2.1 analysing linear motion. introduction how fast ? (speed / velocity) does it change its speed ?...
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2.1
ANALYSING
LINEAR MOTION
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INTRODUCTION
How fast ?(Speed / velocity)
Does it change its speed ?(Acceleration / deceleration)
How would you describe the motion in word ?
How far does it travel ? (distance/displacement)
Information required:
How fast ?(Speed / velocity)
How far does it travel ? (distance/displacement)
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• A straight line motion
LINEAR MOTION
• Not a straight line motion
NON LINEAR MOTION
• Total path travelled in a given time is the same as the shortest path
• Total path travelled in a given time is different from the shortest path
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• DISTANCE AND DISTANCE AND DISPLACEMENT DISPLACEMENT
• SPEED AND SPEED AND VELOCITY VELOCITY
• ACCELERATION AND ACCELERATION AND DECELERATIONDECELERATION
Learning areaLearning area
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DISTANCE AND DISPLACEMENT
Pontian Kecil Desaru
Johor Bahru
Pontian Kecil
Ayer Hitam
SenaiKota Tinggi
Mawai
Benut
How far is it from Johor Bahru to Desaru ?
Distance = total path length =JB to Desaru via Kota Tinggi
Displacement = shortest path length = JB direct to Desaru
SCALAR
VECTOR
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SPEED AND VELOCITY
start
end
path
Distance =
Displacement =
Average Speed =
Average Velocity =
Time taken =
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ACCELERATION AND DECELERATION
Velocity increases
Constant velocity
Velocity decreases
Acceleration = Rate of change of velocity
= Change of velocity Time= final velocity – Initial velocity Time
a = v – u t
vector
m s-2
Velocity increases = acceleration
Velocity decreases = deceleration
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Carry out Hands-on Activity 2.2
( page 11 of the practical book)
Aim : To differentiate between acceleration and decelerationDiscussion :1. (a) The speed of the trolley increases . (b) The speed of the trolley decreases.
2.Acceleration is the rate of increasing speed in a specified direction.
Deceleration is the rate of decreasing
speed in a specified direction.
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Lesson 2Lesson 2
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RELATING DISPLACEMENT, RELATING DISPLACEMENT, VELOCITY , ACCELERATION VELOCITY , ACCELERATION
AND TIMEAND TIME
a. Using ticker tape
b. Using Equations of Motion
Learning areaLearning area
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ticker timer
ticker tape
A.C. 50 Hz
50 dots made in 1 second
Carbon disc
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Time interval between two adjacent dots = 1/50 s
= 0.02 s
1 tick = 0.02 s
dots
1 tick
Slow movementfaster movementfastest movement
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PREPARING A TAPE CHART (5 -TICKS STRIP)
0 5 10
First 5-tick strip
2nd 5-tick strip
Velocity, v
(cm /s)
Time / s
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INFERENCE FROM TICKER TAPE AND CHART
•Zero acceleration
•constant velocity
• Constant acceleration
• Constant deceleration
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Carry out Hands-on Activity 2.3
( page 13 of the practical book)
Aim : To use a ticker timer to identify the types of motion
Discussion 2.3(A):2. Spacing of the dots is further
means a higher speed.
Spacing of the dots is closer means a slower speed.
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Discussion Hands-on Activity 2.3(B)
( page 13 of the practical book)
Aim : To determine displacement, average velocity and acceleration
Discussion 2.3(B):1. Prepare a tape chart.2. Determine average velocity using v = Total displacement
time3. Determine acceleration using a = final velocity – initial velocity
time
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Lesson 3Lesson 3
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TO DETETMINE THE AVERAGE VELOCITY
EXAMPLE The time for each 5-tick strip = 5 x 0.02 s
= 0.1 sLength / cm
Time / s
0
7
10
1415
22
0.10.2
0.30.4
0.50.6
0.7
= (7 +10 +14 +15 +22 +14 +10) cm= 92 cm
= 7 strips = 0.7 s
Total displacement
Total time taken
Average velocity = displacement Time taken
= 92 / 0.7 = 131.4 cm s-1
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TO DETERMINE THE ACCELERATION
EXAMPLE The time for each 10-tick strip = 10 x 0.02 s
= 0.2 s
5.8 / 0.2 =29 cm s-1
27.3 / 0.2 = 136.5
Initial velocity, u
Final velocity, v
acceleration = v-u t= (136.5 – 29) cm s-1
1.2 s
Length / cm
Time / s
0 0.20.4
0.6 11.2
Time takenTime taken
=(7-1 )strips
= 6 x 0.2 s= 1.2 s
5.8
27.3
1.40.8
= 89.6 cm s-2
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Lesson 4Lesson 4
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ss = Displacement = Displacement
uu = Initial velocity = Initial velocity
vv = Final velocity = Final velocity
aa = Constant = Constant acceleration acceleration
tt = Time interval = Time interval
THE EQUATIONS OF MOTION
v u at 21
2s ut at 2 2 2v u as
2
u vs t
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EXAMPLE
A car travelling at a velocity 10 m s-1 due north speeds up uniformly to a velocity of 25 m s-1 in 5 s. Calculate the acceleration of the car during these five seconds
u = 10 m s-1 , v = 25 m s-1, t = 5 s, a = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using v = u + at
25 = 10 + a(5)
5 a = 15
a = 3 m s-2
Don’t forget the unit
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EXAMPLE
A rocket is uniformly accelerated from rest to a speed of 960 m s-1 in 1.5 minutes. Calculate the distance travelled.
u = 0 m s-1 , v = 960 m s-1, t = 1.5 x 60 = 90 s, s = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using s = ½ (u + v)t
s = ½ (0 + 960) 90
= 43 200 mWhat is the
unit ?
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EXAMPLE
A particle travelling due east at 2 m s-1 is uniformly accelerated at 5 m s-2 for 4 s. Calculate the displacement of the particle.u = 2 m s-1 , a = 5 m s-2, t = 4 s, s = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using s = ut + ½ at2
s = 2(4) + ½ (5)(4)2
= 8 + 40
= 48 m What is the unit ?
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EXAMPLE
A trolley travelling with a velocity 2 m s-1 slides 10 m down a slope with a uniform acceleration. The final velocity is 8 m s-1. Calculate the acceleration.
u =2 m s-1 , v = 8 m s-1 , s = 10 m , a = ?
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using v2 = u2 + 2as
82 = 22 + 2 a (10)
20 a = 64 – 4
= 60
a = 3
What is the unit ?
m s-2
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EXAMPLE teks book pg 27
u =0 m s-1, a = 2.5 m s-2 , t = 10 s v = ? , s = ?
Using v = u + at
= 0 + (2.5)(10)
= 25 m s-1
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
Using s = ut + ½ at2
= 0(10) + ½ (2.5)(10)2
= 125 m
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EXAMPLE teks book pg 27
u = 25m s-1, v = 0 m s-1 , s = 50 m , a = ?
Using v 2 = u2 + 2as
0 = 252 + 2a (50)
0 = 625 + 100a
a = - 625 100
= - 6.25 m s-2
v = u + at
s = ut + ½ at2
v2 = u2 + 2as
s = ½ (u + v) t
The negative sign shows deceleration.