1. This lesson will extend your knowledge of kinematics to two dimensions. 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 You will be able to solve problems involving displacement in two dimensions using two, more accurate methods SPH 3U Grade 11 U Physics

2. By the end of this lesson, you will be able to: What are we going to cover today? Apply the Trig Method SPH 3U Grade 11 U Physics Apply the Comp Method Homework

3.  Right angle triangles (90°)… what math “tools” can we use?  Pythagorean Theorem  Trig ratio’s  S-OH  C-AH  T-OA

4.  Non-right angle triangle's… what options exist?  Sine Law  Cosine Law  C 2 = A 2 + B 2 - 2ABcos  c

5. Allows us to solve problems by looking at parts as right angle triangles

8. Also allows us to solve problems that involve non-right angle triangles Let’s solve yesterdays hiker question using the trigonometric method.

9. SPH 3U Grade 11 U Physics Sol’n: This is the diagram after we apply vector addition 15 km 21 km 24 km 60. o Next we break the vector diagram into Triangles What can we do from here? A B Ex 1: A hiker walks 15 km [N], then 24 km [N60 o W], then 21 km [S]. What is his final displacement? Solve for length A and then solve for length B

10. SPH 3U Grade 11 U Physics We apply cosine law to solve for the unknown side 21 km 24 km 60. o a A C 2 = A 2 + B 2 - 2ABcos  c

11. SPH 3U Grade 11 U Physics We apply sine law to solve for the angles 21 km 24 km 60. o  22.65 km A

12. SPH 3U Grade 11 U Physics We apply cosine law again to solve for the final answer 15 km 21 km 24 km 60. o 53.4 o 60. o 22.65 km 66.6 o A B

13. SPH 3U Grade 11 U Physics We apply sine law again to solve for the final angle 15 km 21 km 24 km 60. o 53.4 o 60. o 22.65 km 66.6 o 21.6 km    d = 22 km [N74 o W] A B

15. Break the VECTOR into its component pieces (a y value and an x value) Remember: vector quantities can be represented by directed line segments

16. “All y, no x”

17. “No y, all x ”

18. “Some y, some x ”

19. 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?

20.  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!

21.  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!

22.  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!

23. We will let r represent our RESULTANT vector (this is the same as our total displacement) Δd total = r = Δd x Or the horizontal displacement = Δd y or the Δ in vertical position

24. What is the only segment of the triangle we automatically know from the start? Ө or 15 º ? ? 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 15 º = opp / 13m Opp = 3.4m Cos Ө = adj / hyp cos 15 º = adj / 13m Adj = 12.6m Use Pythagorean theory to check ans.

25. Ө or 15 º 13m E15ºN 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

26. Unknown Value Horizontal Component

27. “ x -component” of i.e. horizontal Δd Or the adjacent length of the hypotenuse “ y -component” of i.e. vertical Δd Or the opposite length of the hypotenuse 

28.  What happens if we use (or are provided) the complimentary angle? x y

29. 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

30. 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

31. r h1 r h2 Consider the following vectors: A ( r h1 ) = 5.0 km due East B( r h2 ) = 3.0 km East 45 0 North

32. SPH 3U Grade 11 U Physics  hrhhrh Every vector consists of two components. Vector components are  to each other Vector components add together to form the vector r x = r cos  r y = r sin 

33. SPH 3U Grade 11 U Physics Now let’s solve use the component method. Sol’n: First, we break each vector down into its components 60 o ? ? ? Ex 1: A hiker walks 15 km [N], then 24 km [N60 o W], then 21 km [S]. What is his final displacement? 15 km [N] 24 km [N60 o W] 21 km [S]

34. SPH 3U Grade 11 U Physics Then we add the x and y comp’s separately ? ? ? ?

35. SPH 3U Grade 11 U Physics Next we draw the comp’s head to tail and use the Pythagorean Theorem -20.78 km +6.0 km    d = 22 km [W16 o N] ? ? ? ?

36. SPH 3U Grade 11 U Physics Read pgs 68 – 74 Review tutorials before attempting practice! Questions “Practice” 1 – 2 pg 71 1 – 2 pg 74