Physicsgoeasy

Charging by Friction (or Charging by rubbing)

noreply@blogger.com (Rati S.) — Mon, 16 Nov 2020 02:31:00 +0000

We already know that electrical charges are a fundamental or natural property of a fundamental particle and that there are two types of charges, namely negative and positive charges. There are three ways in which electrons are transferred to charge the object, and they are

friction,
conduction, and
induction.

What is charging by friction?

Rubbing, as the term suggests, is pushing two objects back and forth toward each other. The best way to feel the electrical charge is to rub those bodies together. Rubbing or friction allows electrons to travel. This gives a positive charge to one substance and a negative charge to the other. Charges remain on the surfaces of the products until they can flow or discharge. Whenever there is friction between materials that differ in their ability to give up or gain electrons, the electrons can be transferred in this manner, i.e. through friction.

Charging by friction examples

If we run a comb through our hair, the comb gets charged and can attract tiny pieces of paper. This is because the comb may have lost its electrons or gained some electrons when we rub it with the scalp. This comb is now a charged body. The net charge on the comb interacts with the net charge on small pieces of paper resulting in attraction. Many such solid materials are known to attract light objects like light feathers, pieces of paper, straw, etc.

Take another example of a glass rod and a silk fabric. Now if we rub a glass rod with a silk cloth, the silk cloth absorbs electrons from the glass rod and gets negatively charged. As a consequence, the glass rod is positively charged due to the loss of electrons.

Similarly, there are many other materials that are electrically charged when they are rubbed against each other for example, when we rub the woolen cloth with ebonite or very hard rubber rod, then woolen cloth becomes positive and ebonite rod becomes negatively charged.

The interpretation of the presence of the electrical charge on the rubbing is clear. Material bodies consist of a large number of electrons and protons in similar numbers and are thus neutral in their natural state. But when a glass rod is rubbed with a silk cloth, electrons are shifted from a glass rod to silk cloth. The glass rod is positively charged and the silk fabric is negatively charged as it absorbs additional electrons from the glass rod. In this case, bar after rubbing, comb after running through dry hair becomes electrified and this is an example of frictional electricity.

It is important to remember here that only non-metallic bodies can be charged by friction. The explanation is that the charges are free to travel within the material in a metal or conductor. If you apply a charge to one component of the conductive material, the other charges will be automatically re-arranged to neutralize the charge.

In the case of non-metallic or non-conductors, the charges you put on the surface of the material by rubbing remain as there is no re-arrangement taking place. This is because the insulator charges do not pass freely inside the content.

Waves Multiple Choice Questions Answers

noreply@blogger.com (Rati S.) — Sun, 20 Sep 2020 13:08:00 +0000

On this page find conceptual questions for waves. All these are multiple-choice questions and answers are provided in the end. Try not to look at answers before attempting the questions on your own.

Conceptual Questions on Waves

Question 1. Transverse wave velocity in a stretched string depends on
a. frequency of the wave
b. tension
c. length of string
d. linear mass density string

Question 2. A transverse wave travels along the x-axis. The particles of the medium must move
a. Along the z-axis
b. Along the x-axis
c. In the y-z plane
d. Along the y-axis

Question 3. What is true for a standing wave on the string
a. All the particles are never at rest simultaneously
b. In one complete cycle, all the particles cross their mean position simultaneously twice
c. In one complete cycle, all the particles cross their mean position simultaneously once
d. All the particles acquire their positive extreme positions simultaneously once in a cycle

Question 4. Choose the incorrect one
a. When an ultrasonic wave travels from air into water. It bends towards the normal to the air-water interface
b. Any function of the form $y(x,t)=f(vt+x)$ represents a travelling wave
c.the velocity, wavelength, and frequency of wave undergo change when it is reflected from a surface
d. None of the above

Question 5. Match the following (options A to E) with the types of the wave (options P and Q)

A) Thermal radiation received from the sun
B) Sound waves produced by the vibrating string of the guitar
C) Radio waves sent out from the broadcasting station
D) X Rays
E) Waves produced in the air by the vibrating tuning fork

P) longitudinal
Q) Transverse

Question 6. Which of the following functions represent a traveling wave
a. $y=p\cos(qx)sin(rt)$
b. $y=p\sin(qx+rt)$
c. $y=p\sin(qx-rt)$
d. none of the above

Question 7. Which of the following is not a standing wave
a. $y=p\cos(qx)\sin(rt)$
b. $y=p\sin(qx+rt)+p\sin(qx-rt)$
c. $y=p\sin(qx+rt)$
d. None of the above

Question 8. When a wave is refracted into another medium which of the following will change
a. Velocity
b. Frequency
c. Phase
d. Amplitude

Question 9. A pipe closed at one end and open at other will give
a. All even harmonics
b. All odd harmonics
c. All the harmonics
d. None of the harmonics

Question 10. To raise the pitch of a stringed musical instrument, the player can
a. Lossen the string
b. Tighten the string
c. Shorten the string
d. Lengthen the string

Answers

1. b,d

2. c

3. b,d

4. a,c

5. A-Q,B-P,C-Q,D-Q,E-P

6. b,c
7. c

9. c

10. b,c

Numerical Questions On Waves

Question 1.The amplitude of a wave disturbance propagating in the positive direction is given by
$y=\frac{1}{(1+x)}$
at time $t=0$ and by
$y=\frac{1}{[1+(x-5)]}$
at $t=5$ seconds where $x$ and $y$ are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is
a. 1 m/sec
b. 1.5 m/sec
c. .5 m/sec
d. 2 m/s

Question 2.A transverse wave in a medium is described by the equation
$y=Asin^2(\omega t-kx)$.
The magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity.if the value of A is
a.λ/2π
b.λ/4π
c. λ/π
d 2λ/π

Question 3. A plane progressive wave is represented by the equation
$y=\cos(2\pi t-\pi x)$
The equation of the wave with triple of the amplitude and double the frequency
a. y=3cos(4πt-πx)
b. y=3cos(5πt-πx)
c. y=3cos(4πt+πx)
d. y=3cos(3πt-πx)

Question 4.In the above example ,The equation of wave with double of the amplitude and double the frequency but travelling in the opposite direction
a. y=2cos(4πt+πx)
b. y=2cos(5πt-πx)
c. y=2cos(4πt-πx)
d. y=2cos(3πt-πx)

Question 5.The displacement of a particle having wave motion given by
y=cos²(t/4)sin(50t)
This expression may be considered to be a result of the superposition of how many wave motions
a. one
b Two
c. Three
d. Five

Question 6. A wave is represented by the equation
y=(1mm)sin[(60 s^-1)t+(4 m^-1)x]

which one of the following is true
a. Frequency =30/π
b Amplitude=.001mm
c. Maximum Velocity of the Particle 60 mm/sec
d. wave velocity is 100m/s

Question 7.the displacement of the particles in a string streched in the x-direction is represented by y.Among the following expressions for y,those describing wave motion ares
a. cospxsinqt
b p²x²-w²t²
c.cos²(px+wt)
d. cos(p²x²-w²t²)

Question 8.A transverse wave on a string,the string displacement is described as
y(x,t)=1/1+(x-at)²

where a is negative constant
which of the following is true
a. The Shape of the string at t=0 is y=/1+x²
b. The shape of the waveform does not change as its move along the string
c Waveform moves in the -x direction
d. The speed of the waveform is |a|

Answer
1.Ans a

2.Ans b

3.Ans a

4. Ans a

5.Ans 3

6.Ans a and c

7.Ans a

8.Ans a,b,c,d

What unit of measurement is used for mass?

noreply@blogger.com (Rati S.) — Sun, 20 Sep 2020 12:07:00 +0000

We measure objects not only in physics or chemistry labs but in our day to day life. Whenever we are measuring any object we must measure it according to some predefined units which are standards for industry and scientific community. This article is about units of measurements for measuring mass.

What unit of measurement is used for mass?

A simple answer to this question can be $Kg$ or pound. These are the most common units of mass we encounter in our day to day life. It must be noted that Kilogram is the SI unit of mass and a pound is an imperial unit of mass.

To convert from kg to pound,

multiply the kg unit by 2.20462 (1 kg = 2.2 lb)

To convert from pound to kg,

multiply the pound unit by 0.453592 (1 lb = 0.45 kg)

But the question arises are these the only units of measurement that are being used to measure mass?

The answer is no. There are several units of mass used in various systems of units. Also sometimes it depends on the object whose mass we are measuring. For example

To measure the mass of a person we use units like Kg or lbs but if we are measuring the mass of subatomic particles we might employ another unit altogether, for example, $amu$ (atomic mass unit). We can always perform a conversion between units where

amu (atomic mass unit)$=1.6605 \times 10^-27 kg$

Please note that $amu$ is not an SI unit of mass.

Apart from $Kg$, pound, and $a.m.u.$ there are many other units that are used to measure mass. The table given below shows some of the frequently used units of mass

These are all different units used for measurement of mass and different units are employed depending on the object whose mass we are measuring. For example, we often use grams as a unit of measurement of mass when we buy spices and we use Kilogram while buying vegetables. It all depends on the quantity of mass in the object we are measuring.

Metric Ton is a unit of mass that is equivalent to 1,000 Kg. Now, this is a large unit of mass and is used for measuring huge or massive objects. For example, if you want to measure the mass of a truckload of stones for construction of a building Metric Ton might be a better unit to use instead of gram or Kilogram

Second Equation of motion by graphical Method

noreply@blogger.com (Rati S.) — Sat, 22 Aug 2020 04:47:00 +0000

In this article, we will derive the second equation of motion by graphical method.

Formula for the second equation of motion

First equation of motion is given by the relation

$s=ut+\frac{1}{2}at^2$

Where

$v$= final velocity

$u$= initial velocity

$a$= acceleration

$s$= displacement of the object

$t$= time taken

Note: - This equation along with other kinematics equations of motion are valid for objects moving with uniform acceleration.

Derivation of the 2nd equation of motion by graphical method:

To derive the 2nd equation of motion we will make the following assumptions

* Object under consideration is moving with acceleration $a\,\, m/s^2$

* At time $t=0$ object have some initial velocity. Let’s denote it by $u\,\, m/s$

* At time $ts$ object have some final velocity. Let’s denote it by $v\,\, m/s$

* Total displacement of the object in time $t$ seconds is $s$ meters.

Object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time $t = 0$ and $t = t\,\, s$ are $u$ and $v$ respectively. During time $t$, let $s$ be the total displacement of the object.

Figure given below show the velocity-time graph for the object whose initial velocity is $u$ at time $t=0$ and velocity $v$ at time $t$.

We know that area covered under velocity-time graph gives the displacement of the object in given time $t$ So,

Net Displacement = Area under velocity-time graph.

Or,

$s=$ Area of trapezium $OPQS$

$s=$ Area of rectangle $OPRS$ + Area of triangle $PQR$

Thus,

$s=OP\times PR + \frac{RQ\times PR}{2}$

Substituting various values we get

$s=u\times t +\frac{1}{2}(v-u)\times t$

Since $RQ=(v-u)$ and $PR=OS=t$

Or,

$s=u\times t+\frac{1}{2}at\times t$

Since $(v-u)=at$

So, our first equation of motion is

$s=ut+\frac{1}{2}at^2$

How to find vector components from magnitude and angle

noreply@blogger.com (Rati S.) — Fri, 21 Aug 2020 09:00:00 +0000

We already know what are vectors in physics and How to represent vectors graphically. In this article, we will look at the way to find vector components from magnitude and angle.

Finding vector components from magnitude and angle

Let us proceed further with a problem. So our problem is to

find the components of a vector $\vec{v}$ which has a magnitude of 6 units and is directed at an angle of $30^{\circ}$ with respect to the x-axis.

Let us now represent our vector graphically from the information given in the problem.

This vector $\vec{v}$ can be represented by the hypotenuse of this triangle shown below in the figure.

Here $OA=v_x$ represent the x component and $AB=v_y$ represent the y component of the vector $\vec{v}$. In terms of components vector v is written as

$\vec{v}=v_x\hat{i}+ v_y\hat{j}$

where $\hat{i}$ and $\hat{j}$ are the unit vectors along x and y axis respectively.

From trigonometry we know that for any right angled triangle with hypotenuse (H), base (B) and perpendicular (P)

$\sin\theta=\frac{perpendicular}{hypotenuse}$

$\cos\theta=\frac{base}{hypotenuse}$

where $\theta$ is the angle between base and hypotenuse.

Now since triangle $OAB$ is a right angled triangle with

$Base=OA=v_x$

$Hypotenuse=OB=|\vec{v}|$ and

$Perpendicular=AB=v_y$

We have,

$v_x=|\vec{v}|\cos\theta$ and

$v_y=|\vec{v}|\sin\theta$

Here $|\vec{v}|$ is the magnitude of the vector $\vec{v}$.

From our problem $|\vec{v}|=6$ and $\theta=30^{\circ}$ so, we have

$v_x=6\cos30^{\circ}$ $v_y=6\sin30^{\circ}$

Again from trigonometry

$\cos30^{\circ}=\frac{\sqrt{3}}{2}$

and

$\sin30^{\circ}=\frac{1}{2}$

Which gives us

$v_x=6\times \frac{\sqrt{3}}{2}=3\sqrt{3}$

and

$v_y=6\times \frac{1}{2}=3$

So, $(3\sqrt{3},3)$ are the vector components that we have found when we are given the information about magnitude and angle of the vector.

You can verify this answer by finding the magnitude of the vector using the components of the vector we have calculated. This can be done by using Pythagoras theorem for where you are finding the hypotenuse of triangle $OAB$ which is nothing but magnitude of our vector in question.

Solved Example

In the vector $\vec{v}$ as shown below in the figure convert vector from magnitude and direction form into component form.

Solution

Here it is given in the question that magnitude of $\vec{v}$ is 11 and the angle vector makes with the x-axis is $70^{\circ}$. This vector $\vec{v}$ can be represented by the hypotenuse of this triangle shown below in the figure.

Finding x component of $\vec{v}$

$v_x=-|\vec{v}|\cos\theta = -11 \cos (70^{\circ}) \approx -3.76$

Here negative sign appears because our vector is in the second quadrant where all values of x are negative as can be seen clearly from the figure and values of y are all positive.

Finding y component of $\vec{v}$

$v_y=|\vec{v}|\sin\theta = 11 \sin (70^{\circ}) \approx 10.34$

Our answer is

$\vec{v}\approx \left(-3.76,10.34 \right)$

Similarly you can find the vector components from magnitude and angle even if vector lie in third or fourth quadrant.

Further Reading

1. https://www.grc.nasa.gov/www/k-12/airplane/vectors.html

2. https://www.physicsclassroom.com/class/1DKin/Lesson-1/Scalars-and-Vectors

First Equation of motion by graphical Method

noreply@blogger.com (Rati S.) — Thu, 20 Aug 2020 12:11:00 +0000

In this article, we will derive the first equation of motion by graphical method.

Formula for the first equation of motion

The first equation of motion is given by the relation

$v=u+at$

Where

$v$= final velocity
$u$= initial velocity
$a$= acceleration
$s$= distance traveled

Note: - This equation along with other kinematics equations of motion are valid for objects moving with uniform acceleration.

Derivation of 1st equation of motion by graphical method

The first equation of motion can be derived using a velocity-time graph for the moving object with an initial velocity of $u$, final velocity $v$, and acceleration $a$.

Here we are looking to find the velocity of a moving object at any time $t\) when we have knowledge of its initial velocity and acceleration. The reason we are using a velocity-time graph, in this case, is because it relates all these parameters. Also, the slope of the velocity-time graph tells us about the acceleration of the moving object. To find the first equation of motion consider the graph shown below in the figure.

From this graph, we can easily see that
$OP=u$ represents initial velocity of the object at time $t=0$
$OS=v$ represents the final velocity of the object at time $t$
$PR$ is time $t$ during which motion takes place.

Now from this graph
$a=\frac{QR}{PR}=\frac{SP}{PR}$
$SP=a.PR=at$
As
$OS=OP+PS$
Substituting various values we get
$v=u+at$

Note on the first equation of motion

This equation is the relation among initial velocity $(u)$, final velocity $(v)$, acceleration $(a)$ and time $(t)$.
In lots of problems, the object starts motion from rest with zero initial velocity. In such cases, if we have knowledge about two kinematics quantities we can easily find the third quantity.

Representation of vectors

noreply@blogger.com (Rati S.) — Wed, 19 Aug 2020 06:44:00 +0000

We already know about vectors and scalars in physics. In this article, we will explore how to represent vectors graphically.

Now we already know vectors have

Magnitude and
Direction

To represent them graphically we must take account of both magnitude and direction of a vector quantity.

How to represent vectors graphically?

Vectors are represented graphically by directed line segments.

They are represented such that the length of the line segment is the magnitude of the vector and the direction of the arrow marked at one end represents the direction of the vector.

The length of the line shows the magnitude or size of the vector and the direction of the arrowhead shows the direction of the vector.

The starting point of the vector is called the “tail” of the vector and its ending point is known as “head” of the vector.

In this figure point, $A$ is the initial point or tail of the vector, and $B$ is called the terminal point, tip, or head of the vector.

This directed line segment with its tail at point A and head at point B is written as $\vec{AB}$ or AB. So, vectors are also represented symbolically either by the arrow on top of the symbol or in bold type.

Here in this figure shown above $\vec a =\vec{AB}$. The magnitude of this vector is \[\left|\vec{a}\right|=\left|\vec{AB}\right|=AB\]

which is the distance between the head and tail of the vector.

It must be noted that the magnitude of a vector is always a non-negative real number.

Every vector has the following three characteristics

Length of the vector:

The length of the vector $\vec a =\vec{AB}$ is denoted by $\left|\vec{AB}\right|=AB$

Support of the vector:

The line of unlimited length of which $AB$ is a segment is called the support of the vector $\vec{AB}$.

Sense:

The sense of the vector $\vec{AB}$ is from point $A$ to point $B$ and the sense of the vector $\vec{BA}$ is from $B$ to $A$. The sense of ant vector is from its initial point to its final point.

Further Reading

1. https://www.physicsclassroom.com/class/vectors/Lesson-1/Vector-Resolution

2. How to find vector components from magnitude and angle

Vectors and Scalars in Physics

noreply@blogger.com (Rati S.) — Thu, 16 Jul 2020 08:27:00 +0000

This article about vectors and scalars in physics gives a basic introduction of both these quantities. Here, we have defined both these quantities and created a list containing examples of both vector and scalar quantities. In this article, you will also get to know the differences and some similarities between both scalar and vector quantities.

Vectors are one of the most important concepts of mathematics. Vectors find a wide variety of applications in fields like

Geometry
Mechanics
Applied mathematics
Engineering
Physics
Computer Science etc.

We will discuss vectors here in the context of Physics. While studying

Contents

Introduction
Scalar Quantity Definition
Scalar Quantity Examples
Vector Quantity Definition
Vector Quantity Examples
Difference between scalar and vector quantities

Introduction

physics we come across various physical quantities. These quantities generally are of two types:

Scalar Quantities: They only have magnitude.
Vector Quantities: They have both magnitude and direction.

Scalar Quantity Definition

Those quantities which only have magnitude and does not relate to any fixed direction in space are called Scalar Quantities.

To represent a scalar quantity, we assign a real number to it which gives the magnitude of the quantity under consideration. We also attach the unit to this quantity for example number 20 can mean anything but if we associate this number with length, we must associate a unit with it. This unit shows the order of magnitude of the quantity under consideration as a length of 20 Km is greater than the length of 20 m and a length of 20 m is greater than 20 cm.

So, you must specify a unit of Scalar quantity under consideration otherwise, the quantity does not have a clear meaning.

Scalar quantity examples

We encounter lots of scalar quantities while studying physics. Given below is a scalar quantity list of some commonly used scalars

Mass
Volume
Density
Work
Speed
Energy
Power
Area
Volume
Time
Temperature
Distance
Current
Potential difference
Resistance
Charge etc.

Vector Quantity Definition

Physical quantities which have both magnitude and direction are known as vector quantities.

It is important to note here that in addition to magnitude and direction, two vector quantities of the same kind must compound according to the parallelogram law of vector addition. If they fail to compound according to parallelogram law of vector addition, then they will not be treated as vectors.

For example, rotations of a rigid body through finite angles have both magnitude and directions but they do not satisfy parallelogram law of addition of vectors.

Vector quantity examples

The list given below shows some of the examples of vector quantities in physics

Displacement
Velocity
Acceleration
Force
Weight
Impulse
Pressure
Momentum
Gravity
Torque
Electric field
Magnetic field
Angular velocity
Angular momentum
Electric flux
Angular acceleration etc.

List of scalar and vector quantities in tabular form

Difference between scalar and vector quantity

Similarities between Scalars and vectors

Despite being different there are several similarities between both vectors and scalars

They both express certain physical quantities.
Both these quantities can be measured. That is these quantities can be quantified.
Both of these physical quantities have dimensions and units

Third Equation of Motion by Graphical Method

noreply@blogger.com (Rati S.) — Mon, 13 Jul 2020 17:18:00 +0000

Here in this article, we will derive the third equation of motion by graphical method.

Formula for the third equation of motion:

Third equation of motion is given by the relation

$v^2=u^2+2as$

Where,

$v=$ final velocity

$u=$ initial velocity

$a=$ acceleration

$s=$ distance travelled

Note: - This equation along with other kinematics equations of motion are valid for objects moving with uniform acceleration.

Derivation of 3rd equation of motion by graphical method:

To derive 3rd equation of motion we will make following assumptions

Object under consideration is moving with acceleration $a\,\, m/s^2$
At time $t=0$ object have some initial velocity. Let’s denote it by $u\,\, m/s$
At time $ts$ object have some final velocity. Let’s denote it by $v\,\, m/s$
Total distance covered by object in time $t$ seconds is $s$ meters.

Object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time $t = 0$ and $t = t\,\, s$ are $u$ and $v$ respectively. During time $t$, let $s$ be the total distance travelled by the object.

Figure given below show the velocity-time graph for the object whose initial velocity is $u$ at time $t=0$ and velocity $v$ at time $t$.

We know that the distance covered by the object moving with uniform acceleration is given by the area under the velocity-time graph. Here area under the velocity-time graph is equal to area of trapezium OPQS

∴ Area of trapezium OPQS \[= \frac{1}{2}\text{(Sum of Parallel Slides + Distance between Parallel Slides)}\]

Or, \[s=\frac{OP+SQ}{2}\times PR\]

We have knowledge about acceleration of the moving object. So,

Acceleration \[a=\frac{\text{Change in velocity}}{time}=\frac{QR}{PR}\]

Or,

\[PR=\frac{QR}{a}\]

From graph we can see that

\[QR=v-u\]

\[t=PR=\frac{v-u}{a}\]

From above figure we can see that

\[OP=u\]

\[SQ=v\]

\[OP+SQ=u+v\]

Substituting these values, we get

\[s=\left(\frac{u+v}{2}\right)\times\left(\frac{v-u}{a}\right)=\frac{v^2-u^2}{2a}\]

Rearranging it we get

\[v^2=u^2+2as\]

Which is third equation of motion.

How to find displacement on a velocity time graph

noreply@blogger.com (Rati S.) — Tue, 16 Jun 2020 16:48:00 +0000

Finding displacement on a velocity-time graph is extremely easy. The area under a velocity-time graph gives the measurement of the displacement of the object under consideration.

A velocity-time graph is a graph between the time taken and the velocity acquired by the moving object during that time. Here time taken is plotted along x-axis and velocity acquired is plotted along the y-axis

We know that displacement of an object is the product of velocity and time i.e.,

$d=v\times t$

In this article, we will explore how displacement can be calculated using the velocity-time graph. Let us look at the following cases

Case I: When velocity is uniform (or constant)

To understand this, consider a truck is moving along a straight line with a uniform velocity of 50 Km/hr. Velocity time graph for this man moving with uniform velocity is

From the above figure, we can see that graph for truck moving with a uniform velocity of 50 Km/h is s straight line parallel to the time axis.

Displacement covered by this truck in between time $t_1=2 h$ at $P$ to $t_2=8 h$ at $Q$ is given by

Displacement $d=50 \times (t_2-t_1)$

Or, $d=50 \times (8-2)=50 \times 6 = 300 Km$

From the above graph, we can clearly see that

$PS=50 Km/h$ and $PQ=t_2-t_1=6h$

Hence

Displacement $=PS \times PQ = \text{Area of rectangle PQRS}$

As shown in the shaded region.

Case II: When acceleration is uniform or constant

For this case let us consider the table given below which shows the change in velocity of a truck at regular intervals of time.

Time in (s)	Velocity of truck (m/s)
0	0
10	4
20	8
30	12
40	16
50	20
60	24

The table given above shows the velocity-time relationship of the truck. We get the following graph when we plot this information graphically

From the graph, we can see that it is a straight line which shows velocity is increasing by equal amounts in equal intervals of time. So, this motion represents a motion with uniform acceleration.

The area under this line OC on a velocity-time graph is equal to the displacement of the truck (or object under consideration).

In this case, the displacement covered in any time interval can be found by drawing the perpendiculars on the time axis (for example AB and DC showed in the above figure) at given times. The area between these perpendiculars under the graph gives the displacement.

Let us now find the distance covered by truck between time intervals 10 s and 60 s.

\begin{align*} \text{Distance covered } &= \text{Area of trapezium ABCDA} \\ &=\frac{AB+DC}{2}\times AD \\ &=\frac{4+24}{2} m/s \times (60-10)s \\ &= 14 m/s \times 50 s \\ &= 700 m \end{align*}

Alternatively, for finding displacement from this v-t graph, we can break the shape of the graph into simple geometric shapes. The total area under the line can be calculated by adding the areas of those shapes. For example, here for finding displacement of the truck between time intervals 10s and 60s we can break this graph into two shapes

1. Triangle BCE and

2. Rectangle ABED

We can then find the areas of both these shapes and add them to get the net displacement between given time intervals.

Case III: When velocity and acceleration both are variable or non-uniform

The figure given below shows the velocity-time graph of the body moving with variable velocity.

For such cases, we will divide time interval $t_1$ and $t_2$ into small intervals of time $\Delta t$, as there would be a negligible change in velocity during this time interval $\Delta t$ and hence velocity can be taken as constant.

Therefore for this small time interval displacement $\Delta s = v\Delta t$ or area of the red strip.

For a complete time interval between $t_1$ and $t_2$ displacement would be equal to the area enclosed between the curve and time axis between $t_1$ and $t_2$ or the area of the portion $PQRS$.

Is area under velocity-time graph distance or displacement

It is particularly important to note that

The area under a speed-time graph is the distance, not displacement.
The area under a velocity-time graph is the displacement.

How to find average velocity on a position time graph

noreply@blogger.com (Rati S.) — Mon, 01 Jun 2020 07:53:00 +0000

Average velocity is defined as the displacement divided by the time during with the change in position of the particle takes place. Average velocity is a vector quantity and its SI unit is meter per second $(m/s)$. So,
\[v_{avg}=\frac{\Delta \vec{x}}{\Delta t}=\frac{x_2-x_1}{t_2 -t_1}\]
Consider the figure given below

Figure 1:- Linear position-time graph

On a position vs time graph, the average velocity is the slope of the secant line joining the position at the beginning and end of $\Delta t$. That is we have to find the value of $x_1$ at time $t_1$ and $x_2$ at time $t_2$. Here in case of the above figure secant line is line $AB$ in red color.

NOTE:- Please note that in position vs time graph we plot independent quantity on the x-axis and dependent quantity on the y-axis. Here time is our independent quantity and position of the particle which is changing with the passage of time is our dependent quantity.

To find velocity on the position-time graph you can follow the following steps:-

Find the positions on the graph that represent the initial position and final position.
Draw secant line joining these points.
Calculate the slope of the secant \[Slope(m)=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\]

The most important thing when you write your answers do not forget to mention the unit and the sign (whether the quantity is positive or negative). If the quantity you are calculating is a vector quantity then also check and mention the direction.

Average velocity for linear graph

Let us again consider Figure 1 which is a linear position-time graph. Now we will find the average velocity of the particle during time interval $t_1$ and $t_2$. For finding the average velocity of particle we have to find the slope of secant $AB$ in this case.
The slope of secant line $AB$ would be,
\[v_{avg}=\text{slope of AB}=\frac{x_2-x_1}{t_2 -t_1}\]

Average velocity for a curved graph

Consider the figure given below which shows a curved position-time graph. We will now look at how to find average velocity on a position-time graph which is curved and not linear.

Figure 2:- Curved position-time graph

In this case, if we want to find the average velocity of the particle during the time interval $t_1$ and $t_2$ then we would have to find points on the graph that represent initial and final position. Our initial position is at time $t_1$ and the final position is at time $t_2$.
Their positions are at point $A$ and $B$ respectively. Let $x_1$ and $x_2$ be the initial and final positions of the particle. So initially our particle is at point $A$ at time $t_1$ and at point $B$ at time $t_2$ . The line $AB$ (red line) is the secant and slope of this secant represent the average velocity of the particle during time interval $t_1$ and $t_2$.
So, our average velocity would be \[v_{avg}=\text{slope of AB}=\frac{x_2-x_1}{t_2 -t_1}\]

Key Takeaway:- The slope of the straight line joining two points on the position-time graph gives the average velocity of the particle between these two points.

How to find displacement on velocity-time graph

Difference between speed and velocity

noreply@blogger.com (Rati S.) — Sun, 24 May 2020 12:36:00 +0000

Contents

1. Speed Vs. Velocity
2. Tabular Form
3. Graphical Form

The current article is dedicated to finding the difference between speed and velocity. Here we will look at the speed vs. velocity comparison in tabular form along with the differences between them.

Speed and velocity are some of the basic concepts of physics. These concepts are introduced under kinematics. These are the terms that people use almost interchangeably in their day to day life.

In physics, both the terms speed and velocity have differences between them which cannot be ignored.

Speed Vs. Velocity

Velocity is defined as the rate of change of displacement whereas speed is defined as the rate of change in distance with respect to time.
Velocity is a vector quantity, both magnitude and direction (same as that of displacement) are required to define it whereas speed is a scalar quantity. Only the magnitude is required to define the speed.
SI unit of velocity and speed is m/s. Speed and velocity both have the same unit.
Now sometimes students ask questions like

Is average speed = average velocity?

The answer is no, the average speed is not equal to the average velocity. This is because

Average Speed = Total distance/Total time

Average Velocity = Total displacement/Total time

Just like both displacement and distance are different, average speed and average velocity are also different for example- Motion around a circular track.

Speed Vs. Velocity tabular form

S.No.	Speed	Velocity
1	Rate of change of distance	Rate of change of displacement
2	Scalar quantity	Vector quantity
3	Can never be negative or zero	Can be negative, zero or positive
4	Velocity without direction	Velocity is the speed with direction
5	It may or may not be equal to the velocity	A body may possess different velocities but the same speed
6	Never decreases with time. For a moving body, it is never zero	Velocity can decrease with time. For a moving body it can be zero.
7	SI unit is m/s	SI unit is m/s

Speed Vs. Velocity Graphical form

I hope you are now aware of the basic difference between speed and velocity.

Acceleration Formula with Mass and Force

noreply@blogger.com (Rati S.) — Sun, 10 May 2020 13:31:00 +0000

In this article, we will look at the acceleration formula with mass and force. We already have discussed the acceleration formula with velocity and time.

In this article, we will look at the formula for acceleration with mass and force. We use the acceleration formula with mass and force when we do not have any knowledge of the velocity of the moving body and time.

In this case, we only have information about

Force (f) acting on the body or object and
Mass (m) of the body or object

According to Newton’s second law of motion, force is mass times acceleration we can use this relation to find the acceleration of the moving object. Of course, we need to have force acting and mass.

Acceleration formula with mass and force

We know that force formulaaccording to Newton’s second law of motion is \[f=m\times a\] Now if we want to find acceleration from this force formula then we would have to rearrange the above equation.

So, rearranging above equation we get \[a=\frac{f}{m}\] Here,

acceleration $a$ is in $m/s^2$
force $f$ is in N (or $Kg.m/sec^2$)
mass $m$ is in $Kg$

We can also use the force formula triangle to find the third quantity if any two of force, mass, and acceleration are given.

Solved questions based on acceleration formula with mass and force

Question 1 A car of mass 1000 Kg is moving with velocity 10 m/s and is acted upon by a forward force of 1000 N due to engine and a retarding force of 500N due to friction. What will be its velocity after 10 seconds?

Solution Here it is given that

$m=1000 Kg$

$u=10ms^{-1}$

$t=10 s$

We have to find velocity $v$ after 10 seconds.

Net forward force,

$F= \text{Forward force} - \text{Retarding force}$

$F=1000 N-500 N=500 N$

Acceleration

$a=\frac{F}{m}=\frac{500}{1000}=\frac{}{2}ms^{-2}$

From kinematic equation of uniformly accelerated motion

$v=u+at$

So we have

$v=10+\frac{1}{2}\times 10 = 15 ms^{-1}$

Question 2 A force of 72 dynes is inclined to the horizontal at an angle of $60^{\circ}$. Find the acceleration in the mass of $9g$, which moves in a horizontal direction.

Solution Here it is given in the question that $m=9g$, $F=72 dyne$, $\theta=60^0$

The horizontal component of the force is

\[F_x=F\cos \theta= 72 \times \cos 60^0 \]

$F_x=72\times .5=36 dynes$

Acceleration,

$a=\frac{F_x}{m} = \frac{36}{9}=4 cm.s^{-2}$

Further Reading

Acceleration Formula with Velocity and Time

noreply@blogger.com (Rati S.) — Sun, 10 May 2020 03:06:00 +0000

In this article, we will look at the acceleration formula with velocity and time.
Before going any further let us define what is acceleration.

Acceleration is change in velocity over time

There are two ways to find the acceleration

acceleration formula with velocity and time
acceleration formula with mass and force

In this article, we will look at the formula for acceleration with velocity and time. We use the acceleration formula with velocity and time when we do not have any knowledge of force acting on the body. We only have information about

Change in velocity
Time

Acceleration formula with velocity and time

As mentioned earlier acceleration is the change in velocity over time. So acceleration is the change in velocity divided by time. Mathematically,

If,

$v_1$ is the initial velocity
$v_2$ is the final velocity and
$t$ is the time taken to reach ending or final velocity from starting or initial velocity then

Here,

acceleration $a$ is in $m/s^2$
velocities $v_f$ and $v_i$ are in $m/s$
$\Delta v$ is short form for change in velocity.
and time $t$ is in $seconds$

Acceleration formula important facts

Acceleration s a vector quantity. Acceleration has both magnitude (a value) and direction
The direction of acceleration can be different from the direction of velocity.
Unit of acceleration is $m/s^2$
Acceleration can be negative and positive. It is negative when velocity decreases with time. It is positive when velocity increases with time.

Solved Questions on acceleration formula with velocity and time

Question 1. A bus decreases its speed from $80\,\,m\,s^{-1}$ to $60\,\,m\,s^{-1}$ in 5 seconds. Find the acceleration of the bus.

Solution. Here it is given that

initial velocity,

$v_i=80ms^{-1}$

and final velocity,

$v_f=60ms^{-1}$

Time $t=5\,s$

We have to calculate the acceleration from this data. Now from accelration formula we have

\[a=\frac{v_f-v_i}{t}\]

putting in the respective values

\[a=\frac{(60-80)ms^{-1}}{5s}=-4ms^{-2}\]

Note that the answer is negative. This negative sign indicates that velocity is decreasing with time.

Effect Of Force

noreply@blogger.com (Rati S.) — Thu, 07 May 2020 14:59:00 +0000

Contents

1. Effect of force on state of motion
2. Effect of force on shape of object

In this article, we will learn about the effect of force. We would learn what happens when force cats on various bodies or objects. In our daily life, we can see a lot of examples of the application of force. Force is everywhere.

Know more about Force Push and Pull.

The force is applied by an interaction between objects. So, we can say that force occurs due to interaction. his interaction between objects might be push or pull.

Know more about Force due to interaction with examples.

Force also acts on objects due to phenomena like gravity, magnetism etc.

Effect of Force

Let us now have a look at the effect caused by the application of force on objects.

The effect of force on any object is to change its state of rest or motion. When an object is in state of motion
- Force applied on an object of certain mass can cause the object to accelerate.
- Force applied on an object can change the direction of motion of moving objects.
Force applied on an object can also change its shape.

Effect of force concept map

Force can change the state of motion

Force acting on an object can make an object at rest to move. For example, if you kick a stationary ball it starts moving.
Force can change the position of rest. For example, If you pick up a book and move it from table to chair kept nearby then you have changed the position of the book from table to chair. This is done by applying the force to lift the book.
Force can change the speed of a moving object. In a game of football players can kick the moving ball in the same direction to make it move fast. Similarly, when a goalkeeper stops the ball it applies force in a direction opposite to the direction of the motion of the ball. This force applied in the opposite direction to the direction of motion slows down the ball and eventually stops it in the end.
Force can change the direction of motion of moving objects. For example, While playing tennis games if you hit a tennis ball coming towards you with a tennis bat then it begins to move in a different direction. Here you have applied force on the ball and this force has changed the direction of motion of the tennis ball.

Force can change the shape of an object

Force can change the shape and size of an object. Some of the examples include
Squeezing a sponge changes its shape.
When we pull a rubber band or spring they become longer than their usual size. If we stop pulling them then they regain their original size.
You can change the shape of a ball of dough by rolling it using a rolling pin.

Newton's Law of Motion Multiple Choice Questions (MCQs)

noreply@blogger.com (Rati S.) — Sun, 19 Apr 2020 10:35:00 +0000

Question 1 Friction force can be reduced to a great extent by

(a) lubricating the two moving parts

(b) using ball bearing between two moving parts

(d) all the above

Question 2 Force exerted on a body can change its

(a) kinetic energy

(b) direction of motion

(d) momentum

Question 3 A block is released from the top of a smooth inclined plane. Another block of the same mass is allowed to fall freely from the top of the inclined plane. Both blocks reach the bottom of the plane

(a) with equal speed

(b) in equal interval of time

(d) with equal momentum

Question 4 Which of the following expression does not represent the force acting on the body of mass m

(a) GM_em/(R_e)²

(b) ma

(d) 6ηπrv

Question 5 A body of mass m is resting on the floor. A minimum amount of force is applied on the body and the force continues to act on the body as it moves. If the coefficient of static and dynamic friction are 0.4 and 0.3 respectively, then the acceleration of the body is

(a) 0

(b) 3.98

(d) 4.98

Answer

(1) d.

(2) a, b, c, d

(3) a and c

(4) d

(5) c

Circle of reference in Simple Harmonic Motion

noreply@blogger.com (Rati S.) — Thu, 02 Apr 2020 09:09:00 +0000

Circle of reference is a graphical representation which facilitates the understanding of various SHM relationships. Consider the figure given below in which a point Q moves anti-clockwise around a circle of radius A with a constant angular velocity ω (in rad.s^-1).

In this blog many times I had written short notes or questions related to concept of SHM and this article is about yet another important concept of SHM that is Circle of Reference.
Here in the figure
OQ→position of Q relative to O
θ→angle OQ vector makes with the positive x-axis
This vector whose horizontal component represents the actual motion is called a
phasor

Let P be the point representing the projection of point Q (known as the reference point) on the horizontal diameter of the circle (known as reference circle). As the reference point revolves, the point P moves back and forth along the horizontal diameter always keeping below or above the point Q. The motion of this point P is comparable to that of a body moving under the influence of elastic restoring force and executing SHM in the absence of frictional forces.
From the figure given above, displacement of point P at any given time t is the distance OP or x and from the figure
x=Acosθ
The angular velocity of circular motion is

ω = \frac{angle swept}{time-taken} = \frac{θ}{t}

so,
θ=ωt
and
x=Acosωt (1)
The figure given below shows the velocity in SHM using the circle of reference.

Velocity of the reference point Q would be the component of Q's velocity parallel to the x-axis since P is always directed below or above the reference point. So,
v_P=-vsinθ = -ωAsinωt (2)
since v=ωA is the tangential velocity of motion of reference point Q. The negative sign is introduced because the direction of velocity is towards the left. When Q is below horizontal diameter, the velocity of P is towards the right, but since sinθ is negative at such points, the minus sign is still needed.
Figure given below shows the acceleration in SHM using circle of reference.

The acceleration of Q on reference circle is directed radially inwards towards O as can be seen in the figure and its magnitude is equal to

a = \frac{v^{2}}{A} = ω^{2} A

From figure x-component of acceleration is acceleration of point P. So,
a_P=-acosθ= -ω²Acosθ (3)
The minus sign is introduced because the acceleration is towards the left. When Q is to the left of the center, the acceleration of P is towards the right but, since cosθ is negative at such points, the minus sign is still needed.
To prove that motion of point P is simple harmonic, we now use equation 1 in equation 3 then we have
a_P=-ω²x
since ω is a constant, the acceleration of point P at any instant is equal to negative constant times displacement xat that instant and this is just the essential feature of SHM and this proves that motion of point P is indeed SIMPLE HARMONIC.

Physics Equations Kinematics

noreply@blogger.com (Rati S.) — Tue, 24 Mar 2020 07:53:00 +0000

The following are the important kinematics equations list. I will also provide a link to google docs file from where you can download the file as pdf (see at the end of the article).

Physics - Kinematics Equations

Average Velocity and speed

\[v_{avg} = \frac{\Delta s} {\Delta t} \\
\text{Average Speed} = \frac{\text{Total distance}}{\text{time taken}}
\]

Instantaneous velocity and speed

$v = \mathop {\lim }\limits_{\Delta t \to 0} {{\Delta s} \over {\Delta t}} = {{ds} \over {dt}}$
Instantaneous speed or speed is the magnitude of the instantaneous velocity
here $s$ is the displacement of the object and has only one component(out of x, y and z) for motion along straight line and has two components for motion in a plane.

Average acceleration

$${a_{avg}} = {{\Delta v} \over {\Delta t}}$

Instantaneous acceleration

$$a = \mathop {\lim }\limits_{\Delta t \to 0} {{\Delta v} \over {\Delta t}} = {{dv} \over {dt}}$$

Equations of motion (constant acceleration)

$$v = {v_0} + at$$ $$x = {x_0} + {v_0} + {1 \over 2}a{t^2}$$ $${v^2} = {v_0}^2 + 2a(x - {x_0})$$ $$\overline v = {1 \over 2}(v + {v_0})$$

Free fall acceleration

$$v = {v_0} + gt$$ $$x = {x_0} + {v_0} + {1 \over 2}g{t^2}$$ $${v^2} = {v_0}^2 + 2g(x - {x_0})$$

Projectiles

Horizontal distance $$x = {v_x}t$$
Horizontal velocity $${v_x} = {v_{x0}}$$
Vertical distance $$y = {v_{yo}}t - {1 \over 2}g{t^2}$$
Vertical velocity $${v_y} = {v_{y0}} - gt$$
Here,
${v_x}$ is the velocity along x-axis,
${v_{x0}}$ is the initial velocity along x-axis,
${v_y}$ is the velocity along y-axis,
${v_{y0}}$ is the initial velocity along y-axis.
$g$ is the acceleration due to gravity and
$t$ is the time taken.

Time of flight $$t = {{2{v_o}\sin \theta } \over g}$$
Maximum height reached $$H = {{v_0^2{{\sin }^2}\theta } \over {2g}}$$
Horizontal range $$R = {{v_0^2\sin 2\theta } \over g}$$
Here,
$v_{0}$ is the initial Velocity,
${\sin \theta }$ is the component along y-axis,
${\cos \theta }$ is the component along x-axis.

Uniform circular motion

Angular velocity $$\omega = {{d\theta } \over {dt}}$$
where $\theta$ is angle moved in radian's
Relation between linear velocity, angular velocity and radius of circular motion $$v=r\omega$$
Angular acceleration $$\alpha = {{d\omega } \over {dt}}$$
Centripetal acceleration $${a_c} = {{{v^2}} \over r}$$ $${{\vec a}_c} = - {\omega ^2}\vec r$$

Get kinematics equation list as pdf (opens in new window)

Questions About Motion With Answers

noreply@blogger.com (Rati S.) — Sun, 22 Mar 2020 14:29:00 +0000

We have written various articles under the topic of kinematics. Now that you are confident with the concepts like position in physics, distance and displacement, speed, velocity, acceleration, etc. you must check your knowledge and understanding by attempting a few questions about motion.

Questions about motion with answers

In this worksheet, you will find miscellaneous motion questions that involve concepts like

Distance and displacement
Uniform and non-uniform motion
Average speed and velocity
Acceleration
Graphical representation of motion
Equations of motion in a straight line with uniform acceleration and
Uniform circular motion

All the questions in this worksheet are of the grade levels 8, 9, 10, 11, and 12. The aim of this worksheet is to test the basics of the concept of motion.

Download this worksheet as pdf :- Motion questions pdf

Conceptual Questions

Question 1 An object has moved through a distance. Can it have zero displacement? If yes, then support your answer with an example.

Question 2 Under what conditions are the magnitude of the average velocity of an object equal to its average speed?

Question 3 Can the magnitude of displacement be more than distance.

Question 4 What is the displacement of motion of the tip of the minute hand of a clock in one hour?

Question 5 Is it possible to have non-zero velocity for a time interval, while the acceleration is zero at any instant within the time interval?

Question 6 What is the nature of distance-time graphs for uniform and non-uniform motion of an object?

Question 7 What do you think about the speed-time graph given below

Question 8 What does the area between the v-t graph give?

Question 9 Is uniform circular motion an example of constant acceleration?

Question 10 An artificial satellite goes around the earth in a perfectly circular orbit with constant speed. Is the motion accelerated? Why?

What is Distance and displacement in physics?

noreply@blogger.com (Rati S.) — Tue, 17 Mar 2020 14:57:00 +0000

In this article learn about distance in displacement in physics. Before learning what is distance and displacement do not forget to check our article on How to find position in physics.
Get more articles in kinematics.

Now that we know how to describe the position of a particle moving along a the straight line at a given time, we will now learn about distance and displacement.
In our day to day language, we use terms distance and displacement interchangeably but in physics, both terms have different meanings.

What is distance?

Length of the actual path between the initial and final position of the moving object in the given time interval is known as the distance traveled by the object.

To explain this further let us take our example where a man starts moving towards the right of the the tree we are using as a reference point.

Now at t = 10 minute he is at 10 meters from the reference point O. Let’s call this point as A. He then starts to move towards left and at t = 8 minute he is 5 meters from the origin. Let's call this point as B
So, distance traveled by a man going from O to A is equal to 10 meters and distance traveled in going from A to B is equal to 5 meters.
The total distance traveled is equal to OB. So,
$OB = OC + AB$
Putting in the values we get
$OB= 10\,\, meter + 5\,\, meter =15 \,\, meters$
So, here we are considering the actual length of the path traveled by the man between initial point O and final point B.

What is Displacement?

Let us now define term displacement.

Displacement is the shortest distance between the initial position and the final position of a moving object in the given interval of time along with its direction.

Let us now find the displacement of the man moving from his initial position O to position A and then to position B.
Here to find the displacement of the man from the initial position to final position we must have knowledge of

How far the final position is from the initial position, which is nothing but the straight-line distance between initial and final positions.
The direction of the final position as seen from the initial position.

Now going back to our example let us now find the displacement of man from his initial position.
Displacement is equal to straight line distance between points O and B which is equal to OB. Here O is the initial position and B is the final position.
From the above figure, displacement is equal to

$OB= 5\,\,meters$ towards the right of origin O.

This straight-line distance between initial and final positions of a particle is called the magnitude of the displacement
Also note that we have mentioned the direction in our example which is ‘towards the right’ or ‘right direction’ of the reference point or origin.
So, displacement has both magnitude and direction whereas distance traveled has magnitude but no direction.

Difference between distance and displacement

Let us now look at the differences between distance and displacement.
The first one is the displacement of an object may be zero but the distance traveled by the object is never zero.
To explain this further again consider that in our example the man travels to point A, then to point B and finally returned to reference point O then
$Distance = OA + OB + BO$
$= 10\,\, meters + 5\,\, meters + 5\,\, meter= 20\,\, meters$
Now displacement, in this case, would be zero since O is the initial as well as the final position of the man.
Again, if any object is moving along a circular path, starts at point A and return back to point A then total distance covered would be $2 \pi r$ but the net displacement of the object would be zero.
So, distance traveled by a moving body cannot be zero but the final displacement of a moving body can be zero.
We have also discussed earlier that displacement of an object can be negative but distance traveled by an object can never be negative.
Another difference between distance and displacement is that distance is a a scalar quantity that is it only has magnitude and no direction is specified.
Whereas displacement is a vector quantity as it has both magnitudes as well as direction.
For example, if a car travels a distance of 60 Km then this ’60 kilometer’ is the distance traveled.
Again, if the car is moving in straight line towards west then the displacement of the car of the car would be expressed as ’50 kilometer towards west’.

Watch this video on position, distance and displacement for a clear idea.

How to find position in physics

noreply@blogger.com (Rati S.) — Sun, 15 Mar 2020 02:55:00 +0000

What is position in physics?

In this tutorial, we will learn about the position in physics and how to find position of a particle or object in physics. The concept of position in science or physics is a starting point in the study of physics and is introduced while studying kinematics.

If we want to describe the motion of an object, we measure the position, distance, velocity and other such parameters of the object.
The position of a particle in physics is the place where it is being placed. Generally, this place can be located on a number line when motion we are studying is in one dimension.

The position of an object is measured from some fixed point known as reference point.

For example, if a person says the bus is moving, it means that the bus is changing its position with respect to that person in a given time interval. In this case, the person is the reference point.

So,

A reference point or origin is defined as a fixed point or fixed object with respect to which the given body changes its position.

Finding Position in number line

To describe the position of the particle at a given time we have to specify

Its distance from the origin and whether it is in positive direction or negative direction as seen from the origin.

Let us now consider an example where a man starts walking towards the right of the tree. At $t = 5$ minutes he covered a distance of 10 meters and then start moving towards the left.
Now at time$ t = 8$ minutes, he is at a distance of 5 m from the origin.
We will now mark positions of the man at two given instants of time.

Here in the above figure O is the origin. This position is the starting position of man. Values to the right of the origin are positive and to the left of origin are negative.
Now at t equals to 5 minutes the man is at 10 m from the origin. So, point A in this figure gives the position of the man at this time.

Now he starts moving towards left and at t equal to 8 minutes he is 5 m from the tree or origin. This position is represented by point B.

The position of man can be negative in the sense that if he keeps moving towards left crossed the tree and is at 4m towards the left of the tree. At this time $t = 15$ minutes his position is represented by point C on the line which is negative.

Hence if the position of the man is at 2 m from the tree or origin in a negative direction or towards left then his position is at $x = -2$ meters. If he is at two meters from the origin towards the right or in a positive direction then his position is at $x = 2$ meters. Here quantity x is called the position of the body.

For more information watch this video on position, distance and displacement.

Rest and Motion in Physics

noreply@blogger.com (Rati S.) — Tue, 03 Mar 2020 05:28:00 +0000

Table Of Contents
Introduction
Rest and Motion examples
Define motion and rest in physics
Rest and motion are relative terms

Introduction

The concept of rest and motion is one of the basic concepts of physics and are introduced while studying kinematics. In simple terms, if an object is changing its position then it is in motion and if it does not change its position then it is at rest.
In this article learn about rest and motion in physics. By the end of reading this article, you would be able to

Recognize the difference between rest and motion in physics.
Define terms like rest and motion.
Understand why motion is relative.

Watch this video to get the concept of rest and motion in physics

Rest and Motion examples

Now if you look around then you might find some things moving and some things not moving at all.
Moving things around us include cars, buses, trains, birds, animals, human beings, etc.
All these things move from one place to another. Apart from these examples things are in motion inside of the body too. For example, blood moves through blood vessels, our muscles move when we work or play, heart pumping blood and many more search examples can be found in nature.
We do not see these motions but we know about them from their effect because our life depends on heart pumping blood and we need muscle movement to perform various tasks.
So we can say that

Motion occurs due to movement of a body

Now let's talk about some objects like a tree, house, School, electric pole, etc. These objects are stationary objects. They remain fixed at their position. From these examples, we can see that few things around us are moving and few of them are stationary.

Define motion and rest in physics

Let’s now define terms rest and motion

A body is said to be at rest if its position does not change with the passage of time if the position of an object changes with the passage of time it is said to be in motion.

One common feature of all moving bodies is that they change their position with time.
The position of a body is known by its surroundings.
So for a book kept on a table inside a room, the table, walls, and ceiling of a room make the surroundings.

To be more precise, the state of motion or, state of the rest of the body should be defined with respect to its surroundings.

As an example, look at the book lying on the table in a room. When nobody is near it then the position of the book with respect to tabletop, walls or ceiling of the room is not changing.

So the book as seen from the room is at rest. We can say that the book is at rest with respect to the room.
Now you entered the room pick the book and begin to move out of the room with the book. Now in this case distance of book from the walls changes as time passes. So the book is in motion with respect to the room.

Rest and motion are relative terms

Now let us look at the fact that both rest and motion are relative.
This statement means that any object which is in motion with respect to one person or object might be at rest with respect to another person or object.
For example, consider the case of the book kept on the tabletop. Now you picked up the book and started moving.
Now this book in your hand when you are moving is at rest with respect to you but the same book is in motion or moving with respect to the walls of the room.
To get more clarity on this statement let us take a case of a person observing another person inside a hot air balloon.

Now for a person A who is observing other person B in hot air balloon from the ground, person B is in motion with respect to him or ground. This same person B is at rest with respect to the hot air balloon or another person say person C sitting beside him in a hot air balloon.
This is because both of them are sitting inside the balloon and both are not moving with respect to each other as time passes.
So, both rest and motion are relative and depend on with respect to which object or body we are observing it.
I hope you now understand the concept of rest and motion. You are now aware of the fact that both rest and motion are relative.

How to find displacement on a position time graph

noreply@blogger.com (Rati S.) — Tue, 18 Feb 2020 08:13:00 +0000

In this article, we will learn how to find displacement on a position-time graph. A position-time graph is a graph where the instantaneous position $x$ of a particle is plotted on the y-axis and the time $t$ on the x-axis. The position-time graph shows you where an object is located over a certain interval of time or at any particular instant of time.
Displacement of any object is defined as the change in position of the object in a fixed direction. It is given by the vector drawn from the initial position to the final position of the object.
Displacement is a vector quantity i.e., it has both magnitude and direction. Displacement of an object can be positive, negative and zero.

How to find displacement on a position-time graph

To learn how you can find displacement on a position-time graph consider the figure given below that shows a position-time graph of any object.

Now from the above figure, we can see that our object under consideration starts its journey from the origin at initial time $t=0$ and position $x=0$. It reaches its final position at the time $t=9s$ represented by point $H$ on the position-time graph.
Let us now find the displacement of the object when it reaches point $A$ marked on the graph. Let us now draw a line to get a projection of point $A$ on the y-axis to get the position of the object when it reaches point $A$. The figure given below shows the projection of various points on both the x-axis and the y-axis.

The projection of point $A$ on the y-axis is at point $4m$ from the origin. So,
Position of the object with respect to the origin when it reached point $A$ $= 4m$
Position of the object with respect to the origin when it reaches point $B$ $= 4m$
From the position-time graph, we can see that the projection of $B$ on the y-axis is at position $x=4m$ which means the object is at rest and has not moved with the passage of time.
Similarly, you can find the position of the object at various points with respect to origin $O$ or initial position of the object.
Let us now find the displacement of the object betwen time $t=0\,\,s$ to $t=3.5\,\,s$.
At $t=4s$ our object is at point $P$.
We know that displacement is a change in the position of the object.
Here,
$x_0=0m$
$x_{3.5}=-4m$
So, change in position = displacement =$x_{3.5} - x_0\,\,=\,\,-4-0=-4m$
Note that here displacement is negative. Negative displacement means distance in the negative direction. So, instead of 4 m forward from the origin, object is 4 m backward from the origin.
Let us now find the displacement of the object betwen time $t=0\,\,s$ to $t=4\,\,s$.
At $t=4s$ our object is at point $P$.
Here
$x_0=0m$
$x_4=0m$
So, change in position = displacement =$x_4 - x_0\,\,=\,\,0m$

Similarly, you can find displacement between any time interval using the position-time graph.

Related Articles

How to find displacement on velocity-time graph

How to find average velocity on a position-time graph
How to find instantaneous velocity on a position-time graph

How to find velocity from force and distance

noreply@blogger.com (Rati S.) — Tue, 28 Jan 2020 15:25:00 +0000

Table Of Contents
Can we find velocity using force formula
Using work-energy theorem

Can I find the velocity of a moving car if I have the information about mass, acceleration, and distance through which the car travel?
In this article, we will look at how to find velocity from force and distance. If we look at our problem then we can come to notice that

what physical quantities are being provided to us and
what physical quantities we need to calculate

In our problem, we have to calculate the velocity of an object from force and distance. So,
Known Quantities:- Force and distance
Unknown Quantities:- Velocity

Can we find velocity using force formula

Now when we first think about force the following force formula comes to our mind $$F=ma$$ This formula involves force, mass, and acceleration. Again from acceleration formula with velocity and time, we know that $$a=\frac{v}{t}$$ where $v$ is the velocity of the moving object and $t$ is the time taken.
To find the velocity of the particle using force formula we must have knowledge about

the force acting on the moving object and
time for which force acts

Please note here that distance traveled does not figure in the equation $$v=\frac{Ft}{m}$$. So we can not use force formula to find the velocity of the object from force and distance.

Using work-energy theorem

From the above discussion, it is clear that we need a relationship which could relate

force
distance and
velocity

If you are familiar with the concept of work and energy then you must be aware of the following facts

Kinetic Energy is the energy used for motion. We can say that if an object of mass $m$ is moving with some velocity $v$, it has kinetic energy.
We also know that when a thing moves they do work.

So, when things move they do work and they have kinetic energy. Let us now state the work-energy theorem. It states that

The work done on an object by net force is equal to the change in kinetic energy of the object
$W=\Delta \,\, KE$

Follow this link for more information
Let us now look at how How to Calculate Velocity From Force and Distance. We know that work is done by force $F$ in moving distance $d$ is given by the relation
$W=F\cdot d$
Also, we know that kinetic energy is
$KE=\frac{1}{2}mv^2$
From work-energy relation we have
$ F\cdot d = \frac{1}{2}mv^2 $
rearranging above equation for the velocity we get
\[v=\sqrt{\left(\frac{2Fd}{m}\right)}=\sqrt{2ad}\] where $a=\frac{F}{m}$
From the above relation, we can find the velocity of an object of mass $m$ from force and distance.
So, you can find the velocity of a moving car if you have the information about mass, acceleration, and distance through which the car travel.

Also, read

Force Formula with solved questions

noreply@blogger.com (Rati S.) — Tue, 10 Dec 2019 06:51:00 +0000

Force Formula

According to Newton's Second Law of Motion, the formula of force is mass times acceleration. So,

Force Formula

Unit of force is $Kg.m.s{-2}$ and is also known as $Newton$ written in short as symbol $N$. Force is a vector quantity. It has a magnitude as well as direction.

Force Formula Triangle

Force formula $F=m\times a$ is a simple algebriac expression and if we have information about any two of these quantities we can easily find the third unknown quantity using this force formula expression.

Let us explore the above statement using the force formula triangle. The figure given below shows a force formula triangle that can be used to easily memorize this force formula and how we can use this formula to find the third quantity if any two of force, mass and acceleration are given.

Force Formula Triangle

We can see from this triangle that we have three different possibilities i.e., we can get three different algebraic expressions using the relation between force, mass, and acceleration.

In the first case If we have knowledge of force $F$ and mass $m$ then we can easily calculate the acceleration of the body. So, we have \[a=\frac{F}{m}=F\div m\]
In our second case if we have knowledge of force $F$ and acceleration $a$ then we can easily calculate the mass of the object using our known values of $F$ and $a$. So, we have \[m=\frac{F}{a}=F\div a\]
Our third case in this formula triangle is the case where we have knowledge of mass and acceleration and we can use these quantities to find the force applied to the body. So, we have \[F=m\times a\] this is nothing but our force formula.

Solved Examples on Force Formula

Question 1. What force would be required to produce an acceleration of 2ms^-2 in a ball of 8 Kg?
Solution: Here it is given in the question that
Mass $m=8Kg$
Acceleration $a=2m/s^2$
Force $F=?$ which is to be calculated.
From the relation \[F=ma\] we get \[F=8\times 2 \,\, N\]or, Force \[F=16\,\,N\] which is the required answer.