Physics, asked by AssassinsCreed, 3 months ago

Explain 3 equations of Motions using Calculus Method..
With detailed explanation

Answers

Answered by rudrasethi17
0

Explanation:

There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a). The following are the three equation of motion: First Equation of Motion : v=u+at. ... Third Equation of Motion : v^2=u^2+2as.

Answered by BrainlyLegend2108
25

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Let us Explain the 3 equations of motions using integration method.

\huge \tt\underline{\underline{1st\: Equation\: of \:Motion :}} \green \checkmark

Let the small change in velocity be dv.

Since acceleration is uniform,

➠ a = \dfrac{dv}{dt}

➠ dv = adt

Let us apply Integration on both sides

\huge\sf\red{ \int\limits^v_u  dv } \: =  \int\limits^t_0 {a} \, dt

as "a" is constant acceleration

\huge\sf\red{\int\limits^v_u  dx } \: = a \int \limits^t_0  dt

\implies  v -u \: = a(t-0)

Hence 1st equation of motion is given as

\huge\star \sf{\underbrace\green{v\: = u+at }}

\huge \tt\underline{\underline{2nd\: Equation\: of \:Motion :}} \green \checkmark

dx = vdt

Let us apply integration on both sides

\huge\bf\ \blue{\int\limits^S_0  dx } \: = \int\ vdt

v can substituted as u+at

\huge\bf\ \blue{s\:= \int\ {(u+at)} \,. dt }

s \:  = \int\limits^t_0 {u} \, dt + \int\limits^t_0 {at} \, dt

s \:  = u\int\limits^t_0  dt + a\int\limits^t_0 {t} \, dt

s \:  = u(t-0) + \dfrac{at^{2} }{2}

\huge\blue{ \boxed{S\: = ut+\dfrac{1}{2}at^{2}  }}

\huge \tt\underline{\underline{3rd\: Equation\: of \:Motion :}} \green \checkmark

As we already know that

a =\dfrac{dv}{dt}

\implies a\: = \dfrac{dv}{dt} \times \dfrac{dx}{dx}

\implies a\: = \dfrac{dv}{dx} \times \dfrac{dx}{dt}

\implies a\: = \dfrac{dv}{dx} \times v

\implies a\: = \dfrac{vdv}{dx}

\implies adx\: = {vdv}

Apply integral limits

\huge\bf\ \blue{\int\limits^S_0  adx } \: = \int\limits^v_0 vdv

as\, = \dfrac{v^{2} -u^{2} }{2}

Hence 3rd equation of motion is given as,

\huge \boxed{\boxed {\red{v^{2}-u^{2}\: = 2as  }}}

Hence proved

Hope\: It\: Helps

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