Whole Procedure of Equations of motion.

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KINEMATICKINEMATICKINEMATICKINEMATIC

EQUATIONS EQUATIONS EQUATIONS EQUATIONS

OF MOTIONOF MOTIONOF MOTIONOF MOTION

By: - Amritpal Singh Nafria

+917814080880 & +918559012321

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INDEX

S. NO. DOCUMENTS DETAIL PAGE NO. DATE

1. Equations of motion in detail 06 – 16 --------------

2. Copyrights Certificate 17 22-04-2008

3. Reviews of various Institutes 18 2008

4. RTI TO PSEB 19 – 22 05-12-2012

5. PSEB reply to my RTI 23 – 24 24-12-2012

6. RTI to CBSE 25 15-01-2013

7. CBSE reply to my RTI 26 13-02-2013

8. CBSE reply to my RTI 27 11-03-2013

9. My First Appeal 28 – 33 22-03-2013

10. Proposal letter sent to CBSE 34 06-04-2013

11. My Second RTI to CBSE 35 02-05-2013

12. My Second Appeal 36 21-05-2013

13. CBSE reply to my second RTI 37 07-06-2013

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14 Research paper published in IJERD 38 – 40 23-09-2013

15. CIC reply to my first RTI 41 28-10-2013

16. Research sent to IIT Ropar & IOP through

G-mail 42 31-10-2013

17. Details of video-conference 43 08-11-2013

18. CIC Decision letter. 44 – 45 19-11-2013

19. RTI sent to IIT Ropar 46 06-01-2014

20. RTI sent to IOP (Institute of Physics) 47 06-01-2014

21. Legal notice sent to HRD & others 48 – 58 16-01-2014

22.

COBOSE reply to legal notice

59

20-01-2014

23.

IIT Ropar reply to my RTI

60

24-01-2014

24

IOP reply to my RTI

61

24-01-2014

25.

CBSE reply to legal notice

62

03-03-2014

26.

My clarification & Query on CBSE letter

63 – 64

--------------

27.

HRD reply to legal notice

65

31-03-2014

28.

NCERT’s lawyer reply to legal notice

66 – 67

13-05-2014

4

29.

My lawyer letter to the lawyer of NCERT

to arrange meeting for discussion.

68

04-06-2014

30.

My letter to the lawyer of NCERT to

arrange meeting for discussion.

69 – 70

05-08-2014

31.

My RTI to NCERT to know the status of

my letter.

71

05-09-2014

32.

Meeting fixed through e-mail

conversation with NCERT

72

07-10-2014

33.

Details of conversation with NCERT

73

30-10-2014

34.

NCERT reply to my RTI

74

07-10-2014

35.

Letter to the President

75

19-12-2014

36.

Letter to the Prime Minister

76

19-12-2014

37.

Letter to the HRD Minister

77

19-12-2014

38.

Documents attached with Letters of

President, Prime Minister & HRD Minister

78 – 80

19-12-2014

39.

RTI to the Prime Minister

81

19-02-2014

40.

RTI to the HRD Minister

82

19-02-2014

41.

HRD reply to my letter

83 – 85

09-03-2015

5

42.

PMO reply to my RTI

86 – 88

10-03-2015

43.

HRD reply to my RTI

89

13-04-2015

44.

RTI to CBSE Academic Unit

90

18-05-2015

45.

My letter to PMO Appellate Authority

91

22-05-2015

46.

NCERT letter to me

92

14-05-2015

47.

NCERT letter to me

93

18-05-2015

48.

2 NCERT letters to me

94 – 95

20-05-2015

49.

My first appeal to CBSE

96

28-05-2015

50.

My letter to HRD appellate Authority

98 – 100

30-05-2015

51 Other Documents 101 - ---- -------------

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Facts of the Issue: -

1. In the whole world all education boards have been teaching students that equations of motion are three but in reality equations of motion are five.

2. No sixth equation exists; if anyone derived one more equation of motion then I will take this claim back.

3. All equations are derived from each other.

4. All the problems and numerical regarding equations of motion can solve with only two equations, there is no use of third equation of motion. Choose only one equation from both A and B pool;

Pool-A Pool-B

a = (v –u) ÷ t S = ½ (u + v) t

------------------ S = ut + ½ at2

------------------ S = (v2 – u2) ÷ 2a

------------------ S = vt – ½ at2

5. All the four equations of pool-B are same or in other words all equations of pool-B are used to calculate displacement to make numerical easier, i.e. we use equation ‘S = ut + ½ at2’ to calculate displacement very easily when final velocity is not given, equation ‘S = (v2 – u2) ÷ 2a’ to calculate displacement when time is not given, equation S = vt – ½ at2 to calculate displacement when initial velocity is not given and equation S = ½ (u + v) t to calculate displacement when acceleration is not given?

6. I think only one equation ‘S = (v2 – u2) ÷ 2a’ is not directly derived from velocity-time graph. CBSE experts have already accepted that equation ‘S = (v2 – u2) ÷ 2a’ is usually attached the tag of third equation of motion.

7. In 2001, I had derived ‘S = vt – ½ at2’. At that time when I searched this equation on Google then Google showed no result, so I am also claiming that this equation was first derived by me.

8. In 2008, I took Copyright Certificate of fourth equation of motion (‘S = vt – ½ at2’) from the Govt. of India.

9. In September 2013, we published our research paper named "Equations of motion are five in nature not three" in International Journal of Engineering and Research Development [IJERD].

10. Many people says that they are using equation (‘S = vt – ½ at2’) for long time, but I just want to say that I have derived equation ‘S = vt – ½ at2’ in 2000, if anyone proves that this equation has already published in any authentic book, Website, research journal etc before 2001 OR anyone proves that this equation was derived by any other person after 2000 having valid proofs then I will take that claim back.

11. This is not a key issue that who discovered that equation, by me or by other, the key issue is that why equations ‘S = ½ (u + v) t’ and ‘S = vt – ½ at2’ were not considered as an equations of motion and why ‘S = (v2 – u2) ÷ 2a’ was considered as fundamental equation of motion?

12. A change in number of equations of motion will be a biggest achievement for India. So it can also become a moment of pride for every Indian if we correct that mistake.

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Requisite condition for an equation to be consider as an equation of motion. 1. An equation must consist of kinematic variables i.e. time, displacement, final velocity, initial velocity & acceleration. 2. The value of all the quantities must be correct, so that we can get correct value of the kinematical variable. For e.g. if S = n (here n is any natural number) than in every equation in which we get the value of displacement (S) is equal to “n” is an equation of motion, i.e.

S = (v2 – u2) /2a = ut + ½ at2 = ½ (u + v) t = vt – ½ at2 = n 3. An equation must be in its shortest form.

For e.g. if “a = (v2 – u2) ÷ 2vt – at2” is an equation, then its shortest form is a = (v – u)/t; a = (v2 – u2) ÷ 2vt – at2

a = [(v – u) (v + u)] ÷ [t (2v – at)] a = [(v – u) (v + u)] ÷ [t (2v – v + u)]

a = [(v – u) (v + u)] ÷ [t (v + u)] a = (v – u) ÷ t

4. We use an equation as formulae, so we must remember that LHS variable must not be present in RHS. In case if LHS variable is also present in RHS that means derivation is yet incomplete. After removing LHS variable from RHS, we get an independent kinematical equation of motion. For e.g., u = 2u – v + at, Here initial velocity 'u' is in both sides, so we removing 'u' from RHS by putting u = v – at and get an equation of motion, i.e. u = v – at

u = 2u – v + at u = 2(v – at) – v + at u = 2v – 2at – v + at

u = v – at 5. The equation should not be the opposite view of its own. For e.g. u = v – at is an equation of motion than v = u + at or a = (v - u) ÷ t or t = (v - u) ÷ a are not new equations of

motion.

Now only 5 equations can satisfy above mention all conditions.

i. a = (v – u) ÷ t [Displacement (S) independent equation] ii. S = ut + ½ at 2 [Final velocity (v) independent equation] iii. S = (v2 − u2) ÷ 2a [Time (t) independent equation] iv. S = vt – ½ at 2 [Initial Velocity (u) independent equation] v. S = ½ (u + v) t [Acceleration (a) independent equation]

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Derivation of equations of motion from velocity-time graph

Consider the linear motion of a body with initial velocity u. The body accelerates uniformly and in time t, it acquires the final velocity v. The velocity-time graph is a straight line AB as shown in figure. It is evident from the graph

that: Initial velocity (at t = 0) = OA = CD = u

Final velocity (at t) = OE = BD = v

Figure-1. A velocity–time graph for an object undergoing uniform acceleration

Acceleration of the body (a) = Slope of the line AB

a = BC = BD − CD AC OD

a = (v −−−− u) ÷ t ------------ (1)

From the velocity-time graph shown in Fig-1, the distance ‘S’ travelled by the object in time t, moving under uniform acceleration ‘a’ is given by the area enclosed within the trapezium OABC under the graph. That is,

Distance travelled by a body in time ‘t’ is equal to area of the trapezium OABD. S = ½ (sum of ǁ sides) × (⊥ distance between parallel sides)

S = ½ (OA + BD) × OD S = ½ (u ++++ v) ×××× t ------------ (2)

From first equation of motion “t = (v − u) ÷ a”, we get S = ½ × (u + v) × (v − u) ÷ a

S = ½ × (v2 − u2) ÷ a S = (v2 −−−− u2) ÷ 2a or 2aS = (v2 −−−− u2) ------------ (3)

In Fig-1, the distance travelled by the object is obtained by the area enclosed within OABD under the velocity-time

graph AB. Thus, Distance travelled by a body in time ‘t’ is equal to area of the trapezium OABD.

S = area of rectangle OACD × area of triangle ABC S = (OD × OA) + ½ (AC × BC)

S = (t × u) + ½ [t × at] S = ut + ½ at2 ------------ (4)

Distance travelled by a body in time ‘t’ is equal to area of the trapezium OABD.

S = area of rectangle OEBD – area of triangle ABE S = (OD × OE) – ½ (AE × BE)

S = (t × v) – ½ (at × t) S = (vt) – ½ (at2)

S = vt −−−− ½ at2 ------------ (5)

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Method to use an equation of motion Graphical analysis is an important tool for physicists to use to solve problems. Sometimes, however, we have enough information to allow us to solve problems algebraically. Algebraic methods tend to be quicker and more convenient than graphical analysis. If you were in the vehicle, you would simply use the vehicle’s speedometer to determine the speed of the vehicle. Knowing the speed of your vehicle, you could easily determine how far it would travel in a given time interval using the equation v = S/t. As you can see, the best way to solve a problem is usually determined by the information that is available to you. To be able to solve problems related to motion with uniform acceleration, in which the velocity may change but the acceleration is constant, we need to use algebraic equations to solve the numerical that describe this type of motion. Equations of motion are very useful to locate the position and calculate the final velocity, initial velocity, acceleration and time taken by the object or body in uniform motion. Table-1 shows the five key equations of accelerated motion. You should be able to solve any kinematic numerical regarding equations of motion by correctly choosing one of these five equations. They involve the variables for displacement, initial velocity, final velocity, acceleration, and time interval. In table-1, we see that in each equation one variable is missing. When solving uniform acceleration problems, choose which equation to use based on the given, missing and required variables of the problem.

Table-1. The Five Key Equations of Accelerated Motion.

Our first task is to determine which of the five equations of accelerated motion to use. Usually, you can solve a problem using only one of the five equations. Second task is to identify which equation contains all the variables for which we have given variables, missing variable and the unknown variable that we are asked to calculate. After identifying the correct equation, you can use it to solve the numerical. For example: - 1. A sports car approaches a highway on-ramp at a velocity of 20.0 m/s. If the car accelerates at a rate of 3.2 m/s2 for 5.0 s, what is the displacement of the car? Sol: - Given: u = 20 m/s, a = 3.2 m/s2, t = 5s. Missing: final velocity (v). Required: Displacement (S). Analysis: In table 1, we see that equation-2 has all the given variables, missing variable and required variable. So, we will have to use Equation-2 to easily solve this numerical rather than using another equation to make solution a lengthy procedure.

Solution from Second equation of motion S = ut + ½ at 2

S = (20m/s × 5s) + (½ × 3.2 m/s2 × 5s × 5s) S = (100m) + (40m)

Displacement (S) = 140meter.

S. No.

Kinematic equations of

motion

Variables found in

equation

Missing Variables in

equation 1. a = (v – u) ÷ t a, v, u, t Displacement (S)

2. S = ut + ½ at 2 S, a, u, t Final velocity (v) 3. S = (v2 − u2) ÷ 2a S, a, v, u Time (t) 4. S = vt – ½ at 2 S, a, v, t Initial velocity (u) 5. S = ½ (u + v) × t S, v, u, t Acceleration (a)

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2. A sailboat accelerates uniformly from 6.0 m/s to 8.0 m/s at a rate of 0.50 m/s2. What distance does the boat travel? Sol: - Given: u = 6.0 m/s, v = 8.0 m/s, a = 0.50 m/s2. Missing: Time. Required: Displacement (S). Analysis: In table 1, we see that equation-3 has all the given variables, missing variable and required variable. So, we will have to use Equation-3 to easily solve this numerical rather than using another equation to make solution a lengthy procedure.

Solution from Fourth equation of motion S = (v2 − u2) ÷ 2a

S = [(8.0 m/s)2 − (6.0 m/s)2] ÷ [2 × 0.50 m/s2] S = [64 m2/s2 − 36 m2/s2] ÷ [1 m/s2]

Displacement = 28 meter. 3. A car is suddenly stops in 5s with a retardation of 23m/s2, here final velocity is 0m/s(because car is finally at rest ) calculate the total distance covered by the car. Sol: - Given: v = 0 m/s, a = –23 m/s2, t = 5s. Missing: initial velocity (u). Required: Displacement (D). Analysis: In table 1, we see that equation-4 has all the given variables, missing variable and required variable. So, we will have to use Equation-4 to easily solve this numerical rather than using another equation to make solution a lengthy procedure.

Solution from Fourth equation of motion S = vt – ½ at 2

S = (0m/s × 5sec) – (½ × –23m/s2 × 5sec × 5sec) S = (0m) – (–287.5meters)

Displacement (S) = 287.5meters 4. A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 350 m/s. The dart comes to rest in 0.0050 s. Sol: - Given: u = 350 m/s, v = 0 m/s, t = 0.0050s. Missing: Acceleration (a). Required: Displacement (S). Analysis: In table 1, we see that equation-5 has all the given variables, missing variable and required variable. So, we will have to use Equation-5 to easily solve this numerical rather than using another equation to make solution a lengthy procedure.

Solution from Fifth equation of motion S = ½ (u + v) × t

S = ½ (350 m/s + 0 m/s ) × 0.0050s S = ½ × 350 m/s × 0.0050s

Displacement (S) = .88 meter.

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Today scientific communities have only two options: -

Pool-A Pool-B

a = (v –u) ÷ t S = ½ (v + u) × t

------------------ S = ut + ½ at2

------------------ S = (v2 – u2) ÷ 2a

------------------ S = vt – ½ at2

Option-1. Scientific Community should only two equations [First equation from pool-A and Second equation from any one equation from Pool B] as fundamental equations of motion and other three equations of ‘pool B’ should be consider as additional equations of motion i.e. derive forms of second equation of motion. Option-2. As no sixth equation exists; so Scientific Community should consider all the five equations as equations of motion, so that people could get entire knowledge about equations of motion. Note that: we can’t ignore any equation; all the equations are different in property and important for students. Also sixth equation of motion can’t derive by velocity-time graph. If we consider only one or three or four equations then it will not only confuse for students but also it means you will give incomplete and improper knowledge to the young generation.

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On which basis equation ‘S = (v2 – u2) ÷ 2a’ was considered as fundamental equation of motion?

1. It doesn’t directly derive from velocity-time graph. So, why this equation was considered as

fundamental equation of motion?

2. It is only derived by eliminating time from S = vt – ½ at2 or S = ut + ½ at2 or S = ½ (u + v) t.

So, why a derived equation is considered as fundamental equation of motion?

3. What are the importance of equation ‘S = (v2 – u2) ÷ 2a’. Mention each importance in detail

with example if possible.

4. We can solve all numerical problems with two equations of motion, i.e.

i) a = (v –u) ÷ t and

ii) S = ut + ½ at2

So, why three equation was considered as fundamental equations of motion?

5. If this equation considered as fundamental equation of motion on the basis that it calculate

displacement very easily when time is not given; then why equation S = vt – ½ at2 and equation

S = ½ (u + v) t didn’t consider as fundamental equation of motion as equation S = vt – ½ at2 also

calculate displacement very easily when initial velocity is not given and Equation S = ½ (u + v) t

calculate displacement very easily when acceleration is not given?

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Derivation of other equations from ‘S = ut + ½ at2’

Case-I S = ut + ½ at2

Put t = (v – u) ÷ a in above equation, we get

S = u [(v – u) ÷ a] + ½ a [(v – u) ÷ a]2

S = [(uv – u2) ÷ a] + ½ [(v – u)2 ÷ a]

S = [(uv – u2) ÷ a] + ½ [(v2 + u2 – 2uv) ÷ (a)]

S = [2uv – 2u2 ÷ 2a] + [v2 + u2 – 2uv ÷ 2a]

LCM of RHS is 2a, hence

S = (2uv – 2u2 + v2 + u2 – 2uv) ÷ 2a

S = (v2 – u2) ÷ 2a

Equation S = (v2 – u2) ÷ 2a is derive from Equation S = ut + ½ at 2.

Case-II S = ut + ½ at 2

Put u = v – at in above equation, we get

S = (v – at) t + ½ at2

S = vt – at2 + ½ at2

S = vt – ½ at2

Equation S = vt – ½ at2 is derive from Equation S = ut + ½ at2.

Case-III S = ut + ½ at2

Put a = (v – u) ÷ t in above equation, we get

S = ut + ½ [(v – u) ÷ t] t 2

S = ut + ½ [(v – u)] t

S = ut + ½ (vt – ut)

S = ut + ½ vt – ½ ut

S = ½ vt + ½ ut

S = ½ (v + u) t

Equation S = ½ (v + u) t is derive from Equation S = ut + ½ at2.

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Derivation of other equations from ‘S = (v2 – u2) ÷ 2a’

Case-I S = (v2 – u2) ÷ 2a

Put v = u + at in above equation, we get

S = [(u + at) 2 – u2] ÷ 2a

S = (u2 + a2t 2 + 2uat – u2) ÷ 2a

S = (u2 – u2 + a2t 2 + 2uat) ÷ 2a

S = (a2t 2 + 2uat) ÷ 2a

S = ½ at 2 + ut

S = ut + ½ at 2

Equation S = ut + ½ at 2 is derive from Equation S = (v2 – u2) ÷ 2a

Case-II S = (v2 – u2) ÷ 2a


S = [v 2 – (v – at)2] ÷ 2a

S = [v 2 – (v2 + a2t 2 – 2vat)] ÷ 2a

S = [v 2 – v2 – a2t 2 + 2vat)] ÷ 2a

S = [– a2t 2 + 2vat] ÷ 2a

S = – ½ at 2 + vt

S = vt – ½ at 2

Equation S = vt – ½ at 2 is derive from Equation S = (v2 – u2) ÷ 2a

Case-III S = (v2 – u2) ÷ 2a


S = [(v2 – u2)] ÷ [2(v – u) ÷ t]

S = [(v2 – u2) t] ÷ [2(v – u)]

S = [(v – u) (v + u) t] ÷ [2(v – u)]

S = [(v + u) t] ÷ [2]

S = ½ (v + u) t

Equation S = ½ (v + u) t is derive from Equation S = (v2 – u2) ÷ 2a

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Derivation of other equations from ‘S = vt – ½at2’

Case-I S = vt – ½ at2


S = (u + at) t – ½ at2

S = ut + at2 – ½ at2

S = ut + ½ at2

Equation S = ut + ½ at 2 is derive from Equation S = vt – ½ at2.

Case-II S = vt – ½ at2


S = vt – ½ [(v – u) ÷ t] t 2

S = vt – ½ [(v – u)] t

S = vt – ½ (vt + ut)

S = vt – ½ vt + ½ ut

S = ½ vt + ½ ut

S = ½ (v + u) t

Equation S = ½ (v + u) t is derive from Equation S = vt – ½ at2

Case-III S = vt – ½ at2


S = [v(v – u) ÷ a)] – ½ a[(v – u) ÷ a)]2

S = [(v2 – vu) ÷ a)] – ½ a[(v – u) 2 ÷ a2)]

S = [(v2 – vu) ÷ a)] – ½ a[v2 + u 2 – 2uv ÷ a2)]

S = [(v2 – vu) ÷ a)] – ½ [v2 + u 2 – 2uv ÷ a)]

S = [(v2 – vu – ½v2 – ½u 2 – uv) ÷ (a)]

S = [(v2– ½v2 – ½u 2) ÷ (a)]

S = [(½v2 – ½u 2) ÷ (a)]

S = (v2 – u 2) ÷ 2a

Equation S = (v2 – u 2) ÷ 2a is derive from Equation S = vt – ½ at2

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Derivation of other equations from ‘S = ½ (v + u) t’.

Case-I S = ½ (v + u) t

Put v = u + at in above equation, we get

S = ½ [u + at + u] × t

S = ½ [2u + at] × t

S = ½ [2ut + at2]

S = ut + ½ at2

Equation S = ut + ½at2 is derive from Equation S = ½ (v + u) t

Case-II S = ½ (v + u) t


S = ½ (v + v – at) × t

S = ½ (2v – at] × t

S = ½ (2vt – at2)

S = vt – ½at2

Equation S = vt – ½at2 is derive from Equation S = ½ (v + u) t

Case-III S = ½ (v + u) t


S = ½ (v + u) (v – u) ÷ a

S = ½ (v2 – u2) ÷ a

S = v2 – u2 ÷ 2a

Equation S = v2 – u2 ÷ 2a is derive from Equation S = ½ (v + u) t

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COPYTIGHT CERTIFICATE

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REVIEWS FROM VARIOUS INSTITUTES

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TYPED RTI APPLICATION TO PSEB

RTI Application Form FORM ‘A’

See Rule 3(1) I. D. No…………….. (For Office Use Only)

To The Public Information Officer/ THE SECRETARY, Assistant Public Information Officer P.S.E.B., S.A.S.NAGAR, MOHALI. 1. Full Name of the Applicant : AMRIT PAL SINGH 2. Father Name/Spouse Name : DARSHAN SINGH 3. Permanent Address : HOUSE NO. - 303, STREET NO. - 10, : PREM BASTI, SANGRUR-148001. 4. Correspondence Address : HOUSE NO. - 303, STREET NO. – 10, : PREM BASTI, SANGRUR-148001. 5. Particulars of the Information Solicited a) Subject Matter of Information : EQUATIONS OF MOTION. b) The period to which information relates : 30 DAYS. c) Specific Details of Information required: It has been mentioned in your 9th standard Science book that S = ut + ½ at2 is a second equation of motion and v2 = u2 + 2aS is a third equation of motion where as equation S = ½(v + u)t is not consider as equation of motion. I have noticed that S = ut + ½ at2, v2 = u2 + 2aS and S = ½(v + u)t are derived from each other. Also all numerical regarding equations of motion are solved by using two equations of motion i.e. a =(v-u)/t and S = ut + ½ at2 OR v2 = u2 + 2aS OR S = ½(v + u)t. So in this concern I need all the information that on which basis you teach students that both 2nd & 3rd equations of motion are different, also on which basis 2nd and 3rd equations are consider as equations of motion but S = ½ (v + u) × t is not.

d) Whether information is required by Post or in : person (the actual postal fees shall be included : BY POST in additional fee in providing the information) e) In case by Post (ordinary/registered : REGISTERED POST or speed post) 6. Is this information not made available by public authority under voluntary disclosure? : YES. 7. Do you agree to pay the required fee? : YES. 8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 100/ (If Yes, Details of such deposit) : 9. Whether belongs to below Poverty Line category? : NO. (If yes, you furnished the proof of the same with application?) Place: SANGRUR. Date: 05-DECEMBER-2012 Signature of Applicant

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DOCUMENTS ATTACHED WITH RTI APPLICATION - 1

CASE-I

S = ut +½ at2

Put the value of t = (v – u) ÷ a in above equation, we get

S = u [(v – u) ÷ a] + ½ a [(v – u) ÷ a]2

S = [(uv – u2) ÷ a] + ½ [(v – u)2 ÷ a]

S = [(uv – u2) ÷ a] + [(v2 + u2 – 2uv ÷ 2a)]

S = [2uv – 2u2 ÷ 2a] + [v2 + u2 – 2uv ÷ 2a]

S = 2uv – 2u2 + v2 + u2 – 2uv ÷ 2a

S = (v2 – u2) ÷ 2a

From above derivation is it proved that both equations S = ut +½ at2 and S = (v2 – u2) ÷ 2a are same?

CASE-II

S = (v2 – u2) ÷ 2a

Put the value of v = u + at in above equation, we get

S = [(u + at) 2 – u2] ÷ 2a

S = (u2 + a2t 2 + 2uat – u2) ÷ 2a

S = (a2t 2 + 2uat) ÷ 2a

S = ½ at 2 + ut

S = ut + ½ at 2

From above derivation is it proved that both equations S = (v2 – u2) ÷ 2a and S = ut +½ at2 are same?

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CASE-III

S = ut + ½ at 2

We know that a = (v – u) ÷ t put this value of ‘a’ in above equation, we get

S = ut + ½ [(v – u) ÷ t] t 2

S = ut + ½ [(v – u)] t

S = ut + ½ (vt – ut)

S = ut + ½ vt – ½ ut

S = ½ vt + ½ ut

S = ½ (v + u) ×××× t

From above derivation is it proved that both equations

S = ut +½ at2 and S = ½ (v + u) × t are same or different?

CASE-IV

S = ½ (v + u) ×××× t

We know that v = u + at put this value of ‘v’ in above equation, we get

S = ½ [(u + at) + u] × t

S = ½ [u + at + u] × t

S = ½ [2u + at] × t

S = ½ [2ut + at2]

S = ut + ½ at2

From above derivation is it proved that both equations

S = ½ (v + u) × t and S = ut +½ at2 are same or different?

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CASE-V

S = ½ (v ++++ u) t

Put the value of t = (v – u) ÷ a in above equation, we get

S = ½ (v + u) [(v – u) ÷ a]

S = ½ [(v2 – u2) ÷ a]

S = (v2 – u2) ÷ 2a

From the basis of above derivation is it proved that

S = (v2 – u2) ÷ 2a is a derived form of S = ½ (v + u) t?

CASE-VI

S = (v2 – u2) ÷ 2a

Put the value of a = (v – u) ÷ t in above equation, we get

S = [(v2 – u2)] ÷ [2(v – u) ÷ t]

S = [(v – u) (v + u) × t]] ÷ [2(v – u)]

S = [(v + u) × t] ÷ [2]

S = ½ (v ++++ u) t

From the basis of above derivation is it proved that

S = ½ (v + u) t is a derived form of S = (v2 – u2) ÷ 2a?

23

PSEB REPLY TO MY RTI - 1

24

PSEB REPLY TO MY RTI - 2

25

MY RTI TO CBSE

26

FIRST REPLY BY CBSE TO MY RTI

27

SECOND REPLY BY CBSE TO MY RTI

28

MY FIRST APPEAL TO APPELLATE AUTHORITY

29

TYPED & CORRECTED FIRST APPEAL

To,

The professor / Director,

(Academics, Research, Training & Innovation)

Central Board of Secondary Education

‘Shiksha Sadan’, Institutional Area, Academic Unit,

17, Rouse Avenue, New Delhi – 110002.

Subject: - Appeal to first appellate Authority.

Respected Sir,

I have received the letter of PIO of Academic unit, C.B.S.E. regarding R.T.I. CASE No.-7835 on 15.March.2013. I am not satisfied with the information given to me. I have attached all the documents that justify my point. Now I am appealing to the first appellate Authority of C.B.S.E. (Academics, Research, Training and Innovation) “please provide me correct and reasonable information”.

Thanking You,

DATE = 22.March.2013

Yours Sincerely,

Amritpal Singh

30

DOCUMENTS ATTACHED WITH FIRST APPEAL - 1

S.No. CBSE Experts My query

1. The two equations v = u + at and

S = ut + ½ at2

are regarded as the two

independent kinematical equations of

motion because these can be directly

derived using the basic definitions of

velocity and acceleration.

I disagree that only these two equations [v = u + at and

S = ut + ½ at2]

are independent kinematical equations

of motion and only these can be directly derived using

the basic definitions of velocity and acceleration.

2. The third equation v2 - u2 = 2aS as does

not quite meet the above criterion.

However, it is usually attached the tag

‘third equation of motion’ because of

its usefulness and convenience in

solving a wide variety of useful

problems.

I agree to your statement that it usually attached the

tag of third equation of motion, so now stop attaching

tag of third equation of motion to it. However I

strongly believe that you’re SO CALLED helpful

equation of motion in solving wide variety of problems

is superfluous i.e. to say that first two equations of

motions [v = u + at and S = ut + ½ at2] are more than

enough to solve all kinds of problems and if not then

send me that particular statement [problem] with an

attachment, I can solve it without using third equation

of motion [v2 -

u

2 = 2aS].

3. The expression S = [(u + v) ÷ 2] t, is

just the mathematical form of the

definition of average speed, and is,

therefore, not really an independent

equation of motion.

The whole equation S = [(u + v) ÷ 2]t is not the

definition of average speed, it is the definition of

displacement and is an equation of motion. The

mathematical form of the definition of average speed

is (u + v) ÷ 2.

E.g. The definition of force is F = m [(v - u) ÷ t], what

you people are explaining is that this is not an

equation of force but definition of acceleration

instead. Now you yourself ponder over whether it is an

equation of force or an equation of acceleration.

4. It is not derived using basic definitions

and needs information about the

initial velocity as well as the final

velocity after a certain time.

It is derived using basic definitions i.e. derived from

velocity-time graph and is independent kinematical

equation of motion. It is very useful and convenience

in solving a wide variety of useful problem. I have

attached the documents. SO now is the time for CBSE

to lead the world education and put that change by

considering S = [(u + v) ÷ 2]t as equation of motion.

DOCUMENTS ATTACHED WITH FIRST APPEAL

Derivation of Fifth equation of motion from velocity

Consider the linear motion of a body with initial in time t, it acquires the final velocity v. The velocityin figure. It is evident from the graph that:

Initial velocity (at t = 0) = OA = CD = u

The distance travelled by a body in time “t” is equal to area of the trapezium OABD

S = ½ (sum of

From above derivation it is proved that this equation is independent kinematical equation of motion. Because it is derived directly from velocity

basic definitions of velocities and acceleration.

What is important to notice is that the quantity of acceleration is not present in this equation. We say, therefore, that the equation is

This equation is often useful in kinematics problems where you do not know the the body but still have to work with the

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Derivation of Fifth equation of motion from velocity

Consider the linear motion of a body with initial velocity u. The body accelerates uniformly and in time t, it acquires the final velocity v. The velocity-time graph is a straight line AB as shown in figure. It is evident from the graph that:

Initial velocity (at t = 0) = OA = CD = u

Final velocity (at t) = OE = BD = v


S = ½ (sum of parallel sides) × (⊥ distance between parallel

S = ½ (OA + BD) × OD

S = ½ (u ++++ v) ×××× t

derivation it is proved that this equation is independent kinematical equation of motion. Because it is derived directly from velocity-time graph or in other words is derived using

basic definitions of velocities and acceleration.

ce is that the quantity of acceleration is not present in this equation. We say, therefore, that the equation is acceleration independent.

This equation is often useful in kinematics problems where you do not know the e to work with the velocities, time, and displacement.


Derivation of Fifth equation of motion from velocity-time graph.

velocity u. The body accelerates uniformly and time graph is a straight line AB as shown


distance between parallel sides)

derivation it is proved that this equation is independent kinematical equation of time graph or in other words is derived using

ce is that the quantity of acceleration is not present in this equation. We

This equation is often useful in kinematics problems where you do not know the acceleration of .


Derivation of third equation of motion from velocity

The distance travelled by a body in time ‘t’ is equal to area of the trapezium

S = ½ (sum of parallel

From first equation of motion (1), t

S = (v

Equation (3) represents third equation of motion. What is important to notice is that the quantity of time is not present in this equation. We say, therefore, that the equation is equation kinematical equation of motion. This equation is often useful in kinematics problems

where you do not know the timefinal velocity

32


Derivation of third equation of motion from velocity


OABD

parallel sides) ×××× (⊥⊥⊥⊥ distance between parallel

S = ½ (OA + BD) × OD

S = ½ (u ++++ v) ×××× t ---------- (a)

From first equation of motion (1), t = (v −−−− u) ÷ a

S = ½ × (u + v) × (v − u) ÷ a

S = ½ × (v2 − u2) ÷ a

S = (v2 −−−− u2) ÷ 2a or 2aS = v2 −−−− u2

Equation (3) represents third equation of motion. What is important to notice is that the quantity of time is not present in this equation. We say, therefore, that the equation is equation kinematical equation of motion. This equation is often useful in kinematics problems

time taken by the body but still have to work with the initial velocity, final velocity, acceleration, and displacement.


Derivation of third equation of motion from velocity-time graph.


distance between parallel sides)

(a)

u) ÷ a

--------- (3)

Equation (3) represents third equation of motion. What is important to notice is that the quantity of time is not present in this equation. We say, therefore, that the equation is time independent equation kinematical equation of motion. This equation is often useful in kinematics problems

taken by the body but still have to work with the initial velocity,

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Importance of Fifth Equation of Motion

Question: - if you were asked to solve the below numerical in an entrance examination like NDA or IAS which equation of motion will you use?

A travelling car is travelling with a speed 115m/s is suddenly stops in 5seconds, here final velocity is 0m/s (because the car is finally at rest) calculate the distance covered. Traditional method to solve the above problem

Solution from Second equation of motion S = ut + ½ at2

S = [(115 × 5sec) + (½ × a × 5sec ×5sec)] S = [(575meters) + (½ × a × 25sec2)]

S = [(575meters) + (a × 12.5sec2) ---------- (8)

Solution from Third equation of motion S = (v2 − u2) ÷ 2a

S = [(0m/s) 2 – (115m/s)2] ÷ (2a) S = [(0m/s) 2 – (13225m2/s2)] ÷ (2a)

S = – (13225m2/s2)] ÷ (2a) ----------- (9)

Here we are not able to find exact value of distance in numeric when acceleration is not given,

Solution from Fifth equation of motion S = ½ (u + v) × t

S = ½ (115m/s + 0m/s) × 5sec

S = 57.5m/s × 5sec

S = − 287.5meters Here negative sign shows retardation.

Answer: - I have used Fifth equation of motion to solve the above numerical. Because it is

the easiest & fastest way to solve the numerical when acceleration (a) is not given as compare to other rest of the equations of motion.

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MY PROPOSAL LETTER SENT TO CBSE

35

MY SECOND RTI TO CBSE

36

MY SECOND APPEAL TO CIC

37

CBSE REPLY TO MY SECOND RTI

38

RESEARCH PAPER PUBLISHED IN IJERD - 1

39


40


41

CIC LETTER TO MY SECOND APPEAL

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RESEARCH SENT TO IIT ROPAR THROUGH G-MAIL

RESEARCH SENT TO IOP THROUGH G-MAIL

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DETAILS OF VIDEO-CONFERENCE

On 08.Nov.2013 a video Conference with Ram Shankar held to sort out the issue regarding

number of equations of motion. Before the meeting, I prepared 8 minutes presentation to explain

the deep facts & figures of my research with solution so that this process could be ended. But in

the video conference when I was trying to explain the matter then they cut the video conference

after just 4 minutes conversation rather than to solve the problem. It is not justified to incomplete

discussion with anyone. In our conversation, Mr. Ram Shankar told me, “your RTI doesn’t

comes under RTI act however we gave you the information as this was an educational issue also

CBSE have given you the last and final information hence we can’t give you any more

information”. Look at their reply in the attachment, is information given by CBSE enough and

correct.

CIC DECISION LETTER

44

CIC DECISION LETTER - 1

45

CIC DECISION LETTER - 2

46

MY RTI TO IIT ROPAR

47

MY RTI TO IOP

48

MY LEGAL NOTICE

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COBOSE REPLY TO MY LEGAL NOTICE

60

IIT ROPAR REPLY TO MY RTI

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IOP REPLY TO MY RTI

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CBSE REPLY TO MY LEGAL NOTICE

63

MY QUERY TO THE CBSE LETTER - 1

CBSE letter The set of three equations of motion is derived from a velocity-time graph. i. v = u + at. ii. S = ut + ½ at 2. iii. S = (v2 − u2) ÷ 2a. It is generally accepted by the scientific community that, for a constant acceleration, this set is sufficient to study the particle’s motion at any instance. However, one can present these equations in many other mathematica forms (as it is given in the referred document). Such rearranged equations also lead to a correct description about the particle’s motion. Some websites(e.g.http://www.lakeheadschools.ca/scvi_staff/brecka/Gr11_physics_web/downloadable_con

tent/unit1/text1/phys11_1_5.pdf) also appear to have made reference to this effect.

My Query

Point first: - the web address you give, on this web address it was mentioned that You should be able to solve any kinematics question by correctly choosing one of these five equations. You have seen how the first three are developed. We will leave the others to be developed as an exercise. i. S = ½ (u + v) t. ii. v = u + at. ii. S = ut + ½ at 2. iii. S = (v2 − u2) ÷ 2a. iv. S = vt – ½ at 2. Firstly , on your above said web address S = ½ (u + v) t was mentioned as first equation of motion but the scientific community was not considered as equation of motion, why? Secondly, first three equations mentioned on your mentioned web address are different from three equations accepted by scientific community. Thirdly , it is mentioned on your said web address that equations “S = (v2 − u2) ÷ 2a”, “S = vt – ½ at 2” are developed as an exercise, but actually you consider “S = (v2 − u2) ÷ 2a” as an equation of motion which is developed as an exercise.

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MY QUERY TO THE CBSE LETTER - 2 Point second: - I am strongly said that the set of three equations is wrong; either the set should be consisted of two equations or five equations. E.g. By choosing only one equation from both A and B pool; you can solve all the problems and numerical regarding equations of motion with only these two equations of motion.

Pool-A Pool-B

a = (v –u) ÷ t S = ½ (u + v) t

------------------ S = ut + ½ at2

------------------ S = (v2 – u2) ÷ 2a

------------------ S = vt – ½ at2

Today scientific communities [S.C.] have only two options: -

Option-1 Scientific communities [S.C.] should consider only two equations [First from pool A and Second from Pool B] and other three equations of ‘pool B’ should be considered as derived form of second equation of pool B]. Option-2 Scientific communities [S.C.] should consider all the five equations as equations of motion, so that people could get complete knowledge about equations of motion. Note that: you never ignore any equation; all the equations are different in property and important for students. Also no sixth equation of motion can be derived by velocity-time graph. If you consider only one or three or four equations then it would confuse the students by giving incomplete and improper knowledge to the young generation.

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HRD REPLY TO MY LEGAL NOTICE

66

NCERT LAWYER REPLY TO MY LEGAL NOTICE - 1

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NCERT LAWYER REPLY TO MY LEGAL NOTICE - 2

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MY LAWYER REQUESTING LETTER TO NCERT LAWYER

69

MY LETTER TO NCERT & HRD

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DOCUMENTS ATTACHED WITH MY LETTER 1. You said that a set of mathematically driven equations from a velocity-time graph representing the motion of an object as given in the text books and other reference books is applicable in inertial frames of references. This means that these equations deal motion under constant acceleration. My Query: “S = (v + u) t” and “S = vt – ½ at2” are drive from a v-t graph representing the motion of an object and are applicable in inertial frames of references. These equations also deal motion under constant acceleration.

2. These equations can always be rearranged to generate new mathematical formulations. However, the equations that contain acceleration term are generally chosen in a set of equations of motion. My Query: Here if you says two equations [u = v – at and S = ut + ½ at2] can always be rearranged to generate new mathematical formulations then it makes sense. But you said that set of three equations [i.e. u = v – at, S = ut + ½ at2

and “S = (v2 – u2) ÷ 2a”] can always be rearranged to generate new mathematical formulations and it doesn’t make any sense, because equation “S = (v2 – u2) ÷ 2a” is generated by rearranging u = v – at and S = ut + ½ at2.

S = ut + ½ at2


S = u [(v – u) ÷ a] + ½ a [(v – u) ÷ a]2

S = [(uv – u2) ÷ a] + ½ [(v – u)2 ÷ a]

S = [(uv – u2) ÷ a] + ½ [(v2 + u2 – 2uv) ÷ (a)]

S = [2uv – 2u2 ÷ 2a] + [v2 + u2 – 2uv ÷ 2a]

LCM of RHS is 2a, hence

S = (2uv – 2u2 + v2 + u2 – 2uv) ÷ 2a

S = (v2 – u2) ÷ 2a

3. Now here you said that the equations that contain acceleration term are generally chosen in a set of equations of motion.

My Query: i) There is no individual usefulness of “S = (v2 – u2) ÷ 2a” except it calculate displacement faster than other equation when time is not given.

ii) It is derived by using u = v – at and S = ut + ½ at2, so here if Scientific community consider “S = (v2 – u2) ÷ 2a” as a fundamental equation of motion, then Now Scientific community will have to answer why equations “S = (v + u) t” and “S = vt – ½ at2” are not considered as fundamental equations of motion.

iii) On which criterion equations are considered as fundamental equations of motion.

Please don’t think that if you change the number of equations of motion then people will question that why NCERT had been teaching wrong or improper or incomplete information about equations of motion to them. Moreover people will appreciate you if NCERT will raise this issue in front of international scientific community and make that change. Also it will become a matter of proud for each and every Indian if we correct that mistake and give true and complete knowledge to the world.

Note: At the end international scientific community will have to consider two equations (Acceleration and displacement equation) as fundamental equations of motion and another three equations(Displacement equations) will have to consider as additional equations of motion.

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MY RTI TO NCERT

MEETING FIXED THROUGH E

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MEETING FIXED THROUGH E -MAIL CONVERSATIONMAIL CONVERSATION

73

DETAILS OF CONVERSATION WITH NCERT

On 30.10.2014, meeting held at NCERT campus to sort the issue, but in the meeting I answered all the

doubts/questions of NCERT but on the other hand they didn’t able to answer any of my doubt. Some

main points of the meetings are these: -

1. They didn’t manage to prove the difference between second equation of motion and third equation of

motion instead they only said “we can’t give ‘u’ and ‘a’ independent equations”. To quote their own

wording “YEH HAMARA VERSION HAI, AAP MAANIYE TOH THEEK, NA MAANIYE TOH

THEEK”. Could they explain me the facts that why they can’t give ‘u’ and ‘a’ independent equations”.

2. They agreed that total equations of motion are five and all five equations are equations of motion but

they totally refused to give me written statement. Is that right?

3. Adding to this they also said that they could have published 2 equations or 5 equations but published 3

equations as “per our WISH”.

4. In NCERT book nowhere is written that only these three are equations of motion.

5. They also asked sarcastically what will you do if textbook development committee eradicates the whole

‘Motion’ chapter from NCERT science book.

6. At the end they said that whenever NCERT book will be republished in future then we will place this

matter before the textbook development committee. But till today they forwarded my appeal neither to

textbook development committee (to check whether I am right or not?) nor to the higher concerned

authority (for requesting to reprint NCERT science book).

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NCERT REPLY TO MY RTI

75

LETTER TO THE PRESIDENT

76

LETTER TO THE PRIME MINISTER

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LETTER TO THE HRD MINISTER

78

DOCUMENTS ATTACHED WITH LETTERS OF PRESIDENT, PRIME MINISTER & HRD MINISTER - 1

My particulars are as under: -

Name: - Mr. Amritpal Singh Nafria.

Father Name: - S. Darshan Singh Nafria.

Date of Birth: - 09-August-1985.

Address: - House Number-303, Street Number-10, Prem Basti, Sangrur-148001 (Pb).

E-mail: [email protected].

Mobile: +917814080880, +918559012321.

Current Status: - Researching on laws of motion and Duky’s theory.

Case History: - I filed an application for the request of copyrights certificate on 17.March.2008. On 22.04.2008, Deputy Registrar of copyrights office, New-Delhi issued to me the copyrights certificate of fourth equation of motion [S = vt – ½ at 2].

In the same year I had taken the reviews of various physics professors from Punjab, whether equations [S = ut + ½ at 2 & S = vt – ½ at 2] are equations of motion or not? In their remarks, H.O.D. of various Universities and colleges admitted that these are equations of motion and should be considered as equations of motion.

On May 23, 2012 we had organized press conference at Chandigarh press. Press reporter advised us to send our research to PSEB, CBSE and NCERT to take their reviews. If CBSE & PSEB accepted your research or they didn’t reply then we will highlight this news.

In the same year I sent RTI to PSEB asked for information. [It has been mentioned in your 9th standard Science book that S = ut + ½ at2 is a second equation of motion and v2 = u2 + 2aS is a third equation of motion where as equation S = ½(v + u)t is not consider as equation of motion. I have noticed that S = ut + ½ at2, v2 = u2 + 2aS and S = ½(v + u)t are derived from each other. Also all numerical regarding equations of motion are solved by using two equations of motion i.e. a = (v-u)/t and S = ut + ½ at2 OR v2 = u2 + 2aS OR S = ½(v + u)t. So in this concern I need all the information that on which basis you teach students that both 2nd & 3rd equations of motion are different, also on which basis 2nd and 3rd equations are consider as equations of motion but S = ½ (v + u) × t is not].

PSEB replied that they strictly follow NCERT books and syllabus. In NCERT 9th class book only above said three equations are considered as equations of motion. Teachers are going to teach and follow whatever is written in the prescribed books. If NCERT does any change in the above topic then PSEB is bound to follow that change in text books.

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When PSEB didn’t provide me information then I sent same RTI to CBSE after two months they sent me information but it was not up to the mark. In the information they admitted that 1st & 2nd equations are independent kinematical equations and 3rd equation is usually attached the tag of equation of motion for their usefulness and convenience of solving a wide variety of numerical but they didn’t provide me the usefulness and convenience of 3rd equation of motion till now. Then I filed First appeal with genuine reason that CBSE didn’t provide me the true information. When I didn’t get information in the given time after First appeal then I filed Second appeal. In the response of my second appeal Central Information Commissioner organized a video conference with Ram Shankar on 08-11-2013 at 12:30PM.

I have prepared 8 minutes presentation to explain the deep facts & figures of my research with solution so that this process could be ended. But on 08-11-2013, in the video conference when I was trying to explain the matter then they cut the video conference within 4 minutes conversation, before asking me that have I any doubt regarding the reply of the experts of CBSE. In our conversation, Mr. Ram Shankar told me, “your RTI doesn’t comes under RTI act however we gave you the information as this was an educational issue also CBSE have given you the last and final information hence we can’t give you more information”. Look at their reply in the attachment, is information given by CBSE ok.

When both CBSE & PSEB didn’t provide me the true information then I hired an advocate Tejinderpal Singh from Chandigarh-Haryana High-court. My lawyer sent court notice to HRD ministry, UGC, CBSE, NCERT, COBOSE, Department of School education and Literacy, Department of higher education.

On 31-03-2014, HRD ministry ordered NCERT to provide suitable reply to me. But again NCERT provide me the improper information but they told me that if NCERT reply is not up to the mark then I may visit NCERT campus to solve this issue. Hence on 04-06-2014 my lawyer sent a letter to the advocate of NCERT with valid reason and requesting to arrange a meeting to solve this issue. But they didn’t reply at all.

On 05-08-2014, I have sent a requesting letter to arrange a meeting but they didn’t replied.

On 05-09-2014, I have sent an RTI requesting for information that please give me all the information what were the reasons that you couldn’t arrange the meeting and also provide me information of some dates (with time) on which I may visit NCERT for further discussions.

On 07-10-2014, First meeting fixed on mutually convenient day in the office of Head, DESM, Janaki Ammal Block, NCERT at 11:30M on 30.10.2014.

On 30-10-2014, In the meeting NCERT admitted equations of motion are five on the other side they flatly refused it and said that they could have published 2 equations or 5 equations but published 3 equations as “per our WISH”, “Hamare ghar mein kaun si daal banegi yeh aap thoda bataogey”.

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RTI Information given by CBSE and my Query

S.No. CBSE Experts My query

1. The two equations v = u + at and S = ut + ½ at2 are regarded as the two independent kinematical equations of motion because these can be directly derived using the basic definitions of velocity and acceleration.

I disagree that only these two equations [v = u + at and S = ut + ½ at2] are independent kinematical equations of motion and only these can be directly derived using the basic definitions of velocity and acceleration.

2. The third equation v2 - u2 = 2aS as does not quite meet the above criterion. However, it is usually attached the tag ‘third equation of motion’ because of its usefulness and convenience in solving a wide variety of useful problems.

I agree to your statement that it usually attached the tag of third equation of motion, so now stop attaching tag of third equation of motion to it. However I strongly believe that you’re SO CALLED helpful equation of motion in solving wide variety of problems is superfluous i.e. to say that first two equations of motions [v = u + at and S = ut + ½ at2] are more than enough to solve all kinds of problems and if not then send me that particular statement [problem] with an attachment, I can solve it without using third equation of motion [v2 - u2 = 2aS].

3. The expression S = [(u + v) ÷ 2] t, is just the mathematical form of the definition of average speed, and is, therefore, not really an independent equation of motion.

The whole equation S = [(u + v) ÷ 2]t is not the definition of average speed, it is the definition of displacement and is an equation of motion. The mathematical form of the definition of average speed is (u + v) ÷ 2.

E.g. The definition of force is F = m [(v - u) ÷ t], what you people are explaining is that this is not an equation of force but definition of acceleration instead. Now you yourself ponder over whether it is an equation of force or an equation of acceleration.

4. It is not derived using basic definitions and needs information about the initial velocity as well as the final velocity after a certain time.

It is derived using basic definitions i.e. derived from velocity-time graph and is independent kinematical equation of motion. It is very useful and convenience in solving a wide variety of useful problem. I have attached the documents. SO now is the time for CBSE to lead the world education and put that change by considering S = [(u + v) ÷ 2]t as equation of motion.

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TYPED RTI TO THE PRIME MINISTER



To The Public Information Officer/ Shri Pushpendra Kumar Sharma, Assistant Public Information Officer Under Secretary (RTI), PMO, South Block, New Delhi-110011. 1. Full Name of the Applicant : AMRITPAL SINGH NAFRIA 2. Father Name/Spouse Name : DARSHAN SINGH 3. Permanent Address : HOUSE NO. – 303, STREET NO. – 10, : PREM BASTI, SANGRUR-148001. 4. Correspondence Address : HOUSE NO. – 303, STREET NO. – 10, : PREM BASTI, SANGRUR-148001. 5. Particulars of the Information Solicited a) Subject Matter of Information : Status of my Letter b) The period to which information relates : 30 DAYS. c) Specific Details of Information required: On 20 December 2014 I sent a letter to honourable Prime Minister Mr. Narendra Modi, regarding implementation of my research on equation of motion that are being taught to the Ninth class students in all over India. Now I want to know all the information regarding the current status of my letter as well as action taken by your Honorable ministry on my Letter.

d) Whether information is required by Post or in : person (the actual postal fees shall be included : BY POST in additional fee in providing the information) e) In case by Post (ordinary/registered : REGISTERED POST or speed post) 6. Is this information not made available by public authority under voluntary disclosure? : YES. 7. Do you agree to pay the required fee? : YES. 8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 10 / (If Yes, Details of such deposit) 21F 443606 9. Whether belongs to below Poverty Line category? : NO. (If yes, you furnished the proof of the same with application?)

Place: SANGRUR. Date: 19-FEBRUARY-2015 Amritpal Singh Nafria

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TYPED RTI TO THE PRIME MINISTER



To The Public Information Officer/ Shri Vijay Kumar, Assistant Public Information Officer Room No. 229-C, Shastri Bhawan, C-wing Dr. Rajendra Prasad Road, New Delhi-110001. 1. Full Name of the Applicant : AMRITPAL SINGH NAFRIA 2. Father Name/Spouse Name : DARSHAN SINGH 3. Permanent Address : HOUSE NO. – 303, STREET NO. – 10, : PREM BASTI, SANGRUR-148001. 4. Correspondence Address : HOUSE NO. – 303, STREET NO. – 10, : PREM BASTI, SANGRUR-148001. 5. Particulars of the Information Solicited a) Subject Matter of Information : Status of my Letter b) The period to which information relates : 30 DAYS. c) Specific Details of Information required: On 20 December 2014 I sent a letter to honourable education minister Smt. Smriti Irani, regarding implementation of my research on equation of motion that are being taught to the Ninth class students in all over India. Now I want to know all the information regarding the current status of my letter as well as action taken by HRD ministry on my Letter.

d) Whether information is required by Post or in : person (the actual postal fees shall be included : BY POST in additional fee in providing the information) e) In case by Post (ordinary/registered : REGISTERED POST or speed post) 6. Is this information not made available by public authority under voluntary disclosure? : YES. 7. Do you agree to pay the required fee? : YES. 8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 10 / (If Yes, Details of such deposit) 21F 443605 9. Whether belongs to below Poverty Line category? : NO. (If yes, you furnished the proof of the same with application?)

Place: SANGRUR. Date: 19-FEBRUARY-2015 Amritpal Singh Nafria

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HRD REPLY TO MY LETTER - 1

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PMO REPLY TO MY RTI - 1

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HRD REPLY TO MY RTI

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TYPED RTI APPLICATION TO CBSE ACADEMIC UNIT



To The Public Information Officer/ The PIO, Academic Unit, C.B.S.E. Assistant Public Information Officer Shiksha Sadan, 17 Rouse Avenue, New Delhi-110002. 1. Full Name of the Applicant : AMRIT PAL SINGH NAFRIA. 2. Father Name/Spouse Name : DARSHAN SINGH NAFRIA. 3. Permanent Address : HOUSE NO. - 303, STREET NO. - 10, : PREM BASTI, SANGRUR-148001. 4. Correspondence Address : HOUSE NO. - 303, STREET NO. – 10, : PREM BASTI, SANGRUR-148001. 5. Particulars of the Information Solicited a) Subject Matter of Information : EQUATIONS OF MOTION. b) The period to which information relates : 30 DAYS. c) Specific Details of Information required: On 11.03.2013, I received your letter in response to my RTI application. In your reply it was mentioned that third equation is usually attached the tag of ‘third equation of motion’ because of its usefulness and convenience in solving a wide variety of useful problems. So, I am humbly requesting you to provide me all the information regarding usefulness and convenience of third equation of motion in solving a wide variety of useful problems.

d) Whether information is required by Post or in : person (the actual postal fees shall be included : BY POST in additional fee in providing the information) e) In case by Post (ordinary/registered : REGISTERED POST or speed post) 6. Is this information not made available by public authority under voluntary disclosure? : YES. 7. Do you agree to pay the required fee? : YES. 8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 10/ (If Yes, Details of such deposit) : 21F/ 443607 9. Whether belongs to below Poverty Line category? : NO. (If yes, you furnished the proof of the same with application?)

Place: Sangrur. Date: 17-April-2015 Signature of Applicant

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MY LETTER TO P.M.O. APPELLATE AUTHORITY

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NCERT LETTER TO ME

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NCERT LETTER TO ME

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NCERT LETTER TO ME

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NCERT LETTER TO ME

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MY APPEAL TO APPELLATE AUTHORITY

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DOCUMENTS ATTACHED WITH APPEAL - 1

POINTS THAT WERE NOT CLEARED BY NCERT IN THE MEETIN G

In the meeting they agreed that total equations of motion are five, also on various websites it was clearly mentioned that total equations of motion are five, some website links were attached with this letter.

Basically, only two equations of motion are enough to solve all the numerical problems of equations of motion, then why were four equations of motion published and considered three equations as equations of motion? Is it right to teach students?

The only difference in Second and third equation of motion is that equation ‘S = ut + ½ at2’ is very useful to calculate distance covered when final velocity (v) is not given and equation ‘S = (v2 – u2) ÷ 2a’ is very useful to calculate distance covered when time (t) is not given. I am saying so because equations S = ½ (u + v) t & S = vt – ½ at2 have same specialty, i.e. use equation S = ½ (u + v) t to calculate displacement when acceleration (a) is not given and use equation S = vt – ½ at2 when initial velocity (u) is not given after that these two equation were not considered as equations of motion.

Third equation of motion is not directly derived from velocity-time graph also they didn’t clear what is the extra importance of third equation of motion on which it was considered as third equation of motion and rest of the two equations are not in equation S = ½ (u + v) t & S = vt – ½ at2?

We have only two option and these are: -

1. If you want to teach all the equations of motion then teach all the five equations to students.

2. If you want to teach sufficient equations of motion then teach only two equations of motion, i.e.

All the problems and numerical regarding equations of motion can be solved with only two equations, there is no need of third equation of motion. Choose only one equation from both A and B pool;

Rest of the three equations of motion can be used to solve numerical easily in different situations. For e.g. use Equation ‘S = ut + ½ at2’ to calculate displacement very easily when final velocity is not given, use Equation ‘S = (v2 – u2) ÷ 2a’ calculate displacement very easily when time is not given, use equation S = vt – ½ at2 calculate displacement very easily when initial velocity is not given and use Equation S = ½ (u + v) t calculate displacement very easily when acceleration is not given.

Pool-A Pool-B

a = (v –u) ÷ t S = ½ (u + v) t

------------------ S = ut + ½ at2

------------------ S = (v2 – u2) ÷ 2a

------------------ S = vt – ½ at2

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http://en.wikipedia.org/wiki/Equations_of_motion

http://physicsforidiots.com/physics/dynamics/

http://www.thestudentroom.co.uk/wiki/Revision:Kinematics_-

_Equations_of_Motion_for_Constant_Acceleration

‘MATHEMATICS FOR ENGINEERS’ book by Stephen Lee, Page No. – 29, web-link: -https://books.google.co.in/books?id=x0J9BgAAQBAJ&pg=PA29&lpg=PA29&dq=s+%3D+vt+-

+1/2+at2&source=bl&ots=1GyID_g1i8&sig=BtNfsUWz2UfPXc9O6yg_oW8igG8&hl=en&sa=X&ei=Tu5mVY

jgD9CxuATzp4CYCQ&ved=0CGMQ6AEwCA

‘ENGG MECHANICS: STAT & DYN’ book by A. Nelson, Page 12.18, web-link: -https://books.google.co.in/books?id=6yWf4HOTm10C&pg=SA9-PA138&lpg=SA9-PA138&dq=s+%3D+vt+-+1/2+at2&source=bl&ots=UxjMyVsE22&sig=GcWWMd5WZnxAVA0X-kof07OLlug&hl=en&sa=X&ei=kfFmVZjNGpTkuQSquYGQCA&ved=0CBwQ6AEwADgK#v=onepage&q=s%20%3D%20vt%20-%201%2F2%20at2&f=false

https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=12&cad=rja&uact=8&ved=0CC

MQFjABOAo&url=http%3A%2F%2Fvle.clystvale.org%2Fmod%2Fresource%2Fview.php%3Fid%3D6230&e

i=CvpmVbzhDtjkuQSyw4GQBg&usg=AFQjCNEkRnaAcFIyV1DkTaW7IlBXPcPKSg&sig2=gGTBVOOzYMXyVRj

SMAnLjw&bvm=bv.93990622,d.c2E

https://www.physicsforums.com/threads/problem-on-one-dimentional-motion-bullet-going-through-a-

board.259293/

Time and again, I have warned many those websites which were publishing my equation S = vt – ½ at2 on their websites, about my copyrights and consequently some of them had removed the whole webpage given below.

https://www.lakeheadschools.ca/scvi_staff/brecka/Gr11_physics_web/downloadable_content/unit1/text1/phys11_1_5.pdf

http://www.globalshiksha.com/content/all-physics-formulas-for-10th-grade

http://instruct.tri-c.edu/fgram/web/linear.htm

http://theoreticalphysics.net/Mechanics.htm

http://www.thestudentroom.co.uk/showthread.php?t=15559250

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OTHER DOCUMENTS

Whole Procedure of Equations of motion.

Science

Transcript of Whole Procedure of Equations of motion.