2.1 ANALYSING LINEAR MOTION. INTRODUCTION How fast ? (Speed / velocity) Does it change its speed ?...
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Transcript of 2.1 ANALYSING LINEAR MOTION. INTRODUCTION How fast ? (Speed / velocity) Does it change its speed ?...
2.1
ANALYSING
LINEAR MOTION
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)
• 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
• DISTANCE AND DISTANCE AND DISPLACEMENT DISPLACEMENT
• SPEED AND SPEED AND VELOCITY VELOCITY
• ACCELERATION AND ACCELERATION AND DECELERATIONDECELERATION
Learning areaLearning area
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
SPEED AND VELOCITY
start
end
path
Distance =
Displacement =
Average Speed =
Average Velocity =
Time taken =
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
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.
Lesson 2Lesson 2
RELATING DISPLACEMENT, RELATING DISPLACEMENT, VELOCITY , ACCELERATION VELOCITY , ACCELERATION
AND TIMEAND TIME
a. Using ticker tape
b. Using Equations of Motion
Learning areaLearning area
ticker timer
ticker tape
A.C. 50 Hz
50 dots made in 1 second
Carbon disc
Time interval between two adjacent dots = 1/50 s
= 0.02 s
1 tick = 0.02 s
dots
1 tick
Slow movementfaster movementfastest movement
PREPARING A TAPE CHART (5 -TICKS STRIP)
0 5 10
First 5-tick strip
2nd 5-tick strip
Velocity, v
(cm /s)
Time / s
INFERENCE FROM TICKER TAPE AND CHART
•Zero acceleration
•constant velocity
• Constant acceleration
• Constant deceleration
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.
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
Lesson 3Lesson 3
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
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
Lesson 4Lesson 4
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
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
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 ?
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 ?
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
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
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.