2. Thanks to PowerPoint from Paul E. Tippens, Professor of Physics Southern Polytechnic State University Vectors

3.  Demonstrate that you meet mathematics expectations: unit analysis, algebra, scientific notation, and right-triangle trigonometry.  Define and give examples of scalar and vector quantities.  Determine the components of a given vector.  Find the resultant of two or more vectors.  Demonstrate that you meet mathematics expectations: unit analysis, algebra, scientific notation, and right-triangle trigonometry.  Define and give examples of scalar and vector quantities.  Determine the components of a given vector.  Find the resultant of two or more vectors.

4.  http://www.khanacademy.org/science/physic s/one-dimensional-motion/displacement- velocity-time/v/introduction-to-vectors- and-scalars http://www.khanacademy.org/science/physic s/one-dimensional-motion/displacement- velocity-time/v/introduction-to-vectors- and-scalars

5. Surveyors use accurate measures of magnitudes and directions to create scaled maps of large regions.

6.  You must be able convert units of measure for physical quantities. Convert 40 m/s into kilometers per hour. 40--- x ---------- x -------- = 144 km/h m s 1 km 1000 m 3600 s 1 h

7.  College algebra and simple formula manipulation are assumed. Example:Solve for v o

8.  Let’s do Table talks—I will call on one person at one table. They will tell me their groups answer. I will then call on each table and ask if they agree or disagree and why.

9.  You must be able to work in scientific notation. Evaluate the following: (6.67 x 10 -11 )(4 x 10 -3 )(2) (8.77 x 10 -3 ) 2 F = -------- = ------------ Gmm’ r 2

10.  Turn to your neighbor and explain what the exponent would be in this equation.  Let’s see who agrees and disagrees before you figure out the answer  -8  Now with your table figure out the answer- see if you can do it without a calculator.  F = 6.94 x 10 -9 N = 6.94 nN

11.  You must be familiar with SI prefixes The meter (m) 1 m = 1 x 10 0 m 1 Gm = 1 x 10 9 m 1 nm = 1 x 10 -9 m 1 Mm = 1 x 10 6 m 1  m = 1 x 10 -6 m 1 km = 1 x 10 3 m 1 mm = 1 x 10 -3 m

12.  You are familiar with right-triangle trigonometry. y x R  y = R sin  x = R cos  R 2 = x 2 + y 2

13. We begin with the measurement of length: its magnitude and its direction. Length Weight Time

14. A scalar quantity: Contains magnitude only and consists of a number and a unit. (20 m, 40 mi/h, 10 gal) A B  Distance is the length of the actual path taken by an object. s = 20 m

15. A vector quantity: Contains magnitude AND direction, a number, unit & angle. (12 m, 30 0 ; 8 km/h, N) A B D = 12 m, 20 o Displacement is the straight-line separation of two points in a specified direction.Displacement is the straight-line separation of two points in a specified direction. 

16. Net displacement: 4 m,E 6 m,W D What is the distance traveled? 10 m !! D = 2 m, W Displacement is the x or y coordinate of position. Consider a car that travels 4 m, E then 6 m, W.Displacement is the x or y coordinate of position. Consider a car that travels 4 m, E then 6 m, W. x= +4 x = +4 x= -2 x = -2

17. A common way of identifying direction is by reference to East, North, West, and South. (Locate points below.) 40 m, 50 o N of E EW S N 40 m, 60 o N of W 40 m, 60 o W of S 40 m, 60 o S of E Length = 40 m 50 o 60 o

18. Write the angles shown below by using references to east, south, west, north. EW S N 45 o EW N 50 o S Click to see the Answers... 50 0 S of E 45 0 W of N

19. Polar coordinates (R,  ) are an excellent way to express vectors. Consider the vector 40 m, 50 0 N of E, for example. 0o0o 180 o 270 o 90 o  0o0o 180 o 270 o 90 o R R is the magnitude and  is the direction. 40 m 50 o

20. (R,  ) = 40 m, 50 o (R,  ) = 40 m, 120 o (R,  ) = 40 m, 210 o (R,  ) = 40 m, 300 o 50 o 60 o 0o0o 180 o 270 o 90 o 120 o Polar coordinates (R,  ) are given for each of four possible quadrants: 210 o 300 0

21. Right, up = (+,+) Left, down = (-,-) (x,y) = (?, ?) x y (+3, +2) (-2, +3) (+4, -3) (-1, -3) Reference is made to x and y axes, with + and - numbers to indicate position in space. + + - -

22.  Application of Trigonometry to Vectors y x R  y = R sin  x = R cos  R 2 = x 2 + y 2 Trigonometry

23. 90 m 30 0 The height h is opposite 30 0 and the known adjacent side is 90 m. h h = (90 m) tan 30 o h = 57.7 m

24. A component is the effect of a vector along other directions. The x and y components of the vector (R,  are illustrated below. x y R  x = R cos  y = R sin  Finding components: Polar to Rectangular Conversions

25. x y R  x = ? y = ? 400 m   E N The y-component (N) is OPP: The x-component (E) is ADJ: x = R cos  y = R sin  E N

26. x = R cos  x = (400 m) cos 30 o = +346 m, E x = ? y = ? 400 m   E N Note: x is the side adjacent to angle 30 0 ADJ = HYP x Cos 30 0 The x-component is: R x = +346 m

27. y = R sin  y = (400 m) sin 30 o = + 200 m, N x = ? y = ? 400 m   E N OPP = HYP x Sin 30 0 The y-component is: R y = +200 m Note: y is the side opposite to angle 30 0

28. R x = +346 m R y = +200 m 400 m   E N The x- and y- components are each + in the first quadrant Solution: The person is displaced 346 m east and 200 m north of the original position.

29. First Quadrant: R is positive (+) 0 o >  < 90 o x = +; y = + x = R cos  y = R sin  + + 0o0o 90 o R 

30. Second Quadrant: R is positive (+) 90 o >  < 180 o x = - ; y = + x = R cos  y = R sin  + R  180 o 90 o

31. Third Quadrant: R is positive (+) 180 o >  < 270 o x = - y = - x = R cos  y = R sin  - R  180 o 270 o

32. Fourth Quadrant: R is positive (+) 270 o >  < 360 o x = + y = - x = R cos  y = R sin  360 o + R  270 o

33. Finding resultant of two perpendicular vectors is like changing from rectangular to polar coord. R is always positive;  is from + x axis x y R 

34. 30 lb 40 lb Draw a rough sketch. Choose rough scale: Ex: 1 cm = 10 lb 4 cm = 40 lb 3 cm = 30 lb 40 lb 30 lb Note: Force has direction just like length does. We can treat force vectors just as we have length vectors to find the resultant force. The procedure is the same!

35. 40 lb 30 lb 40 lb 30 lb Finding (R,  ) from given (x,y) = (+40, -30) R   RyRy RxRx R = x 2 + y 2 R = (40) 2 + (-30) 2 = 50 lb tan  = -30 40   = -36.9 o   is S of E  = 323.1 o

36. 40 lb 30 lbR   RyRy RxRx 40 lb 30 lb R   RyRy RxRx 40 lb 30 lb R  RyRy RxRx  40 lb 30 lb R  RyRy RxRx  = 36.9 o ;  = 36.9 o ; 143.1 o ; 216.9 o ; 323.1 o R = 50 lb