒙

�⃗⃗�

�⃗⃗�



PRACTICE: Find the magnitude of the Resultant Vector �⃗⃗� = �⃗⃗� − �⃗⃗� − �⃗⃗� .

𝒚

𝒙 −𝒙

−𝒚

�⃗⃗� �⃗⃗�

�⃗⃗�



CONCEPT: ADDING MULTIPLES OF VECTORS

● When you multiply a vector by a number (𝐴 → 2𝐴 ), the magnitude (length) changes but NOT the direction.

EXAMPLE: Find the magnitude of the Resultant Vector �⃗⃗� = 𝟑�⃗⃗� − 𝟐�⃗⃗� .

𝒚

𝒙

ADDING VECTORS

ADDING MULTIPLES OF VECTORS

�⃗⃗�

�⃗⃗�

● Multiplying by > 1 [ increases | decreases ] magnitude/length

● Multiplying by < 1 [ increases | decreases ] magnitude/length

�⃗⃗� + �⃗⃗� 𝟐�⃗⃗� + 𝟎. 𝟓�⃗⃗� 𝒚

𝒙

Resultant Vector → Shortest Path: (Total Displacement)

_______________

𝒚

𝒙

�⃗⃗� �⃗⃗�

Resultant Vector → Shortest Path: (Total Displacement)

_______________



CONCEPT: VECTOR COMPOSITION AND DECOMPOSITION

● You’ll need to do vector math without using grids/ squares.

- Vectors have magnitude (length), direction (angle 𝜽𝒙), and components (legs).

EXAMPLE: For each of the following, draw the vector and solve for the missing variable(s).

VECTOR DECOMPOSITION

𝑨𝒙 = __________

𝑨𝒚 = __________

VECTOR COMPOSITION

𝑨 = √𝑨𝒙 𝟐 + 𝑨𝒚

𝟐

𝜽𝒙 = ____________

VECTOR COMPOSITION VECTOR DECOMPOSITION

● Use SOH-CAH-TOA to decompose �⃗⃗� →components 𝐴𝑥 & 𝐴𝑦.

- Angle 𝜽𝒙 must be drawn to nearest ________

1D Components → 2D Vector (Magnitude & Direction)

● Components 𝑨𝒙 & 𝑨𝒚 combine → magnitude �⃗⃗�

- Points in direction 𝜽𝒙

a) Ax = 8m, Ay = 6m, 𝑨 = ? θx = ? b) B = 13m, θx = 67.4°, Bx = ? By = ?

+𝒚

+𝒙 3

4

2D Vector (Magnitude & Direction) → 1D Components

θx=53°

5

+𝒚

+𝒙

𝒚

𝒙

𝒚

𝒙



EXAMPLE: A vector A has y-component of 12 m makes an angle of 67.4° with the positive x-axis. (a) Find the magnitude of

A. (b) Find the x-component of the vector.

Vector Composition

(Components→Vector)

Vector Decomposition

(Vector→Components)

𝑨 = √𝑨𝒙 𝟐 + 𝑨𝒚

𝟐

𝜽𝑿 = 𝐭𝐚𝐧−𝟏 (𝑨𝒚

𝑨𝒙)

𝑨𝒙 = 𝑨 𝒄𝒐𝒔(𝜽𝑿)

𝑨𝒚 = 𝑨 𝒔𝒊𝒏(𝜽𝑿)



CONCEPT: VECTOR ADDITION BY COMPONENTS

● You’ll need to add vectors together and calculate the magnitude & direction of the resultant without counting squares.

EXAMPLE: You walk 5m at 53° above the +x-axis, then 8m at 30° above the +x-axis. Calculate the magnitude & direction

of your total displacement.

VECTOR ADDITION

1) Draw & connect vectors tip-to-tail

2) Draw Resultant & components

3) Calculate ALL X&Y components

4) Combine X & Y components according to R equation

5) Calculate R and 𝜃𝑅

Vector Composition




𝑹 = √𝑹𝒙 𝟐 + 𝑹𝒚

𝟐

𝜽𝑿 = 𝐭𝐚𝐧−𝟏 (𝑹𝒚

𝑹𝒙)


𝑨𝒚 = 𝑨 𝒔𝒊𝒏(𝜽𝑿) x y

�⃗⃗�

�⃗⃗�

�⃗⃗� = ______

ADDING VECTORS GRAPHICALLY (WITH SQUARES)

ADDING VECTORS BY COMPONENTS (WITHOUT SQUARES)

+𝒚

+𝒙

+𝒚

+𝒙

�⃗⃗�

�⃗⃗�



EXAMPLE: Vector �⃗⃗� has a magnitude of 10m at a direction 40° above the +x-axis. �⃗⃗� has magnitude 3 at a direction 20°

above the x-axis. Calculate the magnitude and direction of �⃗⃗� = �⃗⃗� − 𝟐�⃗⃗� .

VECTOR ADDITION

1) Draw & connect vectors tip-to-tail

2) Draw Resultant & components

3) Calculate ALL X&Y components

4) Combine X & Y components according to R equation

5) Calculate R and 𝜃𝑅

Vector Composition




𝑹 = √𝑹𝒙 𝟐 + 𝑹𝒚

𝟐

𝜽𝑿 = 𝐭𝐚𝐧−𝟏 (𝑹𝒚

𝑹𝒙)


𝑨𝒚 = 𝑨 𝒔𝒊𝒏(𝜽𝑿)

𝒚

𝒙



CONCEPT: DOING MATH WITH VECTORS IN ANY QUADRANT (MORE TRIG)

● You’ll need to do math with vectors in ALL Quadrants, not just Quadrant 1.

EXAMPLE: Calculate a) the components and b) the absolute angle for the given vector �⃗⃗� :

Signs of Magnitudes & Components of Vectors:

● Magnitudes → Always positive, but Components may be + or − - Positive Components = pointing [ UP | DOWN ] or [ RIGHT | LEFT ]

- Negative Components = pointing [ UP | DOWN ] or [ RIGHT | LEFT ]

When given a Non-Reference Angle:

● Remember: We always use the Reference Angle 𝜽𝒙 to calculate components:

Ax = A cos(𝜽𝒙) Ay = A sin(𝜽𝒙)

- All right angles add up to 90°, so we’ll use this simple equation to get 𝜽𝒙:

Calculating the Absolute Angle (Positive Angle from +x axis) from the Arctangent

● Taking arctan of components [𝜃𝑥 = tan−1 (|𝐴𝑦|

|𝐴𝑥|)] always gives reference angle 𝜽𝒙.

- Remember to always plug in positive value of components!

- To find the Absolute Angle, work your way back to +x-axis (0°)

+𝒚

+𝒙

−𝒚

−𝒙

Quadrant 1 Quadrant 2

Quadrant 4 Quadrant 3

�⃗⃗� =5 �⃗⃗� =5

�⃗⃗� =5 �⃗⃗� =5

𝑨𝒙= 4

𝑨𝒚= 3

𝑩𝒙= 4 𝑩𝒚= 3

𝑫𝒙= 3

𝑫𝒚= 4 𝑪𝒙= 3

𝑪𝒚= 4

+𝒚

+𝒙

−𝒚

−𝒙

10°

�⃗⃗�

________________

+𝒚

+𝒙

−𝒚

−𝒙

�⃗⃗� =5

𝑨𝒙= −4

𝑨𝒚= +3

+𝒚

+𝒙

−𝒚

−𝒙

�⃗⃗� =13

22.6°



PRACTICE: FINDING VECTOR COMPONENTS

Vector F is 65 m long, directed 30.5° below the positive x-axis. (a) Find the x-component, Fx. (b) Find the y-component, Fy.



PRACTICE: VECTOR COMPOSITION IN ALL QUADRANTS

The vector A represented is by the pair of components Ax = -77 cm , Ay = 36 cm. (a) Find the magnitude of vector A. (b)

Find the absolute angle of this vector.



CONCEPT: DESCRIBING DIRECTIONS VECTORS WITH WORDS (MORE TRIG)

● Many problems will use different words to describe the directions of vectors:

1) Counterclockwise angles are [ + / − ]; Clockwise angles are [ + / − ]

- However, reference angle 𝜽𝒙 for component equations is always a positive angle relative to nearest x-axis.

2) Angles North/South/West/East (e.g. 30° north of east): Draw arrow in 2nd direction, curve towards 1st

EXAMPLE: Draw each vector and calculate the x-component

𝑵

𝑬

𝑺

𝑾

�⃗⃗�

�⃗⃗�

+𝒙

𝑵

𝑬

𝑺

𝑾

EXAMPLE: Draw each vector and calculate its components.

a) �⃗⃗� = 5m @ +37° from -x axis b) �⃗⃗� = 5m 53° CW from +y axis

a) �⃗⃗� = 6 @ 30° North of East b) �⃗⃗� = 10 @ 53° West of South



PRACTICE: HELICOPTER TRIP

A small helicopter travels 225 m across a city in a direction 53.1° south of east. What are the components of the helicopter’s trip?



CONCEPT: UNIT VECTORS

● Vectors are sometimes represented using a special notation called Unit Vectors.

● Unit vectors make vector addition very straightforward:

EXAMPLE: Vector �⃗⃗� = 4𝑖̂ + 2𝑗 ̂and �⃗⃗� = −�̂� + 𝟐𝒋.̂ Draw the vectors and calculate �⃗⃗� = �⃗⃗� + �⃗⃗� in unit vector form.

Vector Addition w/ Unit Vectors

�⃗⃗� = 𝐴𝑥 �̂� + 𝐴𝑦𝒋̂ = ______________

�⃗⃗� = 𝐵𝑥 �̂� + 𝐵𝑦𝒋̂ = ______________

�⃗⃗� = �⃗⃗� + �⃗⃗� = ____________________

�̂� points in ____ direction.

𝒋̂ points in ____ direction.

𝒌 points in ____ direction.

𝒚

𝒙

Graphical Magnitude & Direction Unit Vector

𝒚

𝒙

“5m @ 53°” +𝒚

+𝒙

+𝒛

● special kind of vector that __________ in a direction

- has magnitude/length ____.

“ 3�̂� + 4𝒋̂”



PRACTICE: �⃗⃗� = (4.0 m)�̂� + (3.0 m)𝒋̂ and �⃗⃗� = (−13.0 m)�̂� + (7.0 m)𝒋̂. You add them together to produce another vector �⃗⃗� .

(a) Express this new vector �⃗⃗� in unit-vector notation. (b) What are the magnitude and direction of �⃗⃗� ?



EXAMPLE: Consider the three displacement vectors �⃗⃗� = (3 �̂� − 3 𝒋̂) m, �⃗⃗� = (�̂�̂ − 4 𝒋̂̂) m, and �⃗⃗� = (−𝟐 �̂� + 𝟓 𝒋̂) m.

(a) Find the magnitude and direction of D = A + B + C.

(b) Find the magnitude and direction of E = −A − B + C.



MOTION IN TWO DIMENSIONS

● Motion in 2D (plane) in very similar to 1D (straight line). We can think of it as TWO sets of 1D motion:

- These “two” motions (X & Y) are _________________ & ___________________ (same ___).

- In Physics, whenever anything is 2D, we FIRST decompose it into X & Y components (1D).

● Position, Displacement, Velocity, Acceleration are VECTORS and may need to be decomposed:

POSITION DISPLACEMENT VELOCITY ACCELERATION

O

P

O

B A

v

v = _______ = __________

___ = _______ = ________

___ = _______ = ________

a

a = _______ = __________

___ = _______ = ________

___ = _______ = ________

X Y 2D X Y 2D 2D X Y 2D X Y

EXAMPLE: The diagram shows your position (in meters) in an X-Y plane. You take 10 seconds to go from A to B, while

moving with a constant velocity. You then take 5 seconds to go from B to C, while moving with a different constant velocity.

(a) Draw position vectors (x, y, r) for point A, and displacement vectors (Δx, Δy, Δr) for intervals AB and BC.

(b) Calculate the magnitude and direction of displacement and velocity vectors for intervals AB, BC, and AC.

B

A

16 C

12 B

8

4 A

0 4 8 12 16

1 x 2D 2 x 1D



PRACTICE: MOTION IN A 2D PLANE

PRACTICE 1: An object moves up with a constant 8 m/s while moving to the left with a constant 6 m/s. Calculate the

magnitude and direction of the object’s (total) velocity.

PRACTICE 2: An object is launched with an initial velocity of 50 m/s directed 37 degrees above the horizontal. Calculate the

X and Y components of the object’s initial velocity.

PRACTICE 3: A car is initially at rest at (0, 0) meters on an X-Y plane. The car then accelerates for 4 s with a constant 6

m/s2 directed in the +x-axis. The car then accelerates for 5 s with a constant 5 m/s2 directed at 53o above the +x-axis. Find

the magnitude and direction of the car’s velocity after the two accelerations (after 8 s hint: use v = vo + at).



PROJECTILE MOTION: INTRO & SYMMETRIC LAUNCH

● Projectile Motion occurs when an object is launched and moves freely in 2D space.

- Remember that the first step in any 2D Physics problem is to ____________________ it into X & Y components.

- When decomposing vo, vox goes with the ____ motion, voy with the ____ motion (vo no longer useful).

- So Projectile Motion is _______________ motion WITH _______________________ horizontal motion (ax = 0).

- These “two” motions (X & Y) are ___________________ & _____________________ (same ____).

- Remember: If YA = YC, then A & C are __________________ (1) ______________ and (2) ______________.

- At any point, the object’s velocity vector is _________________ to its path.

aX = ____ ____ Equation(s):

vAx = _____ = _____ = _____

aY = ____ ____ Equation(s)

For every question: (1) Determine AXIS (X or Y) (2) Pick an INTERVAL (3) Pick an EQUATION

EXAMPLE: An object is launched from the ground with 50 m/s at +37o, and later returns to the ground. Find its:

(a) Maximum height

(b) Time to reach max. height

(c) Total time of flight

(d) Horizontal displacement (range)

● Symmetric launches attain max range when Θ = ____. Complementary angles (eg. ____ & ____) attain the same range.



PROJECTILE MOTION: HORIZONTAL LAUNCH (Zero Angle)

● In horizontal launches: ___________ _______________.

EXAMPLE 1: An object is launched horizontally with 30 m/s from a 50 m high cliff. Find its:

(a) time to hit the ground.

(b) range (horizontal displacement).

PRACTICE 1: An object rolls from the top of a hill (shown below) with 20 m/s and takes 4 s to hit the floor.

(a) Find the object’s range.

(b) How tall is the hill?

EXAMPLE 2: An airplane is moving horizontally at 500 m above the ground with an unknown speed. A crate released from

the plane travels a horizontal distance of 4,000 m before striking the ground. Find:

(a) the plane’s horizontal speed.

(b) the crate’s final velocity (magnitude & direction).

UAM EQUATIONS

(1) v = vo + a t

(2) v2 = vo2 + 2 a Δx

(3) Δx = vo t + ½ a t2

*(4) Δx = ½ ( vo + v ) t

PROJECTILE MOTION

(1) Determine Axis (X | Y)

(2) Pick an Interval

(3) Pick an Equation



PRACTICE: HORIZONTAL LAUNCH

PRACTICE 1: When an object that is launched horizontally hits its

target, its velocity has horizontal and vertical components of 80 m/s

and 60 m/s, respectively. Find the object’s range.

PRACTICE 2: In a regulation-sized beer pong table, the ping pong ball is tossed from a horizontal distance of 2.4 meters

and 1.0 meter above the top of its target cup. What horizontal speed must you throw the ball with, so it makes the cup?

PROJECTILE MOTION




UAM EQUATIONS

(1) v = vo + a t

(2) v2 = vo2 + 2 a Δx

(3) Δx = vo t + ½ a t2

*(4) Δx = ½ ( vo + v ) t



PROJECTILE MOTION: NEGATIVE LAUNCH (Negative Angle)

● In negative launch problems, the object starts off with vertical velocity pointing down.

- These usually have just 1 interval. Note all variables are downward (VY, Δy, aY = g).

EXAMPLE: An object slides off an inclined roof (the angle shown is 37o) with 5 m/s, at 3 m above the ground. Find its:

(a) horizontal distance while in the air.

(b) final velocity (magnitude and direction).

PRACTICE: You throw an object from the top of a building with 50 m/s directed at 53o below the horizontal. It covers a

horizontal distance of 80 m while in the air. Find its:

(a) total time of flight.

(b) final velocity (magnitude and direction).

PROJECTILE MOTION






PROJECTILE MOTION: LAUNCH UP *FROM* A HEIGHT

● In these problems, the first part of the motion is Symmetric, but the object drops further.

- When picking intervals, try to include point B to simplify equations (_______).

EXAMPLE 1: You throw an object from the top of a 100 m-tall building with 50 m/s directed at 37o above the x-axis. Find:

(a) the maximum height.

(b) the time to reach the maximum height.

(c) the total range (horizontal displacement).

(d) the magnitude & direction of vB, vC, vD.

● If answering multiple questions, it’s often best to break up the motion (___-___, ___-___).

- But sometimes it’s better / easier / faster to solve in a single step / interval (___-___).

EXAMPLE 2: You throw a rock from a 60-m cliff with 50 m/s at 53o above the +x. Using a SINGLE interval, find:

(a) the vertical component of the velocity at D.

(b) the total time of flight.

PROJECTILE MOTION




A

B

C

D



PROJECTILE MOTION: LAUNCH *TO* A HEIGHT

● In these, we will usually use a combination of the X equation, with the Y equation #3: ΔY = Vot + ½ a t2.

EXAMPLE 1: A fireman, 60 m away from a building, shoots water from a hose at 30 m/s, angled at 53o above the +x. Find:

(a) the time to hit the building.

(b) the height at which it hits the building.

● In some cases, you won’t know if the object hits its target on its way up OR on its way down.

EXAMPLE 2: You want water from hose at 25 m/s and 37o above the +x to reach a building at 10 m above the ground. How

far from the building should the hose be positioned? There are two possible distances, so:

(a) Find the two times it takes for the water to reach the given height.

(b) Find the two distances from the building that would achieve this.



SYMMETRIC LAUNCH: FIND INITIAL VELOCITY (VA, ΘA, or both) ● Symmetric Launch problems where either VA or ΘA is missing require 2-3 equations. - Remember the vector equations. If you have VAx & VAy, you can find VA & ΘA.

● Y equations in the BC interval (voy = 0) give simple combos of the 3 Y vars: Vy, t, H à - If you have any Y variable, you can use these equations to find the other two.

EXAMPLE: An object is launched at an angle from the ground. It reaches a maximum height of 45 m, and returns to the ground after covering a horizontal distance of 240 m. Find its initial velocity (magnitude and direction).



PRACTICE: SYMMETRIC LAUNCH: FIND INITIAL VELOCITY

PRACTICE 1: The range of an object launched from the ground at an angle is 300 m. If the vertical component of its

velocity just before returning to the ground has magnitude 50 m/s, find the magnitude and direction of its initial velocity.

PRACTICE 2: An object is launched from the ground with an initial 40 m/s at an unknown angle. It reaches its maximum

height in 3.5 seconds, before returning to the ground. Find its initial launch angle.

PRACTICE 3: An object is launched from the ground with an unknown initial speed directed at 30o above the horizontal. If it

reaches a maximum height of 40 m before returning to the ground, find its initial speed.



ch 03: kinematics in two or three …lightcat-files.s3.amazonaws.com/packets/admin_physics-3...if...

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