2. 1.What is the initial position of the star? _______________________ 2.What is the final position of the star? _______________________ 3.If the star traveled in a straight line from the initial position to the final position, what distance did it travel? _______________________ 4.What is the displacement of the star? _______________________ Initial position Final position

3. Kinematics Vectors and Scalars

4.  A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities:  Length  Area  Volume  Time  Mass

5.  A vector quantity is a quantity that has both magnitude and a direction in space Examples of Vector Quantities:  Displacement  Velocity  Acceleration  Force

6.  https://youtu.be/bOIe0DIMbI8 https://youtu.be/bOIe0DIMbI8

7.  Vector diagrams are shown using an arrow  The length of the arrow represents its magnitude  The direction of the arrow shows its direction

8. Vectors in opposite directions: 6 m/s10 m/s=4 m/s 6 N10 N=4 N Vectors in the same direction: 6 N4 N=10 N 6 m =10 m 4 m TThe resultant is the sum or the combined effect of two vector quantities

9.  When two vectors are joined tail to tail  Complete the parallelogram  The resultant is found by drawing the diagonal  When two vectors are joined head to tail  Draw the resultant vector by completing the triangle

10. Solution: Two forces are applied to a body, as shown. What is the magnitude and direction of the resultant force acting on the body? CComplete the parallelogram (rectangle) θ TThe diagonal of the parallelogram ac represents the resultant force 5 N 12 N 5 12 a bc d  The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc RResultant displacement is 13 N 67 º with the 5 N force 13 N

11. 45º 5 N 90 º θ Find the magnitude (correct to two decimal places) and direction of the resultant of the three forces shown below. 5 N 5 5 Solution: FFind the resultant of the two 5 N forces first (do right angles first) a b cd 7.07 N 10 N 135º NNow find the resultant of the 10 N and 7.07 N forces TThe 2 forces are in a straight line (45 º + 135 º = 180 º ) and in opposite directions SSo, Resultant = 10 N – 7.07 N = 2.93 N in the direction of the 10 N force 2.93 N

12.  What is a scalar quantity?  Give 2 examples  What is a vector quantity?  Give 2 examples  How are vectors represented?  What is the resultant of 2 vector quantities?  What is the triangle law?  What is the parallelogram law?

13. 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 A common way of identifying direction is by reference to East, North, West, and South. (Locate points below.)

14. 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

15. 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

16. (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

17. 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. + + - -

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

19. 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

20. 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

21. 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

22. 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

23. 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.