Question

Derive : first equation of motion Second
equation of motionThird equation of motion​

Answer 1

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Answer:

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Explanation:

Derivation of First Equation of Motion

The first equation of motion is:

v = u + at

Derivation of First Equation of Motion by Algebraic Method

It is known that the acceleration (a) of the body is defined as the rate of change of velocity.

So, the acceleration can be written as:

a = v − ut

From this, rearranging the terms, the first equation of motion is obtained, which is:

v = u + at

Derivation of First Equation of Motion by Graphical Method

Consider the diagram of the velocity-time graph of a body below:

REFER TO THE ATTACHMENT☺

Derivation of Second Equation of Motion by Algebraic Method

Consider the same notations for the derivation of the second equation of motion by simple algebraic method

Derivation of Second Equation of Motion by Graphical Method

Taking the same diagram used in first law derivation:

REFER TO THE ATTACHMENT☺

Derivation of Third Equation of Motion

The third equation of motion is:

v2 = u2 + 2aS

Derivation of Third Equation of Motion by Algebraic Method

REFER TO THE ATTACHMENT☺

Derivation of Third Equation of Motion by Graphical Method

REFER TO THE ATTACHMENT☺

The total distance travelled, S = Area of trapezium OABC.

So, S= 1/2(SumofParallelSides)×Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= 1/2(u+v)×t

Now, since t = (v – u)/ a

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Answer 2

$\huge \underline \pink{Answer }$

✰ First equation of motion :-

$\bf \red{ \frac{Acceleration \: = \: Change \: in \: velocity \: }{time} }$

$\bf \pink = \pink > \pink a \: \pink= \: \pink \: \frac \pink{v \: - \: u \: }{ \pink t \: }$

$\bf \purple \therefore \displaystyle { \boxed{ \sf { \displaystyle { \purple v \: \purple = \purple u \: \purple + \: \purple a \purple t}}}}$

✰ First Second of motion :-

For uniformly accelerated motion,

$\bf \purple{V_av} \: \purple = \frac{ \purple u \purple+ \purple v}{ \purple2}$

Also, Displacement = (Average velocity) × (Time)

$\bf \green \therefore \: \green s \: \green= \: \green( \frac{ \green u \green+ \green v}{ \green2} \green ) \green \times \green t$

$\bf \orange = \orange > \orange s \: \orange = \: \orange(\frac{ \orange u \orange + \orange a \orange t}{ \orange 2} \orange) \orange t$

$\large \displaystyle { \boxed{ \sf { \displaystyle { \pink s \pink = \pink u \pink \pink \pink + \frac{ \pink1}{ \pink 2} \pink a \pink t \pink ²}}}}$

✰ First Second of motion :-

$\bf \red {F \purple r \pink o\green m } \ (1), \red t \: \green = \frac{ \red v \: \green - \pink u \: }{ \orange a}$

Substituting the value of t in the second equation of motion,

$\bf = > s \: = u( \frac{v - u}{a} ) \: + \: \frac{ 1}{2} a \: ( \frac{v - u}{a})^{2}$

$\bf s \: = > \frac{uv - u ^{2} \: }{a} + \frac{1}{2a} + \frac{ 1}{2} \: ( v^{2} + \: {u}^{2} - 2av \: )$

$\bf = > \frac{2uv - {2u}^{2} + {v}^{2} - 2uv }{2a} = \: \frac{ {v}^{2} \: - {u}^{2} }{2a}$

$\bf \red \therefore \displaystyle { \boxed{ \sf { \displaystyle { \red v ^{ \red2} \: \red = \red {u}^{ \red2} \: \red + \: \red 2 \red a \red s}}}}$

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Thus, the three equations of motion are :

1) v = u + at ‎‎ ‎ this equation does not have s

2) s = ut + $\frac{1}{2}$ at², this equation does not have v

3) v² - u² = 2as this equation does not have t.

The three equations have four physical quantities each.

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