ALGEBRAIC METHOD• This lesson will extend your knowledge of kinematics to

two dimensions.• You will be able to solve problems involving

displacement in two dimensions using two, more accurate methods

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By the end of this lesson, you will be able to:

What are we going to cover today?

Apply the Trig Method

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Comp Method

Homework

Right angle triangles (90°)… what math “tools” can we use? Pythagorean Theorem

Trig ratio’s S-OHC-AH T-OA

Math Review

a2 + b2 = c2 a2 + b2 = c

Non-right angle triangle's… what options exist? Sine Law

Cosine LawC2 = A2 + B2 - 2ABcosc

Math Review

sin a = sin b = sin c A B C

TRIG METHODAllows us to solve problems by looking at parts as right

angle triangles

TRIG METHODAlso allows us to solve problems that involve non-right

angle trianglesLet’s solve yesterdays hiker question using the trigonometric method.

TRIGONOMETRIC METHODS

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Sol’n:This is the diagramafter we apply vectoraddition

15 km

21 km 24 km

60.o

Next we break the vector diagram into TrianglesWhat can we do from here?

A

B

Ex 1: A hiker walks 15 km [N], then 24 km [N60oW], then 21 km [S]. What is his final displacement?

Solve for length A and then solve for length B

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We apply cosine lawto solve for the unknown side 21 km

24 km

60.o

a

oa cos . 2 22 21 24 2 21 24 60

60.o

a .

a .

2 513 0222 65

A

C2 = A2 + B2 - 2ABcosc

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We apply sine lawto solve for the angles

21 km

24 km

60.o

q

60.o

osin sin ..

6021 22 65

o . 53 4

22.65 km

A

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We apply cosine lawagain to solve for the final answer

15 km

21 km

24 km

60.o

53.4o

60.o

oa . . cos . 2 22 15 22 65 2 15 22 65 66 6

a .

a .

2 468 1621 6

22.65 km 66.6

o

A

B

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We apply sine lawagain to solve for the final angle

15 km

21 km

24 km

60.o

53.4o

60.o

22.65 km66.6o

21.6 kmosin sin .

. .

66 6

22 65 21 6

o . 74 2

q

Dd = 22 km [N74oW]

A

B

COMPONENT METHOD

Break the VECTOR into its component pieces (a y value and an x value)

Remember:vector quantities can be represented by directed line segments

“All y, no x”d

avv

“No y, all x”

r

“Some y, some x”

If I gave you 5 different “vectors” or movements… how would you go about finding the TOTAL DISPLACEMENT of all 5?

Ie. 10m N30ºE = V1 13m E15ºN = V2 21m E83ºN = V3 20m W45ºN = V4 11m N72ºE = V5

Perhaps with a graph? But what the heck does that tell us?

If only we knew the total vertical displacement & the total horizontal displacement We could use Pythagorean Theorem

to solve for the total displacement

What we can do is figure out the INDIVIDUAL vertical and horizontal displacement of EACH VECTOR and then add them up!

If only we knew the total vertical displacement & the total horizontal displacement We could use Pythagorean Theorem

to solve for the total displacement

What we can do is figure out the INDIVIDUAL vertical and horizontal displacement of EACH VECTOR and then add them up!

By breaking up each individual vector (ex. 20 km N45ºE) into a y value (ex. Vertical displacement) and x value (horizontal displacement) we eliminate the need to look at its direction – focusing instead on the magnitude only

This effectively turns any set of complicated vectors that might be going in various directions, into a streamlined set of values only going vertical or horizontal (never both)

This allows us to use our kinematics in 1D rules to determine the overall vertical and horizontal displacements – which will always form 90º triangles – Pythagorean Theory - EASY!

r

xr

yr

We will let r represent our RESULTANT vector (this is the same as our total displacement)

Δdtotal = r

= Δdx

Or the

horizontal

displacement

= Δdy

or the Δ in

vertical

position

r

xr

yropp

adj

hyp

What is the only segment of the triangle we automatically know from the start?

Ө or 15º

?

?13m E15ºN

What determines the names of the rest (think location)?Based on this information (hypotenuse and the Ө) how can we figure out the adjacent and opposite lengths?

Adjacent Opposite

Sin Ө = opp / hyp

sin 15º = opp /

13m

Opp = 3.4m

Cos Ө = adj / hyp

cos 15º = adj /

13m

Adj =

12.6m

Use Pythagorean theory to check ans.

r

xr

yropp

adj

hyp

sinyr r

Ө or 15º

13m E15ºN

sin yropphyp r

sin yropp

hyp r

Now that we know how we can use the trig formula’s to solve for the adj or opp lengths we can substitute the variables to represent the vertical and horizontal displacement

Unknown Value

Vertical

Component

r

xr

yr

adj

hyp

cos xradjhyp r

cosxr r

opp

Unknown Value

Horizontal

Component

r

xr

yr

opp

adj

hyp

sinyr r

cosxr r

“x-component” of

i.e. horizontal Δd

Or the adjacent length of the hypotenuse

r “y-

component” of

i.e. vertical Δd

Or the opposite length of the hypotenuse

r

r

q

r

xr

yr

opp

adj

hyp

sinyr r

cosxr r

r

q

What happens if we use (or are provided) the complimentary angle?

x

y

X-Axis Rule

1. Always take your degree from the x axis!I. If you take it from the y-axis you will have

effectively flipped the formula upside, causing cos to represent y (instead of x) and vice versa

2. If the problem involves vectors in multiple directions (i.e. east & west) – determine the degree by counting from the SAME x-axis

I. This will calculate vectors with opposite directions (i.e. east vs west) as a positive vs. negative – ensuring an accurate displacement measurement

E 45º N

E 45º S But if we want our formula to do the thinking for us we should use E 315º S

This will ensure it’s a negative vertical (south) value but still a positive horizontal (east) value

E 315º S

After determining the X & Y values of every vector simply add them up to determine the TOTAL-X & the TOTAL-Y displacement

Vector

x-componentrx

y-componentry

Arh1

(rh1) Cos q = ___ km(rx1)

5.0 cos 0 = 5.0 km

(rh1) Sin q = ___ km(ry1)

5.0 sin 0 = 0

Brh2

(rh2) Cos q = ___ km(rx2)

3.0 cos 45 = 2 .m

(rh2) Sin q = ___ km(ry2)

3.0 sin45 = 2 .1 km

Resultantrh

rx1 + rx2 = rxtotal

7.1 km [E]

ry1 + ry2 = rytotal

2.1 km [N]

Consider the following vectors:                                         A (rh1) = 5.0 km due East                                         B(rh2) = 3.0 km East 450 North  

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q

rh

Every vector consists of two components. • Vector components are

to each other

• Vector components add together to form the vector

rx = r cos q

ry = r sin q

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Now let’s solve use the component method.

Sol’n:

First, we break each vector down into its components

60o

x

y

d

d km

1

1

0

21

x

y

d . km

d km

2

2

20 78

12

x

y

d

d km

3

3

0

15

cosxr r

sinyr r

?

?

?

Ex 1: A hiker walks 15 km [N], then 24 km [N60oW], then 21 km [S]. What is his final displacement?

15 km [N]24 km [N60oW]

21 km [S]

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Then we add the x and y comp’s separately

Rx x x xd d d d

( . )

. km

1 2 3

0 20 78 020 78

Ry y y yd d d d

( ) ( ) ( )

km

1 2 3

21 12 156

??

??

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Next we draw the comp’s head to tail and use the Pythagorean Theorem

Rd ( . )

. km

2 26 20 78

21 63

-20.78 km

+6.0

km q

Rd

o

tan.

620 7816

Dd = 22 km [W16oN]

??

?

?

HOMEWORKS

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Read pgs 68 – 74

Review tutorials before attempting practice!

Questions “Practice”1 – 2 pg 711 – 2 pg 74