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 pointdetermine 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

o-6 m +6 m

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

a. OQ = 12 mQ is 12 m east of O

QO

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

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

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

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

POP QUIZ

1

2 3

4

565

7

8

9

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 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

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

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

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

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

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

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

EVALUATION

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.)

(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

(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

(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.

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.