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

Cikgu Jasmin’s house is 500m away. Cikgu Jasmin’s house is 500m due east.

CHAPTER 9.1 THE CONCEPT OF DISPLACEMENT

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

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

Displacement of Upin and Ipin 6 metres to right or to the left of point O Example 1

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

OQ = 12 m Q is 12 m east of O Q O 12 m Solutions

b. OR = -5 m R is 5 m west of O Q O 5 m

c. OS = 0 m The displacement 0 refers to 0 m from O, meaning that S is at O O S

9.1.2 Displacement, s of a particle can be represent as a function of time, t. For example: DETERMINE DISPLACEMENT OF PARTICLE

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

(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

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

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

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

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

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

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

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

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

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.)

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

(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

(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.

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.

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

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