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Vectors (10)

Explore displacement, velocity vector calculations, and modeling techniques for objects in motion with practical examples. Understand how to find average velocity, resultant velocity, and position vectors using vector algebra.

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Vectors (10)

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  1. Vectors (10) Modelling

  2. B r = 6i - 2j A s = -4i + 4j r r r r r r r O A A B B B B B + s = A = + s = (6i - 2j) + (-4i + 4j) = 6i - 2j- 4i + 4j r = (2i + 2j) m B Displacement An object with the position vector (6i - 2j)m is displaced by (-4i + 4j)m, what is it’s final position vector?

  3. Velocity Vector Calculation (1) r r + s = r r A A B B s = - s A r = 5i + 5j If a ship has a position vector 5i + 5j, 3 seconds later it has the position vector 2i - 4j What is the average velocity in the time? A Firstly, what is the displacement s? O r B = 2i - 4j s = (2i - 4j) - (5i + 5j) B s = 2i - 4j - 5i - 5j s = -3i - 9j

  4. Velocity Vector Calculation (2) average velocity = change in position vector time taken = (-3i - 9j) m average velocity s 3 sec. = (-i - 3j) ms-1 average velocity A r = 5i + 5j If a ship has a position vector 5i + 5j, 3 seconds later it has the position vector 2i - 4j What is the average velocity in the time? A s = -3i - 9j O r B = 2i - 4j B Meaning 1 ms-1 West and 3 ms-1 South

  5. Magnitude of the Velocity = (-i - 3j) ms-1 average velocity Magnitude of velocity = (32 + 12) = 10 = 3.2 ms-1 s 3 1

  6. Resultant Velocity (1) Something like this If the boat sails straight across, which direction will it actually go? A River 5ms-1 2ms-1 Direction of flow

  7. Resultant Velocity (2) a 5ms-1 The resultant  2ms-1 5ms-1 Direction tan  = opp/adj = 2/5  = tan-1(0.4) = 21.8o 2ms-1 or a = 90 - 21.8 = 68.2o measured from the river bank The RESULTANT VELOCITY can be found by a triangle of velocities Resultant = 5i + 2j Magnitude = (52 + 22) = 29 = 5.4 ms-1

  8. Resultant Velocity (3) = 5i - 2j NE Wind = 3i + 3j A plane flys with velocity 5i - 2j The wind blows with velocity 3i + 3j What is the resultant velocity and direction? Do a sketch

  9. Resultant Velocity (3.5) - sketch resultant velocity v Wind 3i + 3j 5i - 2j Wind 5i - 2j 3i + 3j resultant velocity v A plane flys with velocity 5i - 2j The wind blows with velocity 3i + 3j What is the resultant velocity and direction? You can sketch it either way round. The resultant is identical in both cases.

  10. Resultant Velocity (4) resultant velocity v A plane flys with velocity 5i - 2j The wind blows with velocity 3i + 3j What is the resultant velocity and direction? Wind 5i - 2j 3i + 3j v = 3i + 3j +5i - 2j = 8i + j

  11. Resultant Velocity (5) N 1 b  8 Direction tan  = opp/adj = 1/8  = tan-1(1/8) = 7.1o Bearing b = 90 - 7.1 = 82.9o A plane flys with velocity 5i - 2j The wind blows with velocity 3i + 3j What is the resultant velocity and direction? resultant velocity v v= 8i + j Magnitude = (82 + 12) = 65 = 8.1 ms-1

  12. P 50km O 70km OP = 700i + 500j Modeling Techniques A plane is 700km East and 500km North of the origin (airport). It flies with velocity 300i - 220j km/h. What is it’s Position Vector at time t hours. 1st: Find in terms of i and j the original position vector of the plane.

  13. OP = 700i + 500j Modeling Techniques A plane is 700km East and 500km North of the origin (airport). It flies with velocity 300i - 220j km/h. What is it’s Position Vector at time t hours. P motion O

  14. OP = 700i + 500j PC = 300i - 220j OC = 700i + 500j + 300i - 220j OC = 1000i + 280j Modeling Techniques A plane is 700km East and 500km North of the origin (airport). It flies with velocity 300i - 220j km/h. What is it’s Position Vector at time t hours. P C O 1 hour later

  15. OP = 700i + 500j PC = 600i - 440j OC = 700i + 500j + 600i - 440j OC = 1300i + 60j Modeling Techniques A plane is 700km East and 500km North of the origin (airport). It flies with velocity 300i - 220j km/h. What is it’s Position Vector at time t hours. P C O 2 hours later

  16. OP = 700i + 500j PC = 5(300i - 440j) OC = 700i + 500j + 1500i - 2200j OC = 2200i - 700j Modeling Techniques A plane is 700km East and 500km North of the origin (airport). It flies with velocity 300i - 220j km/h. What is it’s Position Vector at time t hours. P O C 5 hours later

  17. OP = 700i + 500j PC = t(300i - 220)j OC = 700i + 500j + t(300i - 220j) Original Point Parallel Vector Modeling Techniques A plane is 700km East and 500km North of the origin (airport). It flies with velocity 300i - 220j km/h. What is it’s Position Vector at time t hours. P Distance = speed x time C O t hours later

  18. [ ] -3 4 F (0,0) O L (6,-6) OL = 6i - 6j Modeling A Problem: example - 1 An Ocean Liner is at (6, -6) is cruising at 10 km/h in the direction . A fishing boat is anchored at (0,0). A) Find in terms of i and j the original position vector of the liner from the fishing boat • Exercise 17D: ‘Page 407’ Q 3

  19. Modeling A Problem - 2 [ ] -3 4 F (0,0) O L is the direction of the velocity vector Velocity is 10 km/h : v = 2 = (twice as long) [ ] [ ] -6 8 -6 8 [ ] [ ] In t hours; moved by tx -3 4 -3 4 An Ocean Liner is at (6, -6) is cruising at 10 km/h in the direction . A fishing boat is anchored at (0,0). B) Find the position vector of the liner at time t Magnitude = ((-3)2 + 42) = 25 = 5 OL = 6i - 6j+ t(-6i + 8j)

  20. Modeling A Problem - 3 [ ] -3 4 F (0,0) O L OL = 6i - 6j+ t(-6i + 8j) OL = 6i - 6j+ t(-6i + 8j) An Ocean Liner is at (6, -6) is cruising at 10 km/h in the direction . A fishing boat is anchored at (0,0). C) Find the time t when the liner is due East of the fishing boat Due East when j component is zero (-6 + 8t)j = 0 -6 + 8t = 0 8t = 6 t = 6/8 At time 3/4 hour

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