2. A River Problem A crocodile is lurking beside the Platte River. It spots an unsuspecting juggler on the bank of the river exactly opposite.

3. The river flows parallel to its banks at a a velocity of 2.25 m s -1. The maximum velocity of the crocodile in still water is 3.5 m s -1. and the river is 750 m wide. Assume that the crocodile travels at a constant velocity.

4. 1.Draw a sketch to show how the crocodile must swim in order to reach its prey on the opposite bank. 2.Find the resultant velocity of the crocodile 3.Calculate the time taken between the crocodile sliding into the water and the juggler being eaten.

5. 1. Draw a sketch to show how the crocodile must swim in order to reach its prey on the opposite bank. The river Current flows this way Crocodile starts here If the croc swims straight across It will be washed downstream by the current It will travel straight across the river. If the croc swims at the correct angle up stream

6. 1. Draw a sketch to show how the crocodile must swim in order to reach its prey on the opposite bank. Crocodile starts here Velocity of croc in water Velocity of river Resultant velocity of croc Croc ends up here

7. 2. Find the resultant velocity vector of the croc. 3.5 m.s -1 2.25 m.s -1 3.5 2 -2.25 2 = 2.69 m s -1 = Pythagorus: Resultant velocity RESULTANT VELOCITY = 2.69 m s -1 Perpendicular to the bank of the river

8. 3. Calculate the time taken between the crocodile sliding into the water and the juggler being eaten. Since croc. moves at a constant velocity v = 2.69 m s -1 t = d/v ∴ t = 750/2.69 = 278.8 s....

9. current Suppose some kid sets out to swim across the Platte. The kid sees a croc and swims to the nearest sand bank. She starts at A and heads directly across in a direction perpendicular to the bank. Why might she reach the opposite bank but miss the sandbank. Ans:The current could push her downstream (to the left in the diagram). path A x

10. A current In which direction should she try to swim in order to reach the sandbank ? Ans:She should head partly upstream (towards the right ). path We say the velocity of the child in still water is the velocity of the child relative to the water. x To find the actual, or resultant, velocity of the child we must add the velocity of the current to the velocity that the child herself would have in still water.

11. A x current heading path A x current path heading Since velocities are vectors, we can add the velocities by drawing the vectors head-to-tail. path 1.Child heads directly across, perpendicular to the bank. 2.Child heads upstream

12. A x current path heading Velocity Triangle Suppose the child can swim at 1.5 ms -1 in still water and the current flows at 1 ms -1. 1. Child heads straight across river. ( So, we know the direction and magnitude of the relative velocity. ) 1.5 1 v  (relative to the water) We know the magnitude and direction of the current.

13. We use Pythagoras’ Theorem to find v : A x current path heading Velocity Triangle 1·5 1 v  v 2  1 2  1.5 2  v  1.80 ( 3 s.f. ) We have a right angled triangle and we can use any trig ratio to find  Tip: Avoid using the side you calculated in case you made a slip.

14. A x current path heading Velocity Triangle 1.5 1  1.80 tan   1 1.5    33.7  ( 3 s.f. ) The resultant (actual) velocity of the child is 1.80 ms -1 making an angle of 56.3  to the bank. So, the angle made with the downstream bank...  90   33.7   56.3  56·3  It’s convenient to give the angle between the velocity and the bank of the river.

15. 1 1.5 v  2.The child heads upstream at an unknown angle. A current 1.5 resultant x 1 Velocity Triangle Find v and  this time. This line must end so that the 3 rd side of the triangle is in the right direction for the resultant.

16. A current 1.5 resultant x 1 48·2  Pythagoras’ Theorem: v 2  1.5 2  1 2  v  1.12 ( 3 s.f. ) sin   1 1.5    41.8  ( 3 s.f. ) The resultant speed of the child is 1.12 ms -1 The angle made with the upstream bank...  90   41.8   48.2  Velocity Triangle 1 1.5 v  Solution:

17. Velocities are also added when we have an airplane travelling in a wind. The air speed of a plane is the speed relative to the air. This is the speed the plane would have in still air. The ground speed of a plane is the speed at which it covers the ground, so it is the actual or resultant speed.

18. e.g.A plane flying due North needs to fly in a straight line due east. It’s air speed is 350 km h -1. The wind is blowing at 100 km h -1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ? We start by drawing and labelling the vectors separately. Solution:

19. e.g.A plane needs to fly in a straight line due east. It’s air speed is 350 km h -1. The wind is blowing at 100 km h -1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ? The plane is flying due North v We start by drawing and labelling the vectors separately. Solution: Resultant velocity (This is what we want)

20. e.g.A plane needs to fly in a straight line due east. It’s air speed is 350 km h -1. The wind is blowing at 100 km h -1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ? v We start by drawing and labelling the vectors separately. Solution: Resultant velocity Just write the magnitude of this vector as it’s quite tricky to spot its direction. Air speed = 350

21. 100 We start by drawing and labelling the vectors separately. Solution: Wind Air speed = 350 v Resultant velocity e.g.A plane needs to fly in a straight line due east. It’s air speed is 350 km h -1. The wind is blowing at 100 km h -1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ? Plane flying due North

22. e.g.A plane needs to fly in a straight line due east. It’s air speed is 350 km h -1. The wind is blowing at 100 km h -1 from the south. The plane is fling due North. What is the resultant velocity of the plane and what bearing must the pilot choose ? 100 We start by drawing and labelling the vectors separately. Solution: We can now draw the triangle. Air speed = 350 v Resultant velocity Wind

23. v 100 v v (i) (ii) (iii) 350 100 Air speed = 350 v Resultant velocity Wind Which of the following diagrams is correct ? Ans: (iii) is correct. Not head-to-tail. Not the sum of the other vectors. The two vectors which are equivalent to the resultant must be head-to-tail. Wrong direction

24. v 100 350 What is the resultant velocity of the plane and what bearing must the pilot choose ? v 2  350 2  100 2 v  335 ( 3 s.f. ) The resultant velocity is 335 km h -1 due east.

25. 100 350 335  What is the resultant velocity of the plane and what bearing must the pilot choose ?

26. 100 350 335  What is the resultant velocity of the plane and what bearing must the pilot choose ?

27. 100 350 335 We can use  to find the bearing.  cos  100 350  73.4  The pilot needs to set a bearing of 107  ( nearest degree ). 180   73.4   06.6  What is the resultant velocity of the plane and what bearing must the pilot choose ?

28. Instead of giving the vectors as magnitudes and directions, the velocities can be given using i and j e.g. A boat is rowed across a river. In still water the velocity of the boat is m s -1.  0.5  0.3 ) i j A current is flowing with a velocity of m s -1.    0.2 ) i j What is the resultant speed of the boat ? Solution: The resultant velocity v is given by the sum of the velocities of the boat and current   j v  0.5  0.3 ) + (  0.2 ) jii i = 1.5  0.5 j The speed is v  1.58 m s -1 ( 3 s.f. )  v  1.5 2  0.5 2 v

29. SUMMARY  To find the resultant velocity of a boat or swimmer in water or an aeroplane in the wind we the resultant velocity is given by closing a triangle, we always use a double headed arrow for the resultant and check that this vector is the sum of the other two.  The relative velocity of a swimmer, boat or plane is the velocity in still water or still air. or by drawing the vectors head-to-tail. add the relative velocity and the velocity of the wind or water j either by adding the i and components,

30. Page 321 If the athlete is assisted by a wind of 1 ms -1, his speed will be 7 ms -1. If the athlete runs into a head wind of1 ms -1, his speed will be 5 ms -1.

31. Page 321 1.2 ms -1 0.6 ms -1

32. Page 321 1.2 ms -1 0.6 ms -1 1.34 ms -1 Mary’s actual velocity is about 1.34 ms -1 in the direction of 26.6° to the left of the intended line.

33. Page 321 1.2 ms -1 0.6 ms -1 Mary should aim to swim 30° to the right of Q.

34. Page 321 1.2 ms -1 0.6 ms -1

35. HOMEWORK Page 322 (3 – 5)