Linear Motion 1

Moyin Sir (M. Sc). Cell Phone: 01716169640

LINEAR MOTION

Distance: Distance is the length of the route between two points.

There can be many routes between two points thus there can be many

distance between two points. Distance is a scalar quantity. The unit of

distance in SI unit is meter.

Displacement: is the distance between two points in a straight line.

It is also the shortest distance between two points.

Displacement between two points is always the same but the distance can

be any route and different. In the following route map, the distance between

the two points A and B can be any route taken by a moving object, but the

displacement is the only line AB though there is no route.

AAAA

BBBB


Speed: The rate of change of distance is called ‘Speed’. It is a scalar

quantity and the unit is m/sec. In case of a straight line the speed and the

velocity are the same.

�� = ��

In everyday life speed means average speed, as the speed is very difficult to

keep constant because we always live with different types of forces and

obstacles. Thus the average speed is defined as

�� = ��

Velocity: The rate of change of displacement with time is called the

‘Velocity’. It is a vector quantity. It means that it has both magnitude and

direction. Unit in SI system is meter/second (m/sec).

�� = �� …………….(3)

Or �� = � ………………… (4)

(Where s = distance)

Constant velocity means that in every unit time the moving object covered

the same unit of displacement. Velocity is only constant when the net force

is zero.

Practically the constant velocity is difficult to achieve. Velocity can only be

changed when a ‘Force’ is applied. Velocity will be changed, either in

magnitude (value) or in the direction. If the force is applied in the same

direction of the velocity then it will cause acceleration, and if the force is

applied in the opposite direction of the velocity then it will cause

deceleration. And if the force is perpendicular then direction of the velocity

will be changed.

�� = �� + �� ……………………………..(5)


If the acceleration is constant then the average velocity can be calculated

by:

�� = � � �� = �

………………….(6)

Force

Velocity

= Acceleration

Force Velocity = Deceleration

Velocity

Force

= Direction Change


Acceleration: The rate of change of velocity is called the acceleration. To

produce acceleration there must be net force acting on the moving object.

No NET forces no acceleration. Acceleration is a vector quantity, and its

unit is m/sec2

i) Initial velocity = �

ii) Final Velocity = �

iii) Velocity change = ( � – �)

iv) Velocity Change in unit time = ( � " � )

Thus the acceleration � = ( � " � ) ………………….(7)

� = �� ………………………..(8)

� = ��


Equations of Motion:

� = � …………………(1)

� � �

� = � ………………(2)

� = � + � …………………. (3)

Putting the value of v from equation (3) into the equation (2):

And by making d as subject:

(�� )� �

� = �

� = � + � �

� ……………………(4)

Again from equation (2), by making t subject:

= ( �"� )�

And putting this value in equation (2):

� � �

� = �

(��)

� = �

(�"�)

�(�"�)

� = �

�� = �� + �� ………………… (5)

Thus the equations of motion are:


� = � ………………… (1)

� � �

� = � ……………… (2)

� = � + � …………………. (3)

� = � + � �

� ……………….. (4)

�� = �� + �� ………….. (5)


Motion Graph: Distance-time graph.

Distance cannot be negative as it is scalar, and time cannot be negative as

well, thus there is only one quadrant. In distance-time graph the gradient

indicates the speed of the object. And the area under the graph has no

physical significance.

Gradient = Speed

i) No gradient = No speed.

The car A is at rest. The car A is 15meter away from the starting point.

START Positive direction Negative direction

Car A

15 m

Time/sec

Distance/m

Car A 15


ii) Constant Gradient = Constant Speed.

Higher Constant Gradient = Higher constant Speed.

a) All the cars C and D and Car E have constant speed, as they have

constant gradient.

b) Car C has higher Speed than Car D as it has higher gradient.

c) The Car E is coming back to the starting point with a constant

speed.

d) The car C and D is moving away from the starting point with a

constant speed

Time/sec

Distance/m

Car D

Car C

Car E


Car D

Car C Car E


e) All the cars have different constant speed.

iii) Increasing Gradient = Increasing Speed, which is ACCELERATION

iv) Decreasing Gradient = Decreasing Speed, which is

DECELERATION.

a) The gradient of this car F is increasing every moment constantly, thus

car F is accelerating.

b) The gradient of the car G is decreasing every moment constantly, thus

car G is decelerating.

Car G

Time/sec

Distance/m

Car F

i. Constant Gradient = Constant Speed/Velocity.

ii. Greater constant gradient = Greater constant Speed/Velocity.

iii. No gradient = no speed (stopped/rest)

iv. Increasing gradient = Increasing speed/velocity (acceleration).

v. Decreasing gradient = Decreasing speed/velocity (deceleration)


Displacement-Time graph.

Displacement is a vector quantity thus it can be negative and it has negative

axis but time is a scalar quantity and it cannot be negative. So there will be

two quadrant and negative means ‘Opposite direction’. The gradient of

displacement time graph is ‘Velocity,

GRADIENT = VELOCITY

i) No gradient = No Velocity.

Car D


Car A

Time/sec o

-

+

Displacement/m

Car D

Car A


a) Both the car A and D are stopped (At rest), but their positions are in

the opposite of the starting point thus they have opposite direction.

ii) Constant Gradient = Constant Velocity


Car C Car E

Car D

Car M

Car N

Time/sec

Car M

Car E

o

Car D

Car C

-

+

Displacement/m

Car N


Car C: Moving with a constant velocity, Positive direction, Highest velocity

(It has the highest gradient)

Car D: Moving with a constant velocity, but it is less than Speed of all the

cars (It has the lowest gradient).

Car E: Moving with a constant velocity, in the negative region, moving

towards Starting position. Graph approaches towards origin.

Car M: Moving with a constant velocity, In positive region, moving to the

starting point, graph approaches to the origin

Car N: Moving with a constant velocity, in the negative region, going away

from the starting position

All the above cars have different velocities. Their velocities are constant as

the graphs are straight lines with constant gradient. The graph also shows

the direction of their motion as well.

Q1. Explain the motion of the cars from the following graphs:

P

Time/sec O

Displacement/m

Q

R

P

Time/sec O

Displacement/m

Q

R

Graph-1 Graph-2


iii) Changing Gradient = Changing Velocity (Acceleration

or deceleration)

Car S and T are both accelerating as their gradient is increasing with time.

They are moving in the opposite direction. Car S accelerates in the positive

direction (To the right) and the car T is accelerating in the opposite direction

(To the left)

Car T

Car S

Time/sec o

-

+

Displacement/m


The car P and Q are both decelerating as their gradient gradually decreasing.

They are decelerating in the opposite direction.

Car Q

Car P

Time/sec o

-

+

Displacement/m


Speed time graph: The gradient means ‘Acceleration’ and area under the

graph means ‘distance travelled’. Speed cannot be negative as it is scalar

and time cannot be negative as well.

Gradient = Acceleration

Area under the graph with time axis = Distance travelled

i) No gradient = no acceleration ( Constant speed or at rest)

The car A1 is moving with a constant speed of 15ms-1. As the gradient of the

graph is zero. And the car A is not moving at all it has also ‘Zero’ gradient.

The shaded area shows that the distance travelled by the car A1 during

40second

a) This graph is not showing which direction the car was moving.

Time/sec

Speed/ms-1

Car A1 15

Car A 40


b) If several cars are moving in different directions, but in the graph,

they will be plotted in the same quadrant.

ii) Constant positive gradient = Constant acceleration.

Higher gradient = Higher acceleration.

a) Both the cars M and N are moving with a constant acceleration, as

they have constant positive gradient.

b) The car N has higher acceleration than car M as N has more gradient

than M.

c) The shaded area shows the distance travelled by the car M after time

‘t’.

Time/sec

Speed/ms-1

Car M

Car N

t


iii) Negative gradient = Deceleration (the line moves towards

the time line)

a) Both car P and Q are decelerating as their gradients are negative.

b) Car P has more deceleration than Q as it has higher negative gradient

than Q.

c) The shaded area is the distance travelled by the car Q during

deceleration up to time‘t’.

Time/sec

Speed/ms-1

Car Q

Car P

t


iv) Increasing gradient = Increasing acceleration.

v) Decreasing gradient = Decreasing acceleration

a) The car T is moving with increasing acceleration. Example: When the

rocket moves upward its acceleration increases as the net force

gradually increases.

Car S

Time/sec

Speed/ms-1

Car T


Velocity time graph:

Gradient = Acceleration.

Area under the graph with time axis = Displacement.

The velocity- time graph also shows the direction.

■ Both the cars M and N are moving with constant velocity. The graphs

have no gradient.

■ Car N has higher constant Velocity than Car M.

■ The area under the graph shows the displacement travelled

Car M

Car N

Time/sec o

-

+

Velocity/ms-1


■ Both the car J and K are moving with constant Acceleration. They are

moving in the opposite direction.

■ Car J has higher acceleration than car K as J has higher gradient than

k.

■ The area is the displacement of the two cars.

o

Car K

Car J

Time/sec

-

+

Velocity/ms-1


■ Both the cars L and M are decelerating, they are moving in the

opposite direction.

■ The deceleration of the car L is higher than car M.

■ Area = displacement moved while decelerating.

Car L

o

Car M

Time/sec

-

+

Velocity/ms-1


■ Car X is moving with a constant deceleration, as it has negative

gradient.

■ Car Y is moving with a constant acceleration as it has positive

gradient.

■ The area under the graph shows the displacement travelled by the two

cars.

■ The two cars are moving in the opposite direction.

Car X

o

Car Y

Time/sec

-

+ Velocity/ms-1


Calculating Motion:

Speed, acceleration or deceleration can be calculated by different ways:

1) Light Gates.

2) Motion Sensors.

3) Ticker-timer and tape.

1) Light Gates:

A narrow beam of light or radiation falls from a source on a sensor

connected to a data logger to a computer, if this beam is intercepted then

the time of the interception is recorded in computer. And the rate of change

of distance or velocity can be measured.

There is a card (interrupter) attached with the glider and the length of the

card is measured, these data are fed in the computer and the computer can

accurately measure the motion of the glider.


Advantage of using light-gate:

1) The readings are human-error free.

2) Large number of data can be obtained.

3) The data can be analyzed directly with computer.

4) Time saving.

Disadvantages:

1) Needs accurate setup.

2) Depends on power supply.

3) Depends on software.

4) Needs training to use.


Thinking Distance: The distance travelled by a car before the brakes are

applied, while the Driver is still reacting.

● The car moves with a constant velocity now.

● The thinking distance depends on the performance of the driver, the

reaction time.

The reaction time of a Driver depends on:

� Tiredness of the Driver.

� Poor weather conditions effecting visibility of the driver.

� The effect of alcohol or drugs.

AS the car moves with a constant velocity the thinking distance can be

calculated by the following equation:

Thinking Distance = Speed of the car at that moment × Reaction time

d1 = V × t …………………(a)

The braking distance: The distance moved by the car after the brakes

have been applied.

● This time the car is decelerating, as the car is brought to rest, the final

velocity is ‘ZERO’.

● The initial velocity is the velocity at which the car was moving.

The Braking distance depends on: (The brakes may take longer to stop

the car)

� A wet or icy road.

� A heavy load in the car.

� Worn Brakes or Tyres.

� The initial velocity of the car


As the car is decelerating the braking distance must be calculated with

average velocity equation:

Braking distance = Average velocity × time

…………..(b)

Stopping Distance: The total distance needed to stop a car completely. It

is the addition of the ‘Thinking distance’ and the ‘Braking distance’.

Thus:

Stopping distance = Thinking distance + Braking Distance

D = d1 + d2

The following chart shows the stopping distances for a car at different speed.

This is an average figure for a dry road; the actual stopping distance may be

greater.

tvu

d+

=2

2

Thinking Distance Braking Distance

30m/s

25m/s

20m/s

15m/s


This graph shows the velocity time graph of stopping distance. The car

moved with a constant velocity that is indicated in the blue part, and when

the brakes are applied the car decelerated and that is indicated in the yellow

part. The total area is the stopping distance.

i) Blue area = Thinking distance.

ii) Yellow area = Braking distance.

Time/s

Velocity/ms-1

O 2

4

6

8


Q1: The motion of a moving object is shown in the following graph:

i) Explain the motion of the object for the first 2seconds.

ii) The total distance travelled by the object is 150m. Calculate the

value of the maximum velocity ‘v’.

iii) Calculate the acceleration of the object,

iv) How long the object was in the constant velocity?

v) Calculate the deceleration of the object.

vi) Calculate the distance travelled during deceleration.

vii) Calculate the distance travelled during acceleration.

viii) Calculate the distance travelled by the object during constant

velocity.

ix) The mass of the object is 1350kg; calculate the braking force of

the object.

x) On another occasion, the braking force is doubled; calculate the

stopping time when the object was moving with the same

velocity.

v

Velocity/ms-1

O 2

4

6

8

Time/sec


Q2: An object of mass 250g is thrown vertically upward with a velocity

45m/s.

Calculate:

i) The maximum height gained by the object.

ii) The time to reach that height.

iii) The total time to fall to the hand again.

iv) Draw a speed/time and velocity/time graph of the object of the

whole Journey (From the throwing to the catch, you must include

the value).


Bouncing Ball: A rubber ball is dropped from a height it will bounce several

times on a solid hard surface. The motion of the ball can be plotted on a

graph paper.

Distance time graph:

The distance-time graph would be like this. The centre of gravity is moving

and as the centre never touches the ground the graph is not touching the

time axis.

Distance/m

Time/s


Velocity –time graph:

Velocity is a vector; the direction is indicated by positive or negative sign. It

is considered that upward is positive and downward is negative.

i) The object is dropped and it is falling downward so it is moving to

the negative direction with a constant acceleration, so it is a graph

in the negative part with a constant gradient.

ii) In the part OA the ball is falling. At A the ball touches the ground.

In the part AB there is an impact force on the ball and it is

decelerating, thus the graph goes to the time line.

iii) In the part BC the ball expands and there is an upward force

(Normal reaction) on the ball, and the ball accelerates upward (as

P

Q

Velocity/ms-1

Time/s O

A

B

C

D

E

F

G


long as the ball is in contact with the ground) thus the graph goes

away from the time line.

iv) From CD the ball is moving upward and it is now facing deceleration

due to gravitational force. Thus the graph moves to the time line.

The gradient of line OA and CD are same and the gravitational

acceleration and deceleration are same in magnitude.

v) At the point D the ball is in the top point after one bouncing

vi) The area under the graph is the height fallen by the ball. The green

area is the first dropping height and the yellow area is the bouncing

height (the height gained by ball as it goes up).

vii) The first dropping height is more than the height gain after the

impact as some energy is converted into heat and sound.

viii) The time, like PF is the impact time, the first impact time is very

less than the second impact time. As in the first impact the ball

dropped from a greater height, and in the second impact the height

is less than before so the impact time is also less than before.

AAAA


Acceleration –time graph:

i) Only the acceleration is plotted. No deceleration is plotted.

ii) The negative and positive indicates the direction of the acceleration.

iii) The acceleration due to gravity is constant as 9.81, and as

acceleration takes places in the downward so it is in the negative

region.

(-) 9.81

Acceleration/ms-2

Time/s


iv) The upper positive acceleration is the due to the reaction force. And

initially it is more as the ball is dropped from a greater height so

impact is more and the magnitude of the acceleration is more, and

as the ball is bouncing, the height reduces and the impact force

reduces and the acceleration also.

v) In the first impact the force was greater so the acceleration was

more and the impact time was less #$ ∝∝∝∝ %& . In the second impact

the force was less (as it fell from less height) and the acceleration

was less as well, and the impact time was more than before.

vi) AS the dropping height decreases the time for the ball to drop

decreases, thus the duration of the g decreases, but the magnitude

remains the same.


Calculation of the Gravitational acceleration ‘g’ (3B)

To calculate the value of gravitational acceleration ‘g’ in lab:

i) An object is dropped from a known height and the falling time is

calculated.

ii) If the dropping height is short then it is difficult to calculate the

dropping time with a stop watch, as human reaction time cannot be

ignored in this case, and if the dropping height is very long then the

air friction acting on the falling object will be significant that the

value of ‘g’ will be different than actual value.

iii) To get a more accurate result for ‘g’ electronic milli-timer timed

with electromagnet and controlling switch is used.

iv) When the switch is open the electromagnet will lose magnetism and

the iron sphere will start to fall simultaneously the electronic timer

will start, and when the iron sphere will hit the contact plate, the

Iron sphere

Electronic timer

Contact plate

Electromagnet

Ruler

Switch

Stand

h


plate will lose the contact and the circuit will be stopped and the

timer will be stopped (show the time required to drop the height

‘h’) and will show the dropping time.

v) This experiment is free from the errors occurred due to human

reaction time and more accurate. To avoid further errors due to air

friction height should be as small as possible.

As

2

2

1gtuth += Here u = 0. Thus

2

2

1gth= or

2

2

t

hg =

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Creation Date: 02-Jul-12 12:47:00 PM

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Last Saved On: 19-Mar-13 12:16:00 PM

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Linear Motion 1

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