Linear Linear MotionMotion

TYPES OF QUANTITIESTYPES OF QUANTITIES

SCALAR QUANTITIES / UNIT

VECTOR QUANTITIES / UNIT

Distance / m Displacement / m

Speed / ms-1 Velocity / ms-1

Acceleration / ms-2

Linear Motion

Linear motion is a motion in a straight line

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

50 m

Distance = 50 m

Displacement = 50 m in the south-west direction

Example 1

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 :

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 :

30 m

40 m

Distance = 30 +40 = 70 m

Displacement = m503040 22

Example 4

in the south-west direction

30 m

Distance = 50 + 30 = 80 m

Displacement =

40 m 50 m

m403050 22

Example 5

in the south direction

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

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

Speed is a scalar quantity and it has magnitude but no direction

but

Velocity is vector quantity which has both magnitude and direction

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

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

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 :

Example 5

An athlete runs a 100 m track in 10 s. what is his velocity?

Solution :

110

10

100

,

sm

t

svvelocity

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

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

THE DEFINITION OF THE ……

ACCELERATION : the rate of change of the velocity

DECELERATION : the rate at which the velocity of an object decreases

The car was moved with acceleration

The car was moved with deceleration

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

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

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

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

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

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

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

v = 0 m s-1

u = max

gtuv )1(

gsuv 2)2( 22

2

2

1)3( gtuts

a = - g

Thrown vertically upwards

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

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

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

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

A car starts from rest with a constant acceleration of 2 m s-2. what is its velocity after 5 s?

Practice 4

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

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

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

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

STUDY OF LINEAR MOTION

USING OF TICKER TIMER

TICKER TIMER

Power supply

Trolley

Ticker-timer

Ticker tape

Track

Wooden block

• 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

•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.

Ticker tapes for a fast-moving and a slow-moving object are depicted below.

• 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.

Ticker tapes for objects moving with a constant velocity and an accelerated motion are shown below.

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

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

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

5

dot tick

Referring to figure:The time interval for 8 tick = 8 x 0.02 s= 0.16 s

• 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

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

• Refferring to figure:

5

125.31

02.08

5

cms

takentime

ntdisplacemevelocity

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

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

PRACTICEDirection of motion

Calculate the acceleration of the object

Direction of motion

Calculate the acceleration of the object

Constant velocityDirection of motion

accelerationDirection of motion

decelerationDirection of motion

Constant velocity then acceleration

Constant velocity then deceleration

Acceleration then constant velocity

Length/cm

Time

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

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

PRACTICELength/cm

Time

FIND:(a) Displacement(b) Time interval of the motion(c) Average velocity(d) acceleration