Add Math (F5) Motion Along A Straigh Line Subtopic 9.1
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Transcript of Add Math (F5) Motion Along A Straigh Line Subtopic 9.1
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Cikgu Jasmin’s house is 500m
away.
Cikgu Jasmin’s house is 500m
due east.
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CHAPTER 9.1
THE CONCEPT OF DISPLACEMENT
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At the end of the lesson, students should be able to:identify the direction of the displacement of a particle from a fixed pointdetermine the displacement of a particle from a fixed point
LEARNING OUTCOMES
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• 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
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• Displacement of Upin and Ipin 6 metres to right or to the left of point O
Example 1
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o-6 m +6 m
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Describe the position of each of the following points with respect to O, taking east as the positive direction.
a. OQ = 12 mb. OR = -5mc. OS = 0 m
Exercises
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a. OQ = 12 mQ is 12 m east of O
QO
12 m
Solutions
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b. OR = -5 m
R is 5 m west of O
Q O
5 m
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c. OS = 0 m
The displacement 0 refers to 0 m from O, meaning that S is at O
O
S
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DETERMINE DISPLACEMENT OF PARTICLE9.1.2
Displacement, s of a particle can be represent as a function of time, t.
For example:
23 tts tts 23
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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.
32 2 tts
Example 2
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Time,t 1 2 3 4
Displacement,s
4 3 0 -5
(a)
(b)
s (m)0 1 2 3 4-5 -3 -2-4 -1
t=4 t=3 t=2 t=1
Solutions
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POP QUIZ
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1
2 3
4
565
7
8
9
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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
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4m of the southof O
4m of the westof O
4m of the northof 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
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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
DB
C
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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
DB
C
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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 = 3t2 – 18. Find the displacement of the particle
before it starts to move.
A
DB
C
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27m of the left9m of the right
27m of the right9m 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 = 3t2 – 18. Describe the position of the particle after 3
seconds.
B
A
D
C
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7m of the left4m of the left
4m of the right7m 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 = t2 – 3t- 4. Describe the position of the particle before
it starts to move.
A
B D
C
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5m of the right32m of the right
32m of the left5m 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 = 3t2 – 3t – 4 . Find
the displacement of the particle at 4 seconds.
A
DB
CA
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EVALUATION
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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 = 10t – 2t2
(Take the right direction from point O as positive direction.)
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(a) Find the displacement of the particle at t=1.5.
s = 10t – 2t2
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
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(b) the time at which the particle is 12 m on the left side of point O
s= -1210 t – 2 t2 = -12
Rearrange - 2 t2 + 10 t +12 = 0Multiply -1 2 t2 – 10 t – 12 = 0
t2 – 5t – 6 = 0Factorize (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
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(c) When the particle returns to point O.
When s = 010t – 2t2 = 0
Factorize 2t (5 - t) = 0 2t = 0 or 5 – t = 0 t = 0 or t = 5
Therefore, the particle returns to point O att = 5.
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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.
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