Kinematics in Two Dimensions

Section 1: Adding Vectors Graphically

• Adding Vectors Graphically• Remember vectors have magnitude

(length) and direction. • When you add vectors you must maintain

both magnitude and direction • This information is represented by an

arrow (vector)

• A vector has a magnitude and a direction– The length of a drawn vector represents

magnitude.– The arrow represents the direction

Larger Vector Smaller Vector

Graphical Representation of Vectors

• Given Vector a:

Draw 2a Draw -a

Problem set 1:

1. Which vector has the largest magnitude?

2. What would -b look like?

3. What would 2 c look like?

a

bc

Vectors

• Three vectors

a

bc

• When adding vectors graphically, align the vectors head-to-tail.

• This means draw the vectors in order, matching up the point of one arrow with the end of the next, indicating the overall direction heading.

• Ex. a + c • The starting point is called the origin

a

bc

a

c

origin

• When all of the vectors have been connected, draw one straight arrow from origin to finish. This arrow is called the resultant vector.

a

c

origin

a

bc

• Ex.1 Draw a + b

a

bc

• Ex.1 Draw a + b

origin

Resultant

a

bc

• Ex. 2 Draw a + b + c

a

bc

• Ex. 2 Draw a + b + c

origin

Resultant

a

bc

• Ex. 3 Draw 2a – b – 2c

a

bc

• Ex. 3 Draw 2a – b – 2c

origin

Resultant

a

bc

Section 2: How do you name vector directions?

Vector Direction Naming

• How many degrees is this?

W

S

E

N

Vector Direction Naming

• How many degrees is this?

W

S

E

N

90º

Vector Direction Naming

• What is the difference between 15º North of East and 15 º East of North?

W

S

N

E

Vector Direction Naming• What is the difference between 15º North

of East and 15º East of North? (can you tell now?)

W

S

E

N

15º North of East

W

S

E

N

15º East of North

Vector Direction Naming

W

S

N

15º North of what?

15º

Vector Direction Naming

W

S

N

E15º

15º North of East

W

S

E

15º East of What?

15º

W

S

E

N

15º East of North

15º

___ of ___

This is the baseline. It is the direction you look at first

This is the direction you go from the baseline to draw your angle

N E

Describing directions• 30º North of East

– East first then 30º North

• 40º South of East– East first then 30º South

• 25º North of West– West first then 30º North

• 30º South of West– West first then 30º South

Problem Set #2 (Name the angles)

20º

20º

30º

30º

45º

Intro: Get out your notes

1. Draw the resultant of

a – b + c

2. What would you label following angles

a. b.

3. Draw the direction 15º S of W

a b c

28º

18º

Section 3: How do you add vectors mathematically (not

projectile motion)

The Useful Right Triangle

• Sketch a right triangle and label its sides

Ө

a: opposite

b: adjacent

c: hypotenuse

The angle

• The opposite (a) and adjacent (b) change based on the location of the angle in question

• The hypotenuse is always the longest side

Ө

a: opposite

b: adjacentc: hypotenuse

• The opposite (a) and adjacent (b) change based on the location of the angle in question

• The hypotenuse is always the longest side

Ө

a: opposite

b: adjacentc: hypotenuse

To figure out any side when given two other sides

• Use Pythagorean Theorem

a2 + b2 = c2

Ө

a: opposite

b: adjacent

c: hypotenuse

The angle

Sometimes you need to use trig functions

Ө

a: opposite

a: adjacent

c: hypotenuse

Sin Ө = _____

Cos Ө = _____

Tan Ө = _____Opp

Hyp

Adj

Hyp

Opp

Adj

Sometimes you need to use trig functions

Ө

a: opposite

a: adjacent

c: hypotenuse

Sin Ө = _____

Cos Ө = _____

Tan Ө = _____Opp

Hyp

Adj

Hyp

Opp

AdjSOH CAH TOA

More used versions

Sin Ө = _____

Cos Ө = _____

Tan Ө = _____

Opp

Hyp

Adj

Hyp

Opp

Adj

Opp = (Sin Ө)(Hyp)

Adj = (Cos Ө)(Hyp)

Ө = Tan-1 _____Opp

Adj

• To resolve a vector means to break it down into its X and Y components.

Example: 85 m 25º N of W• Start by drawing the angle

25º

• To resolve a vector means to break it down into its X and Y components.

Example: 85 m 25º N of W• Start by drawing the angle• The magnitude given is always the hypotenuse

25º

85 m

• To resolve a vector means to break it down into its X and Y components.

Example: 85 m 25º N of W• this hypotenuse is made up of a X component (West)• and a Y component (North)

25º

85 m

West

North

In other words:

I can go so far west along the X axis and so far north along the Y axis and end up in the same place

85 m

25º

West

North

origin origin

finish finish

• If the question asks for the West component: Solve for that side– Here the west is the adjacent side

Adj = (Cos Θ)(Hyp)

25º

85 m

West or Adj.

• If the question asks for the West component: Solve for that side– Here the west is the adjacent side

Adj = (Cos Θ)(Hyp)

Adj = (Cos 25º)(85) = 77 m W

25º

85 m

West or Adj.

• If the question asks for the North component: Solve for that side– Here the north is the opposite side

Opp = (Sin Θ)(Hyp)

25º

85 mNorthorOpp.

• If the question asks for the North component: Solve for that side– Here the west is the opposite side

Opp = (Sin Θ)(Hyp)

Opp = (Sin 25º)(85) = 36 m N

25º

85 mNorthorOpp

Resolving Vectors Into Components

• Ex 4a. Find the west component of 45 m 19º S of W

Resolving Vectors Into Components

• Ex 4a. Find the west component of 45 m 19º S of W

• Ex 4a. Find the south component of 45 m 19º S of W

• Ex 4a. Find the south component of 45 m 19º S of W

5 m/s forward

velocity = 30 m/s down

Hypotenuse = Resultant speed

5 m/s

30 m/s

Remember the wording. These vectors are at right angles to each other.

Redraw and it becomes

Right angle

Section 4 (Solving for a resultant)

• Ex. 6 Find the resultant of 35.0 m, N

and 10.6 m, E.

• Start by drawing a vector diagram• Then draw the resultant arrow

• Ex. 6 Find the resultant of 35.0 m, N

and 10.6 m, E.

Then draw the resultant vector and angle– The angle you find is in the triangle closest to the origin

• Now we use Pythagorean theorem to figure out the resultant (hypotenuse)

 

• Then inverse tangent to figure out the angle

• The answer needs a magnitude, angle, and direction

 

 

 

 

 

 

 

 

 

Problem Set 3: Resolve the following vectors

1) 48m, S and 25m, W

2) 12.5m, S and 78m, N

Problem Set #3

1) 48m, S and 25m, W

Section 4: How does projectile motion differ from 2D motion

(without gravity)?

Projectile Motion

• Projectile- Object that is launched by a force and continues to move by its own inertia

• Trajectory- parabolic path of a projectile

• Projectile motion involves an object moving in 2D (horizontally and vertically) but only vertically is influenced by gravity.

• The X and Y components act independently from each other and will be separated in our calculations.

X and Y are independent

• X axis has uniform motion since gravity does not act upon it.

X and Y are Independent

• Y axis will be accelerated by gravity -9.8 m/s2

The equations for uniform acceleration, from unit one, can be written for either x or y variables:

• If we push the ball harder, giving it a greater horizontal velocity as it rolls off the table, the ball would take _________ time to fall to the floor.

Horizontal and vertical movement is independent

• If we push the ball harder, giving it a greater horizontal velocity as it rolls off the table, the ball would:– Y axis: take the same time to fall to the floor. – X axis: It would just go further.

Solving Simple Projectile Motion Problems

• You will have only enough information to deal with the y or x axis first

• You cannot use the Pythagorean theorem since X and Y-axes are independent

• Time will be the key: The time it took to fall is the same time the object traveled vertically.

• dx = (vx)(t) is the equation for the horizontal uniform motion.

• If you don’t have 2 of three x variable you will have to solve for t using gravity and the y axis

Equations Solving Simple Projectile Motion Problems

• Do not mix up y and x variables• dy – height (this is negative if falling down)

• dx – range (displacement x)

For all projectile motion problems

• Draw a diagram• Separate the X and Y givens• Something is falling in these problems

X Givens Y Givens

dX = a = -9.8 m/s

vX = …t = …

…

Example Problem 8

• A stone is thrown horizontally at 7.50 m/s from a cliff that is 68.4 m high. How far from the base of the cliff does the stone land?

Write out your x and y givens separately

• A stone is thrown horizontally at 7.50 /s from a cliff that is 68.4 m high. How far from the base of the cliff does the stone land?

X givens Y givens

• A stone is thrown horizontally at 7.50 m/s from a cliff that is 68.4 m high. How far from the base of the cliff does the stone land?

X givens Y givens

Ex. 9

A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18, to the plate before it is caught.a) How long does the ball stay in the air?b) How far does it drop during its flight?

• A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18, to the plate before it is caught.– How long does the ball stay in the air?– How far does it drop during its flight?

X givens Y givens

• A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18, to the plate before it is caught.– How long does the ball stay in the air?– How far does it drop during its flight?

X givens Y givens

• A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18, to the plate before it is caught.– How long does the ball stay in the air?– How far does it drop during its flight?

X givens Y givens

Example10

1. What is the initial vertical velocity of the ball?

voY = 0 m/s

Same as if it was dropped from rest

2. How much time is required to get to the ground?

Since voY = 0 m/s use

2(-10)

-10t = 1.4 s

3. What is the vertical acceleration of the ball at point A?

aoY = -10 m/s2 always

4. What is the vertical acceleration at point B?

aoY = -10 m/s2 always

5. What is the horizontal velocity of the ball at point C?

vX = 5 m/s (does not change)

6. How far from the edge of the cliff does the ball land in the x plane?X givens

vX = 5 m/s

t = 1.4

dx = ?

dx = (vX)(t)

dx = (5)(1.4) = 7m

• What will happen if drops a package when the plane is directly over the target?

• The package has the same horizontal velocity as the plane and would land far away from the target.

Section 5: What do you do different if you have projectile motion and V0Y is not equal to 0

Projectile Motion Concepts

Arrows represent x and y velocities (g always = 10 m/s2 down)

Key points in a projectiles path

• When a projectile is at its highest point its vfy = 0. This means it stopped moving up.

• Use vfy = 0 in a question that asks you to predict the vertical distance (how high)

VoY = 0 m/s

Key points in a projectiles path

• If an object lands at the same height its vertical velocities final magnitude equals its initial but is in the opposite direction (down)

VoY = +30 m/s

VfY = -30 m/s

VoY = +30 m/s

VfY = -30 m/s

• The time it takes to rise to the top equals the time it takes to fall.– Givens to use to find time to the top:

VoY = +30 m/s VfY = 0 m/s

– Givens to use to find time of entire flight:VoY = +30 m/s VfY = -30

m/s

Key points in a projectiles path

• If a projectile lands below where it is launched the vfy

magnitude will be greater than voy and in the reverse direction

It stays constant during the entire flight (no forces acting in the x direction)

It accelerates (the force of gravity is pulling it to Earth)

Ex. 11 A ball of m = 2kg is thrown from the ground with a horizontal velocity of 5 m/s and rises to a height of 45 m.

1. What happens to velocity in the x direction? Why?

2. What happens to velocity in the y direction? Why?

3. Where is the projectile traveling the fastest? Why?

4. Where is the projectile traveling the slowest? What is its speed at this point?

5. Where is the acceleration of the projectile the greatest? Why?

A and E (has the largest VY component)

C (has only VX component VY=0)

All (g stays -10m/s2)

6. What is the acceleration due to gravity at point B?

7. What is the initial vertical velocity the ball is thrown with?

All (g stays -10m/s2)

Must solve

aY = -10m/s2

d = 45m

vo = ?

Vf = 0

vf2 = vo

2 + 2ad

vo = √(vf2 – 2ad)

vo = √(02 – 2(-10)(45)

vo = 30 m/s up

8. What is the time required to reach point C if thrown from the ground? Must solve Y givens

aY = -10m/s2

vo = +30 m/s

Vf = 0 m/s

t = ?

VfY = -30 m/s

VoY = +30 m/s

vf = vo + at

t = (vf – vo)

a

t = (0 – 30)

-10

t = 3 s

9. From point C, what is the time needed to reach the ground?

Same as time it took to get to the top

t = 3 s

10. What is the horizontal velocity at point A?

11. What is the horizontal acceleration of the ball at point E?

5 m/s (never changes horizontally while in the air)

ax = 0 m/s2 (they asked for acceleration no horizontal acceleration)

vx stays 5 m/s

12. What is the vertical acceleration due to gravity at point E?

aY = -10 m/s2

13. How far in the x plane (what is the range) does the ball travel?

Must solve

X givens

t= 6 seconds total in air

vX = 5 m/s

dX = ?

dX = (vX)(t)

dX = (5)(6) = 30 m

14. What would happen to the problem if the objects mass was 16 kg

Nothing would change. The acceleration due to gravity is the same for any mass

• More complex projectile motion problems require you separate a resultant velocity vector into its components using soh-cah-toa

• A stone is thrown at 25 m/s at a 40º angle with the horizon. Start with the finding the vx and voy

• Then solve the problem like we have

voy

Example

• The punter on a football team tries to kick a football with an initial velocity of 25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it travel?

Example

• The punter on a football team tries to kick a football with an initial velocity of 25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it travel?

Example• The punter on a football team tries to kick a football with an initial velocity of

25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it travel?

• The punter on a football team tries to kick a football with an initial velocity of 25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it travel?

• The punter on a football team tries to kick a football with an initial velocity of 25.0 m/s at an angle of 60.0º above the ground, what range (dx) does it travel?

45º will get you the greatest range

• Range is dx

• Horizontal displacement

Besides 45º, two sister angles will give you the same range

• 45º is would give you the greatest dx

• Any similar degree variation on either side of 45º would give you the same dx

• Ex these would give you the same dx.

• 40º and 50º• 30º and 60º

• 15º would give you the same range as what? ___________

Classwork/Homework

• 2D motion Packet• Pg 2 Exercise 10-16

• Honors Addition:• Book Pg 79 #16,17,18,20,22,27,31• Try 35