2.1 analysing linear motion
TRANSCRIPT
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Linear Linear MotionMotion
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TYPES OF QUANTITIESTYPES OF QUANTITIES
SCALAR QUANTITIES / UNIT
VECTOR QUANTITIES / UNIT
Distance / m Displacement / m
Speed / ms-1 Velocity / ms-1
Acceleration / ms-2
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Linear Motion
Linear motion is a motion in a straight line
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THE DEFINITION OF THE ……
DISTANCE : the length of the actual path it has followed
DISPLACEMENT : the straight line from the start to the end of movement
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50 m
Distance = 50 m
Displacement = 50 m in the south-west direction
Example 1
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A B C3 m 4 m
(a) Distance = 3 +4 = 7 m
(b) Displacement = 3 + 4 = 7 m in the east direction of A
Example 2
Calculate :(a)Distance(b)displacement
Solution :
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A B C3 m 4 m
(a) Distance = 3 +4 + 4 = 11 m
(b) Displacement = 3 + 4 - 4 = 3 m in the east direction of A
Example 3
Calculate :(a)Distance(b)displacement
Solution :
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30 m
40 m
Distance = 30 +40 = 70 m
Displacement = m503040 22
Example 4
in the south-west direction
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30 m
Distance = 50 + 30 = 80 m
Displacement =
40 m 50 m
m403050 22
Example 5
in the south direction
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Speed
)(,
(m) sDistance,
sttimespeed
Speed is the distance travelled per unit time
Or Speed can also be define as the rate of change of distance
The average speed is calculated by:
takentime
travelleddistance totalspeed average
SI unit: m s-1
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Velocity
)(,
(m) snt,displaceme
sttimevelocity
Velocity is the speed in a given directionvelocity can also be define as the rate of change of displacement
The average velocity is calculated by:
takentime
ednt travelldisplaceme total velocityaverage
SI unit: m sSI unit: m s-1-1
OrOr
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Speed is a scalar quantity and it has magnitude but no direction
but
Velocity is vector quantity which has both magnitude and direction
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A B C5 m in 2 s 4 m in 1 s
(a) Distance = 5+4 = 9 m
(b) Displacement = 5 + 4 = 9 m
Example 1
Calculate :(a)Distance (c) speed(b)Displacement (d) velocity
Solution :13
3
9)(
sm
speedc
13
3
9)(
sm
velocityd
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A B C4 m in 2 s 4 m in 2 s
(a)Distance
= 4 + 4 + 4 + 4 = 16 m
(b) Displacement
= 4+ 4 – 4 – 4 = 0 m
Example 2
Solution :
Calculate :(a)Distance (c) speed(b)Displacement (d) velocity
12
8
16)(
sm
speedc
10
8
0)(
sm
velocityd
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30 m in 7 s
40 m in 3 s
(a) Distance
= 30 +40 = 70 m
(b) Displacement
= m503040 22
Example 4
Calculate :(a)Distance (c) speed(b)Displacement (d) velocity
17
10
70)(
sm
speedc
15
10
50)(
sm
velocityd
Solution :
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Example 5
An athlete runs a 100 m track in 10 s. what is his velocity?
Solution :
110
10
100
,
sm
t
svvelocity
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Acceleration
Acceleration is defined as theAcceleration is defined as the rate of change of velocityrate of change of velocity
t
uva
takentime
velocityinitial- velocityfinal
takentime
velocityofchangeonaccelerati
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Acceleration is a vector quantity
Negative acceleration is also called deceleration
SI unit: m s-2
The acceleration is positive if the velocity increases with time.
The acceleration is negative if the velocity decreases with time
The acceleration is positive if the velocity increases with time.
The acceleration is negative if the velocity decreases with time
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THE DEFINITION OF THE ……
ACCELERATION : the rate of change of the velocity
DECELERATION : the rate at which the velocity of an object decreases
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The car was moved with acceleration
The car was moved with deceleration
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Using the information below, try to calculate the value of acceleration :
• Initial velocity, u = 0 ms-1
• Final velocity, v = 50 ms-1
• Time taken, t = 10 s
Solution:
Example 1
t
uva
25
10
050
sm
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Using the information below, try to calculate the value of acceleration :
• Initial velocity, u = 40 ms-1
• Final velocity, v = 0 ms-1
• Time taken, t = 10 s
Solution:
t
uva
24
10
400
sm
Example 2
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A van accelerated uniformly from a velocity of 15 m s-1 to 20 m s-1 in 2.5 s. What was the acceleration of the van?
Solution:
t
uva
20.2
5.2
1520
sm
u = 15 m s-1
v = 20 m s-1
t = 2.5 s
Example 3
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A runner accelerates at a constant rate from rest and reaches a velocity of 10 m s-1 after 5.0 s. what is his acceleration?
Example 4
Solution:
t
uva
20.2
5
010
sm
u = 0 m s-1 ( rest )v = 10 m s-1
t = 5.0 s
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A motorcycle, travelling at 40.0 m s-1, takes 10 s to stop. What is its deceleration?
Example 5
Solution:
t
uva
20.4
10
400
sm
u = 40 m s-1
v = 0 m s-1 (stop)
t = 10 s
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atuv )1(
asuv 2)2( 22
tvus )(2
1)3(
2
2
1)4( atuts
u = initial velocityv = final velocityt = times = displacement or distancea = acceleration
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u = 0 m s-1
v = max
gtuv )1(
gsuv 2)2( 22
2
2
1)3( gtuts
a = g
g = gravitational acceleration
Fall towards the ground
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v = 0 m s-1
u = max
gtuv )1(
gsuv 2)2( 22
2
2
1)3( gtuts
a = - g
Thrown vertically upwards
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A school bus accelerates with an acceleration of 4.0 m s-2 after picking up some student at a bus stop. If bus travelled after 5 s.Calculate the:(a) velocity(b) distance
10.20
50.40
)(
sm
atuva
u = 0 m s-1
a = 4.0 m s-2
t = 5 s
m
atutsb
50
542
150
2
1)(
2
2
Practice 1
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In the long jump event, Ahmad was running at a velocity of 5 m s-1 towards the long jump pit. He needed to achieve a velocity of 10 m s-1 after covering a distance of 4.5 m before lifting himself off the ground from the jumping board. Calculate the required acceleration for Ahmad to do so?
Practice 2
u = 5 m s-1
v = 10 m s-1
s = 4.5 m2
22
22
3.8
5.42510
2
sma
a
asuv
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A train moving at 20 m s-1 takes 10 s to accelerate to 25 m s-1.(a) calculate its acceleration.(b) calculate the distance it travels while it accelerates.
Practice 3
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u = 20 m s-1
v = 25 m s-1
t = 10 s
25.0
10
2025
)10(2025
)1(
sm
a
a
atuv2
2
1)4( atuts
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A car starts from rest with a constant acceleration of 2 m s-2. what is its velocity after 5 s?
Practice 4
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A cyclist riding at a velocity of 4 m s-1 braked with uniform deceleration and stopped in 3.6 m. How long did he take to stop.
Practice 5
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A car accelerates at a uniform rate from 5 m s-1 to 30 m s-1. The acceleration of the car is 5 m s-2. Find;(a) time taken of car(b) distance car travelled
Practice 6
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A stone is thrown vertically upwards at a velocity of 20 m s-1. If the gravitational acceleration is 10 m s-2 and the air resistance is neglected, calculate:(a) the time it will take stone to reach the maximum height(b) the maximum height reached
by the stone.
Practice 7
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A tap drips into a sink which is 50 cm below the end of the tap. If the gravitational acceleration is 10 m s-2 . How long does it take for each drop to fall?
Practice 8
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STUDY OF LINEAR MOTION
USING OF TICKER TIMER
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TICKER TIMER
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Power supply
Trolley
Ticker-timer
Ticker tape
Track
Wooden block
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• A common way of analyzing the motion of objects in physics labs is to perform a ticker tape analysis.
• A long tape is attached to a moving object and threaded through a device that places a tick upon the tape at regular intervals of time.
• As the object moves, it drags the tape through the "ticker," thus leaving a trail of dots.
• The trail of dots provides a history of the object's motion and is therefore a representation of the object's motion
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•The distance between dots on a ticker tape represents the object's position change during that time interval.
• A large distance between dots indicates that the object was moving fast during that time interval.
•A small distance between dots means the object was moving slow during that time interval.
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Ticker tapes for a fast-moving and a slow-moving object are depicted below.
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• The analysis of a ticker tape diagram will also reveal if the object is moving with a constant velocity or with a changing velocity (accelerating).
• A changing distance between dots indicates a changing velocity and thus an acceleration.
• A constant distance between dots represents a constant velocity and therefore no acceleration.
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Ticker tapes for objects moving with a constant velocity and an accelerated motion are shown below.
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The ticker timer can be used to determine the following variables:
1.Time interval of motion2.Displacement of the object3.Velocity of the object4.Acceleration of the object5.Type of motion of the object
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Time Interval of Motion
• The time interval between one carbon and the next one on the ticker tape is known as 1 dot space of time or 1 tick.
• The ticker timer vibrates at frequency of 50 Hz. Therefore, 50 dots will be marked on the ticker tape in motion in 1 second.
• One dot space or one tick is the distance travelled by an object in second = 0.02 second 50
1
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To determine the time interval of motion of the object:
To determine the time interval of motion of the object:
Time interval = Number of tick X 0.02 sTime interval = Number of tick X 0.02 s
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5
dot tick
Referring to figure:The time interval for 8 tick = 8 x 0.02 s= 0.16 s
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• In the analysis of motion using the ticker timer, the displacement of the object is determined by measuring the length of the ticker tape that pulled through the ticker timer.
displacement = length of ticker tape
5
Referring to figure:The displacement for 8 tick= length of the first dot to last dot= 5 cm
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Velocity of the object
• With the quantities of time interval and displacement, velocity can be calculate using the following equation:
s
velocity
02.0tickofNumber
tapetickerofLength
takenTime
ntDisplaceme
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• Refferring to figure:
5
125.31
02.08
5
cms
takentime
ntdisplacemevelocity
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Acceleration of the object
1
1
9.0
90
02.0
8.1
sm
scm
t
sv
1
1
4.0
40
02.0
8.0
sm
scm
t
su
s
t
1.0
02.016
Direction of motion
2
2
5
500
1.0
4090
sm
scm
t
uva
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Direction of motion
1
1
4.0
40
02.05
4
sm
scm
t
su
1
1
8.0
80
02.05
8
sm
scm
t
sv
s
t
2.0
02.0515
2
2
2
200
2.0
4080
sm
scm
t
uva
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PRACTICEDirection of motion
Calculate the acceleration of the object
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Direction of motion
Calculate the acceleration of the object
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Constant velocityDirection of motion
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accelerationDirection of motion
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decelerationDirection of motion
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Constant velocity then acceleration
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Constant velocity then deceleration
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Acceleration then constant velocity
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Length/cm
Time
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Calculation:
(a) displacement = 4+8+12+16+20+24
= 84 cm
(b) Time of the motion, t = 6 x 10 x 0.02
= 1.2 s
(c) Average velocity
170
2.1
84
scm
timetotal
ntdisplaceme
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Calculation: (d) acceleration
1
1
2.0
20
02.010
4
sm
scm
t
su
1
1
20.1
120
02.010
24
sm
scm
t
sv
s
t
0.1
02.01016
2
2
0.1
100
0.1
20120
sm
scm
t
uva
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PRACTICELength/cm
Time
FIND:(a) Displacement(b) Time interval of the motion(c) Average velocity(d) acceleration