Whole Procedure of Equations of motion.

1
KINEMATICKINEMATICKINEMATICKINEMATIC
EQUATIONSEQUATIONSEQUATIONSEQUATIONS
OF MOTIONOF MOTIONOF MOTIONOF MOTION
By: - Amritpal Singh Nafria
+917814080880 & +918559012321

2
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

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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

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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.
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).
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).
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
+ a2
t 2
+ 2uat – u2
) ÷ 2a
S = (u2
– u2
+ a2
t 2
+ 2uat) ÷ 2a
S = (a2
t 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
+ a2
t 2
– 2vat)] ÷ 2a
S = [v 2
– v2
– a2
t 2
+ 2vat)] ÷ 2a
S = [– a2
t 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

15
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

16
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

18
REVIEWS FROM VARIOUS INSTITUTES

19
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,
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

20
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
+ a2
t 2
+ 2uat – u2
) ÷ 2a
S = (a2
t 2
+ 2uat) ÷ 2a
S = ½ at 2
+ ut
S = ut + ½ at 2
S = (v2
– u2
) ÷ 2a and S = ut +½ at2
are same?

21
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
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
S = ½ (v + u) × t and S = ut +½ at2
are same or different?

22
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?

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
- 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.

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 velocity
in figure. It is evident from the graph that:
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
31
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:
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 time
final 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
of time is not present in this equation. We say, therefore, that the equation is
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)
of time is not present in this equation. We say, therefore, that the equation is time independent
taken by the body but still have to work with the initial velocity,

33
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.

34
MY PROPOSAL LETTER SENT TO CBSE

37
CBSE REPLY TO MY SECOND RTI

38
RESEARCH PAPER PUBLISHED IN IJERD - 1

39

40

41
CIC LETTER TO MY SECOND APPEAL

42
RESEARCH SENT TO IIT ROPAR THROUGH G-MAIL
RESEARCH SENT TO IOP THROUGH G-MAIL

43
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

59
COBOSE REPLY TO MY LEGAL NOTICE

62
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.

64
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.

65
HRD REPLY TO MY LEGAL NOTICE

66
NCERT LAWYER REPLY TO MY LEGAL NOTICE - 1

67
NCERT LAWYER REPLY TO MY LEGAL NOTICE - 2

68
MY LAWYER REQUESTING LETTER TO NCERT LAWYER

70
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.

MEETING FIXED THROUGH E
72
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).

76
LETTER TO THE PRIME 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: dukyalways4u@gmail.com.
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.

79
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”.

80
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.

81
TYPED RTI TO THE PRIME MINISTER
FORM ‘A’
See Rule 3(1)
I. D. No……………..
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
3. Permanent Address : HOUSE NO. – 303, STREET NO. – 10,
4. Correspondence Address : HOUSE NO. – 303, STREET NO. – 10,
a) Subject Matter of Information : Status of my Letter
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.
or speed post)
8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 10 /
(If Yes, Details of such deposit) 21F 443606
application?)
Place: SANGRUR.
Date: 19-FEBRUARY-2015 Amritpal Singh Nafria

82
TYPED RTI TO THE PRIME MINISTER
FORM ‘A’
See Rule 3(1)
I. D. No……………..
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
3. Permanent Address : HOUSE NO. – 303, STREET NO. – 10,
4. Correspondence Address : HOUSE NO. – 303, STREET NO. – 10,
a) Subject Matter of Information : Status of my Letter
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.
or speed post)
8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 10 /
(If Yes, Details of such deposit) 21F 443605
application?)
Place: SANGRUR.
Date: 19-FEBRUARY-2015 Amritpal Singh Nafria

90
TYPED RTI APPLICATION TO CBSE ACADEMIC UNIT
FORM ‘A’
See Rule 3(1)
I. D. No……………..
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,
4. Correspondence Address : HOUSE NO. - 303, STREET NO. – 10,
a) Subject Matter of Information : EQUATIONS OF MOTION.
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.
or speed post)
8. Have you deposited application fee? : INDIAN POSTAL ORDER / Rs. 10/
(If Yes, Details of such deposit) : 21F/ 443607
application?)
Place: Sangrur.
Date: 17-April-2015 Signature of Applicant

91
MY LETTER TO P.M.O. APPELLATE AUTHORITY

97
MY APPEAL TO APPELLATE AUTHORITY

98
DOCUMENTS ATTACHED WITH APPEAL - 1
POINTS THAT WERE NOT CLEARED BY NCERT IN THE MEETING
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

99
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=0CBwQ6A
EwADgK#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

100

Whole Procedure of Equations of motion.

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