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# Add Math (F5) Motion Along A Straigh Line Subtopic 9.1

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• 1. Cikgu Jasmin’s house is 500m away. Cikgu Jasmin’s house is 500m due east.
• 2. MOTION ALONG A STRAIGHT LINE
• 3. CHAPTER 9.1 THE CONCEPT OF DISPLACEMENT
• 4.
• At the end of the lesson, students should be able to:
• identify the direction of the displacement of a particle from a fixed point
• determine the displacement of a particle from a fixed point
LEARNING OUTCOMES
• 5.
• Displacement: The distance from a fixed point measured towards a specific direction
• Fixed point: Reference point i.e: point O
9.1.1 Identifying the direction of displacement
• 6.
• Displacement of Upin and Ipin 6 metres to right or to the left of point O
Example 1
• 7. o -6 m +6 m
• 8.
• Describe the position of each of the following points with respect to O, taking east as the positive direction.
• OQ = 12 m
• OR = -5m
• OS = 0 m
Exercises
• 9.
• OQ = 12 m
• Q is 12 m east of O
Q O 12 m Solutions
• 10.
• b. OR = -5 m
• R is 5 m west of O
Q O 5 m
• 11.
• c. OS = 0 m
• The displacement 0 refers to 0 m from O, meaning that S is at O
O S
• 12. 9.1.2 Displacement, s of a particle can be represent as a function of time, t. For example: DETERMINE DISPLACEMENT OF PARTICLE
• 13.
• A particle moves along a straight line, passing through a fixed point O . Its displacement, s m, from O is given by
• where t is the time in seconds after passing through O .
• a) Find the displacement of a particle at t =1, t =2, t =3, and t =4.
• b) Illustrate the displacement of the particle on a number line with respect to O for every second from t =0 to t =4.
Example 2
• 14. (a) (b) Solutions Time, t 1 2 3 4 Displacement, s 4 3 0 -5 s (m) 0 1 2 3 4 -5 -3 -2 -4 -1     t =4 t =3 t =2 t =1
• 15. POP QUIZ
• 16. 1 2 3 4 5 6 5 7 8 9
• 17. 9m of the south of O 9m of the east of O 9m of the north of O 9m of the west of O 9m of the south of O 9m of the east of O Describe the position of the point OE= 9m with respect to O, taking the east as the positive direction. A B D C
• 18. 4m of the south of O 4m of the west of O 4m of the north of O 4m of the east of O Describe the position of the point OE= -4m with respect to O, taking the north as the positive direction. A B D C
• 19. 5m of the east of O 5m of the west of O 5m of the south of O 5m of the north of O Describe the position of the point OE= -5m with respect to O, taking the east as the positive direction. A D B C
• 20. 25m of the east of O 25m of the west of O 25m of the south of O 254m of the north of O Describe the position of the point OE= 25m the east as the positive with respect to O, taking direction. A D B C
• 21. 16m to the left 16m to the right 18m to the left 18m to the right A particle moves along a straight line, passing through a fixed point O. It’s displacement, s m, from O, t s after passing through O is given by s = 3 t 2 – 18. Find the displacement of the particle before it starts to move. A D B C
• 22. 27m of the left 9m of the right 27m of the right 9m of the left A particle moves along a straight line, passing through a fixed point O. It’s displacement, s m, from O, t s after passing through O is given by s = 3 t 2 – 18. Describe the position of the particle after 3 seconds. B A D C
• 23. 7m of the left 4m of the left 4m of the right 7m of the right A particle moves along a straight line, passing through a fixed point O. It’s displacement, s m, from O, t s after passing through O is given by s = t 2 – 3 t - 4. Describe the position of the particle before it starts to move. A B D C
• 24. 5m of the right 32m of the right 32m of the left 5m of the left A particle moves along a straight line, passing through a fixed point O. It’s displacement, s m, from O, t s after passing through O is given by s = 3 t 2 – 3t – 4 . Find the displacement of the particle at 4 seconds. A D B C A
• 25. EVALUATION
• 26. A particle moves along a straight line and passes through a fixed point O . its displacement, s m, from point O , t s after passing through point O is given by s = 10 t – 2 t 2 (Take the right direction from point O as positive direction.)
• 27.
• Find the displacement of the particle at t=1.5.
• s = 10 t – 2 t 2
• when t = 1.5,
• s = 10(1.5) – 2 (1.5) 2 substitute t=1.5
• s = 10.5
• Therefore, the displacement of the particle from point O at t= 1.5 is 10.5 m
• 28. (b) the time at which the particle is 12 m on the left side of point O s= -12 10 t – 2 t 2 = -12 Rearrange - 2 t 2 + 10 t +12 = 0 Multiply -1 2 t 2 – 10 t – 12 = 0 t 2 – 5t – 6 = 0 Factorize (t – 6)(t + 1) = 0 t = 6 or t = -1 Therefore, the particle is 12 m on the left side of point O at t = 6. Not applicable
• 29. (c) When the particle returns to point O. When s = 0 10t – 2t 2 = 0 Factorize 2t (5 - t) = 0 2t = 0 or 5 – t = 0 t = 0 or t = 5 Therefore, the particle returns to point O at t = 5.
• 30. SUMMARY
• When s=0, the particle is at point O.
• When s>0, the particle is on the positive side of point O.
• When s<0, the particle is on the negative side of point O.
• 31. THANK YOU
• 32. BONUS
• 33. WELL DONE
• 34. WELL DONE
• 35. WELL DONE
• 36. WELL DONE
• 37. WELL DONE
• 38. WELL DONE
• 39. WELL DONE
• 40. WELL DONE
• 41. TRY AGAIN
• 42. TRY AGAIN
• 43. TRY AGAIN
• 44. TRY AGAIN
• 45. TRY AGAIN
• 46. TRY AGAIN
• 47. TRY AGAIN
• 48. TRY AGAIN