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Chapter 3 Kinematics in two dimensions

Chapter 3 Kinematics in two dimensions. Up until now we have been working with velocity time etc in 1 dimension ( ie only looking at straight line movement) This chapter extends our work on velocity time into 2 dimensions

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Chapter 3 Kinematics in two dimensions

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  1. Chapter 3Kinematics in two dimensions

  2. Up until now we have been working with velocity time etc in 1 dimension (ie only looking at straight line movement) This chapter extends our work on velocity time into 2 dimensions You will understand how this is done as we work through the questions …..

  3. In order to take the work into 2 dimensions you to Define a starting point – so we can work from this point Define a form a measurement – we use unit vectors i and j Define i and j as being perpendicular to each other Doing this allows us to form equations using time which tell us the across position (horizontal) and the up position (vertical) at a specific point in time … Before we understand this fully lets look at the various notations that we have to understand …..

  4. 1 2 3 Section A

  5. 4 5 6 Section A

  6. Section B

  7. Expressing Vectors - Task Put the cards into groups with some meaning

  8. A new mechanics letter ris the position at time t It is the displacement from a defined reference point (non mechanics people think if it as – distance from a point we made up!)

  9. So now lets look at how this works with the time factor You wont fully understand this until we have got to the end and you have done a few questions

  10. Chapter 3 - Notes Page – This all needs to be copied and vaguely understood!Kinematics in two dimensions Vector i and j (perpendicular directions) Like x and y but not quite the same Resultant – size given by its length (pythag) - Direction given by the angle (trig) As a resultant distance = 5 (because =5) direction = 36.86 (because ) Using this notation in is possible to find the horizontal position and the vertical position at any point in time Eg This considers the postion of i at time t and considers the position of j at time t

  11. There is a potential problem with the angle… what it it? How do you think this is over come? Section C

  12. Page 32/33 In practice we will use the table function on the calculator – but this first time these to it by hand to ensure understanding Once we have completed this we need to work on squared paper with a reference point

  13. Kk remember to describe what this represents

  14. Page 34/35 Table for boat 1 So simple don’t use calc Use calc to save time Table for boat 2 Now complete your Exercise 3a for homework. You need to do all the questions as they change a little as you go along. You can use the table function on the calc to speed you up.

  15. Lets look at the trickier ideas developed in these questions – this is usually how this topic appear in the exam.

  16. Worked Example: A boat moves so that its position, in metres, at time t seconds is given by r= (450 - 5t)i + (4t + 100)j where i and j are unit vectors that are directed east and north, respectively. A rock has position 250i + 200j. (a) i) Calculate the time when the boat is due north of the rock. (ii) Work out the position vector at this point) (b) i) Calculate the time when the boat is due east of the rock. (ii) Work out the position vector at this point)

  17. Task to embed the understanding: Either go through that question again on your own ensuring you fully understand it OR Go onto new questions of this type. Page 35 question 4 4. A boat moves so that its position, in metres, at time t seconds is given by: r= (2t+6)i+ (3t-9)j where i and j are unit vectors that are directed east and north respectively. A lighthouse has position 186i+ 281j. (a) Find the position of the boat, when t = 0, 40, 80 and 120 seconds. (b) Plot the path of the boat. (c) Calculate the time when the boat is due south of the lighthouse and the time when the boat is due east of the lighthouse.

  18. 5. A bullet is fired from a rifle, so that its position, in metres, at time t seconds is given by: r = 180ti + (1.225 - 4.9t2)j where i and j are horizontal and vertical unit vectors, respectively, and the origin is at ground level. (a) Find the initial height of the bullet. (b) Find the time when the bullet hits the ground. (c) Find the horizontal distance travelled by the bullet. Answers a) 1.225m b) 0.5 s c) 90m

  19. 6 A boomerang is thrown. As it moves its position, in metres, at time t seconds is modelled by: r = ( 10t – t2)i+( 2 + t - t2)j whereiand jare horizontal and vertical unit vectors. The origin is at ground level. Find the position of the boomerang when it hits the ground. 16i

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