REST AND MOTION

We see various things around us. Some of them are at rest while others are in motion

Rest:

“A body is said to be at rest, if it does not change its position with respect to its surroundings.”

Surroundings:

Surroundings are the places in its neighbourhood where various objects are present.

Motion:

“A body is said to be in motion, if it changes its position with respect to its surroundings.”

Relative Quantities:

The state of rest or motion of a body is relative. For example, a passenger sitting in a moving bus is at rest because he/she is not changing his/her position with respect to the other passengers sitting in the bus. But to an observer outside the bus, the passengers and objects inside the bus are in motion because they are changing their positions.

TRANSLATORY MOTION

Introduction:

Everything in this universe is in motion. However different objects move differently.

Some objects move along a straight line, some move in curved path, and some move in some other way.

Definition:

“In translational motion, a body moves along a line without any rotation. The line may be straight or curved.”

Examples:

Following are some examples of translatory motion:

Motion of a car in straight line

Motion of electron around the nucleus

Motion of gas molecules

Aero plane moving straight is in translational motion

TYPES OF TRANSLATORY MOTION

There are three types of translatory motion.

Linear Motion (LHR 2014)

Circular Motion

Random Motion (LHR 2013, 2014)

LINEAR MOTION

Definition:

“Straight line motion of a body is known as its linear motion.”

Examples:

Following are some examples of linear motion:

The motion of freely falling bodies.

Motion of a car on a straight and leveled road.

Motion of aeroplanes flying straight in air.

CIRCULAR MOTION

Definition:

“The motion of an object in a circular path is known as circular motion.”

Examples:

Some examples of circular motion are given below:

A stone tied with string, when whirled, it will move along a circular path.

A toy train moving on a circular track.

Motion of individual particle of spinning top.

Earth moving around the sun in solar system

Motion of moon around the Earth.

Motion of a bicycle or a car moving along circular road.

Motion of a rider in Ferris wheel.

RANDOM MOTION

Definition:

“The disordered or irregular motion of an object is called random motion.”

Examples:

The motion of insects and birds.

Brownian motion of gas or liquid molecules along a zig–zag path.

Motion of dust or smoke particles in air

MECHANICS

Definition:

“The branch of physics in which we study motion of objects its causes and effects is called mechanics”

Branches of Mechanics:

There are two branches of mechanics

Kinematics

Dynamics

Kinematics: (GRW 2015)

“The branch of mechanics that deals with the study of motion of an object without discussing the cause of motion is called kinematics.”

Dynamics: (GRW 2015)

“The branch of mechanics that deals with the study of motion of an object and the cause of its motion is called dynamics.”

TYPES OF MOTION

There are three types of motion:

Translatory Motion

Rotatory Motion

Vibratory Motion

Translatory Motion:

“In translational motion, a body moves along a line without any rotation. The line may be straight or curved.”

Examples:

Following are some examples of translatory motion:

Motion of a car in straight line

Motion of electron around the nucleus

Motion of gas molecules

Types of Translatory Motion:

There are three types of translatory motion.

Linear Motion

Circular Motion

Random Motion

Rotatory Motion: (LHR 2013)

“The spinning motion of a body around its axis is called its rotatory motion.”

Examples:

Following are some examples of rotatory motion:

Motion of spinning top

Motion of the Earth around its geographical axis

Motion of wheel and steering wheel around its axis

Motion of a ceiling electric fan

Motion of Ferris wheel

Vibratory Motion: (LHR 2011, GRW 2015)

“To and fro motion of a body about its mean position is known as vibratory motion.”

Examples:

Some examples of vibratory motion are as follows:

Motion of swing back and forth about its mean position

Motion of pendulum of wall clock

Motion of see – saw

Motion of a body attached with a spring.

Motion of hammer of ringing electric bell.

Motion of string of a sitar

Motion of a baby in a cradle moving to and fro

AXIS OF ROTATION

Definition:

“An axis is a line around which a body rotates.”

Position:

In case of rotatory motion the Axis passes through the body while in case of circulatory motion the axis is present out–side the body.

DIFFERENTIATION

Differences between circulatory and rotatory motion are as follows:

A BODY AT REST

When a body does not change its position with respect to its surroundings, then it is said to be in the state of rest.

Example:

A tree standing along a road is in the state of rest with respect to that road.

REST AND IN MOTION AT SAME TIME

If a person is sitting in a moving car, he will be in the state of rest with respect to the other person sitting in the car and he will be in the state of motion with respect to the person standing on the road side at the same time.

TYPES OF MOTION

DIFFERENTIATION

Differences between scalar and vectors are as follows:

MAGNITUDE

Definition:

“The magnitude of a quantity means its numeric value with appropriate unit.”

Examples:

2.5 kg, 40s, 1.8m etc. represent magnitudes of different physical quantities.

NEED OF DIRECTION

It would be meaningless to describe vectors without direction. For example, distance of a place from reference point is insufficient to locate that place. The direction of that place from reference point is also necessary to locate it.

Example of Forces:

Consider a table as shown in figure below:

Two forces F₁ and F₂ are acting on it. It will make lot of difference if the two forces act in opposite direction such as indicated in figure below:

Certainly the two situations differ from each other. They differ due to the direction of the forces acting on the table. Thus the description of a force would be incomplete if direction is not given. Similarly, when we say, we are walking at the rate of 3 kmh^–1 towards north then we are talking about a vector.

REPRESENTATION OF VECTORS

A vector quantity can be represented by two methods

Symbolic Method

Graphical Method

SYMBOLIC REPRESENTATION

To differentiate a vector from a scalar quantity we generally use bold letters to represent vector quantities. Such as F, a, d or a bar or arrow over their symbols such as

GRAPHICAL REPRESENTATION (LHR 2014, GRW 2014)

Graphically, a vector can be represented by a line segment with an arrow head. In figure below, the line AB with arrow head at B represents a vector V. The length of the line AB gives the magnitude of the vector V on a selected scale. While the direction of the line from A to B gives the direction of the vector V.

ADDITION AND SUBTRACTION OF VECTORS

Scalar quantities can be described completely by magnitude only and can be added or subtracted by simple arithmetic rules. Vector quantities in addition to magnitude also need direction for their description. So vectors cannot be added or subtracted by arithmetic rules due to direction.

IMPORTANCE OF VECTOR QUANTITIES

In order to locate a place from a reference point, we will have to describe the distance and direction of that place from reference point. Description of distance along with direction will make up a vector quantity. Hence by using vector quantities we can describe the position (or location) of bodies.

POSITION

Definition:

“The term position describes the location of a place or a point with respect to some reference point called origin”.

Quantity:

Position is a vector quantity. Change in position is called displacement.

Example:

For example you want to describe the position of your school from your home. Let the school be represented by S and home by H. The position of your school from your home will be represented by a straight line HS in the direction from H to S as shown in figure.

ORIGIN

Definition:

“The fixed point that is used as reference point to locate the position of an object or point is called origin.”

Origin is also termed as reference point and it is denoted by “O

DIFFERENTIATION

Differences between distance and displacement are as follows

DIFFERENTIATION

Differences between speed and velocity are as follows:

UNIFORM SPEED

Definition:

“If the speed of a body does not vary and has the same value then the body is said to possess uniform speed.”

OR

“A body has uniform speed if it covers equal distances in equal intervals of time however short the interval may be.”

In this case distance time graph will be a straight line inclined to time Axis.

VARIABLE SPEEED

Definition:

“If a body does not cover equal distances in equal intervals of time, however short the intervals may be, then the speed of the body is said to be variable.”

In this case distance time graph will not be a straight line

UNIFORM VELOCITY

Introduction:

In many cases the speed and direction of a body does not change. In such a case the body possesses uniform velocity. That is the velocity of a body during any interval of time has the same magnitude and direction.

Definition:

“A body has uniform velocity if it covers equal displacement in equal intervals of time however short the intervals may be.”

VARIABLE VELOCITY

Definition:

“If a body does not cover equal displacement in equal intervals of time, however short the intervals may be, then the velocity of the body is said to be variable.”

UNIFORM / VARIABLE VELOCITY

A body moving with uniform speed may have either uniform or variable velocity.

If the direction of the body is not changing then its velocity will also be uniform.

If the direction of the body is changing then its velocity will be variable.

Example 1

A car moving with uniform speed in the straight line will have uniform velocity. If the direction of the body is changing then its velocity will be variable.

Example 2

A car moving with uniform speed in the circular path will have variable velocity because its direction changes at every point on the circle.

VARIABLE VELOCITY ALONG CIRCULAR PATH

A body moving along a circle with uniform speed has variable velocity because its direction is changing at every point on the circular path.

SPEED–O–METER

The speedometer of a car measures only magnitude of velocity not the direction. Therefore, we can say that speedometer of the car does not measure its velocity. It measures only speed.

Acceleration

In many cases the velocity of a body changes due to a change either in its magnitude or direction or both. The change in the velocity of a body causes acceleration in it. If there is no change in the velocity of a body there will be no acceleration in it that is why a body moving with constant velocity does not have acceleration.

ACCELERATION

Definition:

“The rate of change of velocity of a body is known as acceleration.”

Mathematical Form:

If a body is moving with initial velocity ’v_i’ and after some time ‘t’ its velocity becomes ‘v_f’ then change in velocity will be v_f–v_i in time t.

Acceleration =

Acceleration =

So,

Unit:

SI unit of acceleration is meter per second per second (ms^–2).

Quantity:

It is a vector quantity.

UNIFORM ACCELERATION

We know,

Let the time t is divided into many smaller intervals of time. If the rate of change of velocity during all these intervals remains constant then the acceleration a also remains constant. Such a body is said to possess uniform acceleration.

Definition:

“A body has uniform acceleration if it has equal changes in velocity in equal intervals of time however short the interval maybe.”

VARIABLE ACCELERATION

If a body does not have equal changes in velocity in equal intervals of time, however small the intervals may be, then the acceleration of the body is said to be uniform.

POSITIVE ACCELERATION

If the velocity of the body is increasing then acceleration will be positive. The direction of positive acceleration is the same in which the body is moving without change in its direction.

Example:

If a car is moving in straight line and the driver presses the accelerator the velocity of the car starts to increase. So the acceleration of the body will be positive.

NEGATIVE ACCELERATION

If the velocity of the body is decreasing then acceleration will be negative. The direction of negative acceleration is opposite to the direction in which the body is moving. Negative acceleration is also called retardation or deceleration.

Example:

If the driver applies brake, the velocity will start to decrease. So acceleration of the body will be negative and direction of acceleration is opposite to the direction of velocity.

ZERO ACCELERATION

No, a body moving with constant velocity will not have acceleration; its acceleration will be zero because acceleration is defined as the rate of change of velocity. When the body is moving with uniform velocity, the change in velocity will be zero and therefore the acceleration will also be zero.

DIRECTION OF ACCELERATION

Yes, a body moving with certain velocity in the direction of east can have acceleration in the direction of west. It is the case when the velocity of the body decreases. When velocity decreases, acceleration is produced in opposite direction to the direction of motion.

FASTEST ANIMAL

The fastest animal on the Earth is Falcon that can fly at the speed of 200kmh^–1.

LIDAR GUN

A LIDAR gun is light detection and ranging speed gun. It uses the time taken by laser pulse to make a series of measurements of a vehicle’s distance from the gun. The data is then used to calculate the vehicle’s speed. It is being used as motorway speed camera.

TERMINAL VELOCITY

The constant velocity of a body falling down with in gravitational field is called terminal velocity.

EXAMPLE 2.1

STEP # 1 SPECIFICATION OF DIRECTIONS

Draw two lines perpendicular to each other. Horizontal line represents East–West and vertical line represents North–South direction as shown in figure:

STEP # 2 SELECTION OF SUITABLE SCALE

Select a suitable scale to represent the given vector. In this case, we may take a scale which represents 20 N by 1 cm line.

STEP # 3 DRAWING REPRESENTATIVE LINES

Draw a line according to the scale in the direction of the vector. In this case, draw a line OA of length 4 cm along North–East.

STEP # 4 SHOWING DIRECTION

Put an arrow head at the end of the line. In this case, arrow head is at point A. Thus, the line OA will represent a vector i.e., the force of 80 N acting towards North–East.

EXAMPLE 2.2

Given Data:

Total distance = S = 100m

Total time taken = t = 12s

To Find:

Average speed = V_av = ?

Calculations:

Average speed = Total distance moved / Total time taken

V_av = 100m/12s

V_av = 8.33ms^–1

Result:

Hence, the average speed of the sprinter will be 8.33ms^–1.

EXAMPLE 2.3

We can easily deduce given data by drawing figure:

Given Data:

Radius of the circle = r = 318m

Distance covered by the sprinter = S = πr

= (3.14)(318) = 999m

Displacement covered by the sprinter = d = 2r

=2 (318) = 636m

Time taken by the sprinter = t = 1.5 minutes

= 1.5 (60) = 90s

To Find:

Speed of the sprinter = v = ?

Velocity of the sprinter = = ?

Calculations:

Speed= Distance covered/Total time

v =

Putting values

v = 999/90 = 11.1ms^–1

Now we find velocity

Putting values

= 636 / 90 = 7.07 ms^–1

Result:

EXAMPLE 2.4

Given Data:

Initial velocity = v_i = 0

Final velocity = v_f = 20 ms^–1

Time = t = 8s

To Find:

Accleceration = a = ?

Calculations:

As

Or

Result:

Hence the acceleration of the car will be 2.5ms^–2.

EXAMPLE 2.5

Given data:

Initial velocity = v_i = 30ms^–1

Final velocity = v_f = 15ms^–1

Time = t = 3 s

To Find:

Retardation = –a = ?

Calculations:

We know

Acceleration =

So,

Result:

Hence, the retardation in the body will be 3ms^–2.

GRAPH

Definition:

“Graph is a pictorial way of presenting the information about the relation between various Quantities”.

VARIABLES

Definition:

“The quantities between which a graph is plotted are called the variables.”

TYPES OF VARIABLES

Dependent Variables:

The quantities whose values depend on other quantities are called dependent variables. While plotting a graph dependent variable is taken along vertical axis.

Example:

While driving a car distance covered depends on time so distance is a dependent variable

Independent Variable:

The quantity whose value of does not depend on other quantities are called the independent variables. While plotting a graph independent variable is taken along horizontal axis.

Example:

Time is an independent variable.

Uses of Graphs:

Graphs can be used to:

Analyze motion of objects.

Show year–wise growth/decline of export, month–wise rainfall, a patient’s temperature record or runs per over scored by a team and so on.

DISTANCE TIME GRAPH

It is useful to represent the motion of objects using graphs. The terms distance and displacement are used interchangeably when the motion is in a straight line. Similarly if the motion is in a straight line then speed and velocity are also used interchangeably. In a distance–time graph, time is taken along horizontal axis while vertical axis shows the distance covered by the object.

Explanation:

Distance time graphs for different bodies are given below:

OBJECT AT REST

Definition:

“A body is said to be at rest, if it does not change its position with respect to its surroundings.”

In the case the distance moved by the object with time is zero. That is, the object is at rest. Thus, a horizontal line parallel to time axis on a distance–time graph shows that speed of the object is zero.

OBJECT MOVING WITH CONSTANT SPEED

Definition:

“A body has uniform or constant speed if it covers equal distances in equal intervals of time however short the interval may be.”

In this case distance time graph will be a straight line inclined to time Axis.

Consider two points A and B on the graph its slope gives the speed of the object as:

The speed found from the graph is 2ms^–1.

OBJECT MOVING WITH VARIABLE SPEED

Definition:

“If a body does not cover equal distances in equal intervals of time, however short the intervals may be, then the speed of the body is said to be variable.”

In this case distance time graph will not be a straight line

The slope of the curve at any point can be found from the slope of the tangent at that point. For example:

Thus speed of the object at point P is 3ms^–1. The speed is higher at instants where slope is greater and speed is zero at instants where slope is horizontal.

SPEED TIME GRAPH

“The graph that shows the relationship between speed of an object and time taken by it, is called speed time graph.”

In a speed – time graph, time is taken along x – axis and speed is taken along y–axis.

Explanation:

Speed time graph different situations are given below:

SPEED TIME GRAPH FOR CONSTANT SPEED

When speed of an object is constant with time, then the speed – time graph will be a horizontal line parallel to time – axis as shown in figure. In other words, a straight line parallel to time axis represents constant speed of the object.

SPEED TIME GRAPH FOR UNIFORM ACCELERATION

Definition:

“A body has uniform acceleration if it has equal changes in velocity in equal intervals of time however short the interval maybe.”

Let the speed of an object be changing uniformly. In such a case speed is changing at constant rate.

Thus its speed–time graph would be a straight line such as shown in figure below:

A straight line means that the object is moving with uniform acceleration. Slope of the line gives the magnitude of its acceleration.

DISTANCE TRAVELLED BY A MOVING OBJECT

The area under a speed – time graph represents the distance travelled by the object. If the motion is uniform then the area can be calculated using appropriate formula for geometrical shapes represented by the graph.

TO FIND DISTANCE

We can find distance from speed time graph by finding total are under the graph because in speed time graph total area under the graph shows total distance covered by the body.

EXAMPLE 2.6

Below figure shows the distance–time graph of a moving car.

From the graph, find

(a) The distance car has travelled.

(b) The speed during the first five seconds.

(c) Average speed of the car.

(a) Total distance travelled = 40 m

(b) Distance travelled during first 5s is 35 m

(c) Average speed

(d) Distance moved during the last 5 s = 5 m

Result:

EXAMPLE 2.7

On the graph in above figure, point A gives speed of the object as 2 ms^–1 after 5 s and point B gives speed of the object as 4 ms^–1 after 10 s.

As Acceleration = slope of AB

Where slope = change in velocity / time interval

acceleration

Result:

In above speed time graph acceleration of the body is 4ms^–2.

EXAMPLE 2.8

In above figure the graph shows that the speed of the object is decreasing with time. The speed after 5s is 4 ms–1 and it becomes 2ms–1after 10 s.

As acceleration = slope of CD

Result:

Above graph shows that the deceleration of the body is 0.4ms^–2.

EXAMPLE 2.9

A car moves in a straight line. The speed–time graph of its motion is shown in figure below:

From the graph, Find:

(a) Its acceleration during the first 10 seconds.

(b) Its deceleration during the last 2 seconds.

(c) Total distance travelled.

(a) Acceleration during the first 10 seconds,

(b) Acceleration during the last 2 seconds,

(c) Total distance travelled = area under the graph (trapezium OABC)

(d)

Result:

FISRT EQUATION OF MOTION

Consider a body is moving with initial velocity “v_i” in a straight line with uniform acceleration “a”. Its velocity becomes “v_f” after time “t”. The motion of the body is described by speed – time graph as shown in figure.

In this case:

Slope of line AB =

We know that slope of line in speed–time graph gives the magnitude of acceleration.

Acceleration =

As AC = OD and BC = BD – CD

So, a =

As BD = v_f

CD = v_i

OD = t

Hence a =

Or at = v_f – v_i

Therefore, v_f = v_i + at

This is called first equation of motion.

Conclusion:

SECOND EQUATION OF MOTION

Consider a body is moving with initial velocity “v_i” in a straight line with uniform acceleration “a”. Its velocity becomes “v_f” after time “t”. The motion of the body is described by speed – time graph as shown in figure:

In this case:

The total distance “S” travelled by the body is equal to the total area of the under the speed time graph. i.e.

Total Distance Covered = Area of the rectangle OACD + Area of the triangle ABC

Area of the rectangle OACD = (width x length)

= OA x OD

= v_i x t …………. (i)

Area of the triangle ABC = (width x length)

= (BC x AC)

= (BC x OD)

= at x t

–––––––––––––(ii)

Adding (i) and (ii)

S = v_it + at²

This is called Second equation of motion.

Conclusion:

THIRD EQUATION OF MOTION

Consider a body is moving with initial velocity “v_i” in a straight line with uniform acceleration “a”. Its velocity becomes “v_f” after time “t”. The motion of the body is described by speed – time graph as shown in figure.

In this case:

The total distance “S” travelled by the body is equal to the total area of trapezium OABD under the graph.

S =(BD + OA ) (OD)

Or 2S = (BD + OA ) (OD)

Multiplying both sides by , we get

As (BC = BD–CD)

As (= a)

2aS = (BD + OA ) (BD–CD)

As BD = v_f

OA = v_i

BD = v_f

CD = v_i

Putting the values in the in the above equation, we have

2 aS = (v_f + v_i) (v_f – v_i)

2aS = v_f² – v_i²

This is called Third equation of motion.

Conclusion:

EQUATIONS OF MOTION

There are three basic equations of motion of bodies moving with uniform acceleration. These equations relate initial velocity vi, final velocity vf, acceleration a, time t and distance s covered by a moving body. In these equations of motion we suppose the motion of a body is along a straight line. Hence, we consider only the magnitude of displacements, velocities, and acceleration along straight line. For rectilinear motion equations of motion are as follows:

v_f = v_i + at

S = v_it + at²

2aS = v_f² – v_i²

FORMULAE

Formulae for the areas of different shapes are given below:

Area of the rectangle = (width x length)

Area of the triangle ABC = (width x length)

ms–2 TO kmh–2

To convert ms^–2 to kmh^–2 multiply acceleration inms^–2 by {(3600×3600)/1000} = 12960 to get its value in kmh^–2

kmh–2 TO ms–2

Divide acceleration in kmh^–2 by 12960 to get its value in ms^–2.

ms–1 TO kmh–1

Multiply speed in ms^–1 by 3.6 to get speed in kmh^–1

For example:

20ms^–1 = 20×3.6 kmh^–1

=72 kmh^–1

EXAMPLE 2.10

Given Data:

Initial velocity = v_i = 10ms^–1

Acceleration = a = 2ms^–2

Time = t = 5 s

To Find:

Final velocity = v_f = ?

Calculations:

We know

v_f = v_i + at

Putting the values

v_f = (10) + (2) (5)

v_f = 10 + 10

v_f = 20 ms^–1

Result:

EXAMPLE 2.11

Given Data:

Initial velocity = v_i = 80 kmh^–1

Final velocity = v_f = 20 kmh^–1

Acceleration = a = –2ms^–2

To Find:

Time taken = t =?

Calculations:

We know

v_f = v_i + at

Putting the values

5.6 = (22.2) + (–2) (t)

5.6 – 22.2 = –2 t

–16.6 = –2t

t = 16.6 / 2

t = 8.3 s

Result:

EXAMPLE 2.12

Given Data:

Acceleration = a = 1ms^–2

Initial velocity = v_i = 4ms^–1

Time = t = 10s

To Find:

Distance moved = S = ?

Calculations:

We know

S = v_it + ½ at²

Putting values

S = (4) (10) + ½ (1)(10)²

S = 40 + 50

S = 90m

Result:

EXAMPLE 2.13

A car travels with a velocity of 5 ms^–1. It then accelerates uniformly and travels a distance of 50 m. If the velocity reached is 15 ms^–1, find the acceleration and the time to travel this distance. Solution:

Given Data:

Initial Velocity = v_i = 5 ms^–1

Final Velocity = v_f = 15 ms^–1

Distance = S = 50m

To Find:

Acceleration = a =?

Time to travel the distance = t =?

Calculations:

We use 3^rd equation of motion for finding acceleration

2 a S = v_f² – vi²

2 a (50) = (15)² – (5)²

100 a = 225 –25

100 a = 200

a = 200/100

a = 2ms^–2

We can find time to travel by using 1^st equation of motion

As v_f = v_i + at

Putting the values

(15) = (5) + (2) t

15 – 5 = 2 t

10 = 2 t

t = 10/2

t = 5 s

Result:

GRAVITATIONAL ACCELERATION

Definition:

“The uniform acceleration of free falling bodies under the action of force of gravity is called gravitational acceleration.”

Discovery:

Galileo was the first scientist to notice that all the free falling objects have the same acceleration independent of their masses. He dropped various objects of different masses from the leaning tower of Pisa. He noticed that all of them reach the ground at the same time.

Explanation:

If we neglect air resistance, then all the bodies either lighter or heavier will fall down with uniform acceleration. This uniform acceleration of freely falling bodies is known as gravitational acceleration. It is represented by ‘g’. Its value is 9.8ms^–2, but for simplicity we shall use the value of “g” as 10 ms^–2. For bodies falling vertically downward ‘g’ is positive and for bodies moving vertically upward ‘g’ is negative.

FREE FALLING BODIES

Equations of motion can be used for bodies moving under gravity. In such cases we replace ‘a’ by ‘g’ and S by h. so equations of motion for bodies falling freely can be written as,

v_f = v_i + gt

h = v_it + gt²

2gh = v_f² – v_i²

FOR DOWNWARD MOTION

Initial velocity ‘v_i’ of the freely falling body will be zero

Gravitational acceleration will be positive

FOR UPWARD MOTION

Final velocity ‘v_f’ of the body will be zero.

Gravitational acceleration will be negative.

VELOCITY AT HIGHTEST POINT

When a body is thrown vertically upward, it moves against the force of attraction of the

Earth. It slows down gradually and on reaching the highest point it comes to rest. That is why the velocity of a body becomes zero at the highest point.

EXAMPLE 2.14

A stone is dropped from the top of a tower. The stone hits the ground after 5 seconds. Find:

(a) The height of the tower

Given Data:

Initial velocity = v_i = 0

Gravitational acceleration = g = 10ms^–2

Time = t = 5 s

To Find:

Height of tower = S = h = ?

Final Velocity = v_f = ?

Calculations:

We can find height of the tower by using 2^nd equation of motion

h = v_it + ½ gt²

By putting values

h = (0)(5) + ½ (10)(5)²

h = 0 +125

h = 125 m

We can find final velocity of the stone by using 3^rd equation of motion

2gh = v_f² – v_i²

By putting values

2(10)(125) = v_f² – 0

v_f² = 2500

Taking square root on both sides

v_f = 50 ms^–1

Result:

EXAMPLE 2.15

A boy throws a ball vertically up. It returns to the ground after 5 seconds. Find

Given data:

Gravitational acceleration = g = –10ms^–2 (As the ball is moving upward)

Time for up and down motion = t_o = 5 s

Velocity at maximum height = v_f = 0

To Find:

Maximum height reached by the ball = h = ?

The velocity with which the ball is thrown up = v_i = ?

Calculations:

For finding initial velocity first we have to find time to reach maximum height that is

Half of total time of flight (t_o)

So Time to reach maximum height = ½ t_o

t= ½ (5)

t = 2.5 s

Now by using 1^st equation of motion we can find initial velocity

We know

v_f = v_i + at

Putting the values

0 = v_i + (–10)(2.5)

0 = v_i – 25

v_i = 25ms^–1

By using 3^rd equation of motion we can find maximum height reached by the ball

We know

2gh = v_f² – v_i²

By putting values

2(–10)h = 0 – (25)²

–20 h = –625

h = 625/20

h = 31.25 m

Result:

See Q.2 Long Question TOPIC 2.2

Differentiate between the following:

Rest and motion

Circular motion and rotatory motion

Distance and displacement (GRW 2014)

Speed and velocity (LHR 2013, 2015)

Scalars and vectors (GRW 2013, LHR 2014, 2015, 2107)

SPEED

Definition:

The distance covered by an object in unit time is called speed

Symbol:

Speed is represented by “v”

Quantity:

Speed is a scalar quantity. Its S.I unit is metre per second (ms^–1)

Formula:

Speed= Distance covered/Total time

v =

VELOCITY

Definition:

“The displacement covered by an object in unit time is called velocity”

Symbol:

Displacement is denoted by “”

Quantity:

Speed is a scalar quantity. Its S.I unit is metre per second (ms^–1)

Formula:

ACCELERATION

Definition:

“The rate of change of velocity of a body is known as acceleration.”

Mathematical form:

If a body is moving with initial velocity ’v_i’ and after some time ‘t’ its velocity becomes ‘v_f’ then change in velocity will be v_f–v_i in time t.

Acceleration =

Acceleration =

So a =

Unit:

SI unit of acceleration is meter per second per second (ms^–2).

Quantity:

It is a vector quantity.

DISTANCE TIME GRAPH

The distance–time graph is shown below:

The slope of the graph gives speed with the help of the formula

Speed (v) of the object = slope of line AB

= 2 ms^–1

Result:

See Long Questions TOPIC 2.6

Sketch a velocity – time graph for the motion of the body. Calculate total distance covered by the body.

Given Data:

Velocity time graph for the calculation of total distance is given below?

To Find:

Total distance covered=?

Calculations:

By using the given values we plot a graph shown in figure.

Velocity = 48 kmh^–1

Velocity = ^{48 x1000}

¹⁰⁰⁰

Velocity = 13.33 ms^–1

Time taken = 2 minutes

= 2(60)

= 120 s

Again time taken = 5 minutes

= 5(60)

= 300 s

Again time taken = 3 minutes

= 3(60)

= 180 s

We know that area under speed–time graph represents the distance covered by the object.

Total distance covered = Area of trapezium OABC

S (Sum of parallel sides) (Perpendicular distance between parallel sides)

S(600+300) (13.33)

S(900) (13.33)

S= 6000 m

Result:

DIFFERENTIATION

Differences between Rest and Motion are as follows:

DIFFERENTIATION

Differences between circulatory and rotatory motion are as follows:

DIFFERENTIATION

Differences between distance and displacement are as follows:

DIFFERENTIATION

Differences between speed and velocity are as follows:

DIFFERENTIATION

Differences between Linear and Random motion are as follows:

DIFFERENTIATION

Differences between scalar and vectors are as follows:

CONSTANT SPEED AND ACCELERATION

A body moving with constant speed may or may not have acceleration.

It will not have acceleration if the body is moving with constant speed in a straight line that will be case of constant velocity.

That body can have acceleration if its direction of motion changes continuously. For example a body moving with constant speed in a circular path has acceleration.

MOTION OF RIDER

Riders in a Ferris wheel move in a circle without rotation therefore motion of rider in Ferris wheel is translatory not rotatory.

Long question Q. 2 Topic 2.5

Which of the following can be obtained from speed – time graph of a body?

Initial speed (ii) Final Speed

INFORMATION FROM SPEED TIME GRAPH

All the given quantities can be obtained from speed–time graph.

Short question Q. 4 Topic 2.3 & 2.4

ADDITION AND SUBTRACTION

Scalar quantities can be described completely by magnitude only and can be added or subtracted by simple arithmetical rules. Vector quantities in addition to magnitude also need direction for their description. So vectors cannot be added or subtracted by arithmetic rules due to direction.

IMPORTANCE OF VECTOR QUANTITIES

In order to locate a place from a reference point, we will have to describe the distance and direction of that place from reference point. Description of distance along with direction will make up a vector quantity. Hence by using vector quantities we can describe the position (or location) of bodies.

Given Data:

Velocity of train = V_av = 36 kmh^–1 = = 10 ms^–1

Time taken = t = 10 s

To Find:

Distance travelled by train = S = ?

Calculations:

As we know that

S = V_av x t

By putting the values, we have

S = 10 x 10

S = 100 m

Result:

Given Data:

Initial velocity of train = = 0 ms^–1

Distance covered by train = S = 1 km = 1000 m

Time taken by train = t = 100 s

To Find:

Speed of train after 100 s = = ?

Calculations:

First we have to find the acceleration, as we know that

S = t + ½ at²

By putting the values, we have

1000 = 0 x 100 + ½ x a x (100)²

1000 = ½ x a x 10000

1000 = a x 5000

a =

So, a = 0.2 ms^–2

Now from first equation of motion, we have

by putting the values, we have

= 0 + 0.2 x 100

= 20 ms^–1

Result:

Given Data:

Velocity of the car = = 10 ms^–1

Acceleration of the car = a = 0.2 ms^–2

Time taken by car = t = 0.5 min. = 0.5 x 60 = 30 s

To Find:

(a) Distance traveled by car = S = ?

(b) Final velocity of the car = = ?

Calculations:

As we know that

S = t + ½ at²

By putting the values, we have

S = 10 x 30 + ½ x 0.2 x (30)²

S = 300 + 0.1 x 900

S = 300 + 90

S = 390 m

(b) Now, by using first equation of motion, we have

Result:

Given Data:

Initial velocity of the tennis ball = = 30 ms^–1

Time to reach the maximum height = t = 3 s

Gravitational acceleration = g = –10 ms^–2

Final velocity of the ball = = 0ms^–1

To Find:

Maximum height reached by the ball = h = ?

Calculations:

From second equation of motion in vertical motion, we have

h = t + ½ gt²

by putting the values, we have

h = 30 x 3 + ½ x (–10) (3)²

h = 90 – 5 x 9 h = 90 – 45 h = 45 m

As the ball moves with uniform acceleration in vertical motion, so time taken by the ball in both directions will be same.

Total time taken to return the ground = Time taken upwards + Time taken downwards

Total time taken to return the ground = 3 s + 3s

Total time taken to return the ground = 6 s

Result:

Given Data:

For uniform motion:

Uniform velocity = v_av = 40ms^–1

Time for uniform velocity = t =5s

Given Data:

Acceleration of the train = a = 0.5 ms^–2

Initial velocity of the train = = 0 ms^–1

Distance moved by train = S = 100 m

To Find:

Final speed in kmh^–1 = = ?

Calculations:

From third equation of motion, we have

2aS = vf² – vi²

by putting the values, we have

2 x 0.5 x 100 = vf² – (0)²

100 = vf²

by taking square root on both sides, we have

So v_f = 10 ms^–1

Speed In kmh^–1

=

Result:

Given Data:

Velocity = v = 48 kmh^–1

Velocity = v =

Time taken = t = 2 minutes = 2(60) = 120 s

Again time taken = t =5 minutes = 5(60) = 300 s

Again time taken = t= 3 minutes = 3(60) = 180 s

To Find:

Total distance covered= S =?

Calculations:

By using the given values we can plot a graph shown in figure:

We know that area under speed–time graph represents the distance covered by the object.

Total distance covered = Area of trapezium OABC

S (Sum of parallel sides) (Perpendicular distance between parallel sides)

S(600+300) (13.33)

S(900) (13.33)

S= 6000 m

Result:

Given Data:

Final velocity of the ball = = 0 ms^–1

Gravitational acceleration = g = –10 ms^–2

Time in which ball return to ground = t = 6 s

To Find:

Maximum height reached by ball = h = ?

Initial velocity of the ball = v_i = ?

Calculations:

We know that for ball thrown vertically upward in air

Time taken by ball to reach maximum height = Time taken by ball to reach ground from maximum height

time taken by ball to reach maximum height = t = 3 s

From first equation of motion, we have

= + gt

By putting the values, we have

0 = + (–10) x 3

0 = –30

So = 30 ms^–1

Now from second equation of motion, we have

S = t + ½ gt²

By putting the values, we have

S = 30 x 3 + ½ x (–10) x (3)²

S = 90 – 5 x 9

S = 45 m

Result:

Given Data :

Initial velocity of train = = 96 kmh^–1 = =26.67ms^–1

Final velocity of train = = 48 kmh^–1 = = 13.33ms^–1

Distance covered by train = 800 m

To Find:

Distance covered by the train before coming to rest = S = ?

Calculations:

First we have to find

Retardation of the train = –a = ?

From third equation of motion, we have

By putting the values, we have

2 a (800) = (13.33)² – (26.67)²

1600 a = 177.69 – 711.29

1600 a = –533.6

a = – 533.6 / 1600

a = – 0.33 ms^–2

Again For over all motion till trains stops

Initial velocity of train = = 48 kmh^–1 = = 13.33ms^–1

Final velocity of train = = 0 ms^–1

Retardation of train = a = – 0.333 ms^–2

Given Data:

Initial velocity of train = = 96 kmh^–1 = =26.67ms^–1

Final velocity of train = = 0 ms^–1

Acceleration = a = –0.333ms^–2

To Find:

Time taken by the train = t = ?

Calculations:

From first equation of motion, we have

= + at

By putting the values, we have

0 = 26.67 + (–0.333) t

–26.67 = – (0.333) t

t = 26.67/ 0.333

t = 80s

Result: