Discussion & Derivation of Singh's fourth equation of motion.

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

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

Distance travelled ‘S’ = area of the trapezium ABDO

S = area of the rectangle OEBD – area of the triangle ABE

S = (OD × OE) – ½ (AE × BE)

S = (t × v) – ½ [(v – u) × t]

S = vt – ½ [(v – u) × t]

From first equation of motion, v – u = at

S = vt – ½ at × t

S = vt - ½ at² -------- (4)

Equation (4) represents Singh’s fourth equation of motion & is used to calculate distance when initial velocity(u) is not given.


This equation of motion is copyrights on my name by the deputy registrar of copyrights office, New Delhi. Lastly we want that this equation should also be published in all the books with other three equations of motion and student can read it as fourth equation of motion, approx. 90% student don't know about this equation, after publishing this equation with other three equations,all physics student should know completely about equations of motion. if it is not equation of motion then tell why? We answer you.

below are Newton's three equations of motion & importance of Singh's fourth equation of motion.

Derivation of Newton’s first equation of motion from velocity-time graph.

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

a = BC / AC = (BD - CD) / OD .

a = (v – u) ÷ t --- (1)

Equation (1), represent Newton’s first equation of motion.


Derivation of Newton’s second equation of motion from velocity-time graph.

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

Distance travelled ‘S’ = area of the trapezium ABDO

S = area of the rectangle ACDO + area of the triangle ABC

S = (OD × OA) + ½ (AC × BC)

S = (t × u) + ½ [(v – u) × t]

S = ut + ½ [(v – u) × t]

From first equation of motion, v – u = at

S = ut + ½ at × t

S = ut + ½ at²
---(2)
Equation (2) represents Newton’s second equation of motion & is used to calculate distance when final velocity(v) is not given.



Derivation of Newton’s third equation of motion from velocity-time graph.

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

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--- (3)
Equation (3) represents Newton’s third equation of motion & is used to calculate distance when time(t) is not given.


Question: - what is new in Singh’s fourth equation of motion?

Answer:- Till today, no one finds an equation which calculates distance when initial velocity is not given. Singh’s fourth equation of motion is the only equation of motion which calculates distance, when initial velocity is not given.

Question: - All the problems regarding equations of motion are solved by using three equations of motion so, why your equation of motion is consider as equation of motion?

Answer: - Dear Sir, firstly I clear that all the problems regarding equations of motion are solved by using two equations of motion; from Newton’s first and second equations of motion or from Newton’s first and third equations of motion (or from Newton’s first and Singh’s fourth equations of motion), here we consider three equations as equations of motion, instead of two equations because: -

Firstly after deep calculations and many derivations only three different equations are found from the velocity-time graph for a body moving with a constant acceleration, at the time when these are published.

Secondly after considered all the three equations as equation of motion, all people should know completely about equations of motion.

Thirdly however both Newton’s second & third equation of motion calculate distance, but second equation is used to calculate distance when final velocity is not given and third equation is used when time is not given.

Fourthly from my point of view like Newton’s three laws of motion, it considers three equations as equations of motion.

Singh’s fourth equation of motion is consider as equation of motion because this matter is not similar matter like planet Pluto, that many other equations are derived after considering Singh’s equation as fourth equation of motion, I tell you here that this the only equation of motion which calculates distance when initial velocity is not given. If anyone find another equation of motion which calculates distance when initial velocity is not given, then we have no objection that why Singh’s fourth equation of motion is not consider as equation of motion.



Improtance

A travelling car is suddenly stop in 5 seconds with an retardation of 23 m/s2, here final velocity is 0 m/s (because the car is finally at rest),

Traditional method to solve the above problem,

From Newton’s second equation of motion

S = ut + ½ at²

S = [(u × 5 sec) + (½ × 23 × 5 × 5)meters]

S = [(u × 5 sec) + (287.5 meters)]

S = [(u × 5 sec) + (287.5 meters)] ------------- (6)


From Newton’s third equation of motion

S = (v² − u²) / 2a

S = [(0 m/s) 2 − u²] ÷ (2 × 23 m/s²)

S = [(0 m/s) 2 − u²] ÷ (46 m/s²)

S = − [u²] ÷ [46 m/s²] ------------- (7)

Here we are not able to find exact value of distance in numbers when initial velocity is not given.

Therefore, we can use Newton’s first equation of motion,

u = v − at

u = [(0 m/s) − (23 m/s² × 5 sec)]

Initial velocity = − 115 m/s

Substituting the value of ‘u’ in equation (6) and (7), we get

Equation (6),

S = [5u + 287.5] meters

S = [(5 sec × −115 m/s) + (287.5 meters)]

S = (− 575 meters + 287.5 meters)

S = − 287.5 meters

Here negative sign shows retardation.


Equation (7),

S = − (u² ÷ 46 m/s²)

S = − [(115 m/s × 115 m/s) ÷ 46 m/s²]

S = − (13225 m2/s2 ÷ 46 m/s²)

S = − 287.5 meters

Here negative sign shows retardation.


Solution From Singh’s fourth equation of motion

S = vt - ½ at²

S = [0 × 5] - [½ × 23 × (5 × 5)] meters

S = [0 meters] - [½ × 23 × 25] meters

S = [0 meters] - [575 meters ÷ 2]

S = - 287.5 meters ------------- (5)

Here negative sign shows retardation.


Usually we go with three equations of motion which are actually four in number this fact is confusing for some Prof. & students. They cannot understand the difference between laws of motion & equations of motion. So if we consider Singh’s equation as a fourth equation of motion the total number of equations of motion will be four & therefore a great confusion will be ended. This will help to make easier concepts in the mind of a physics student.

1. Singh’s fourth equation of motion makes easy the numerical calculations, Because we finds distance by one equation of motion without knowing final velocity or time or acceleration but no one finds an equation which calculates distance when initial velocity is not given. Singh’s fourth equation of motion, is the only equation of motion which calculates distance, when initial velocity is not given.

2. Some students and prof. have some doubt that on the basis of Newton's three laws we get three equations of motion so it also removes confusion between Laws of motion & equations of motion.

3. It is easy to learn, because both the Newton’s second equation & Singh’s fourth equation of motion are looks like same.

S = ut + ½ at²

S = vt – ½ at²