Derivation of Newton’s Equation

Kinematic Equations for uniformly accelerated motion along straight line

(i) Velocity time equation – {\rm{\upsilon }} = {\rm{u}} + {\rm{at}}

(ii) Displacement time equation – \displaystyle \text{S}=\text{ut}+\frac{1}{2}\text{a}{{\text{t}}^{2}}

(iii)Velocity displacement equation – \displaystyle {{\text{ }\!\!\upsilon\!\!\text{ }}^{\text{2}}}={{\text{u}}^{\text{2}}}+\text{2aS}

(iv)Displacement in nth second – \displaystyle {{\text{S}}_{\text{nth}}}=\text{u}+\frac{\text{a}}{\text{2}}\left( 2\text{n}-1 \right)

Consider a body of mass m having initial velocity u and after time t its final velocity becomes due to uniform acceleration.

We know that

\displaystyle \text{Acceleration}\,\text{=}\,\frac{\text{Change in velocity}}{\text{time taken}}

Let the distance travel by a body be “S”

\displaystyle \begin{array}{l}\text{Distance = Average velocity }\!\!\times\!\!\text{ time}\\\text{S}\,\,=\,\,\frac{\text{u}\,\,+\,\,\text{ }\!\!\upsilon\!\!\text{ }}{2}\times \text{t}\end{array} \displaystyle \text{a}\,\,\text{=}\,\,\frac{\text{ }\!\!\upsilon\!\!\text{ }-\text{u}}{\text{t}-\text{o}}

Substituting \displaystyle \text{ }\!\!\upsilon\!\!\text{ }=\text{u}\,+\text{at} from (i) equation

\displaystyle \begin{array}{l}2\text{S}=\left( \text{u}+\text{u}+\text{at} \right)\,\text{t}\\\text{2S}\,\,=\,2\text{ut}\,\,+\,\text{a}{{\text{t}}^{\text{2}}}\\\text{S}=\,\text{ut}\,\,+\,\,\frac{\text{1}}{\text{2}}\,\text{a}{{\text{t}}^{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\_\_\text{II}\,\text{equation}\end{array}

(ii) We know \upsilon  = {\rm{u}} + {\rm{at}}

\displaystyle \begin{array}{l}\text{ }\!\!\upsilon\!\!\text{ }-\text{u}\,\,=\,\text{at}\\\text{t}\,\,\text{=}\,\,\frac{\text{ }\!\!\upsilon\!\!\text{ }-\text{u}}{\text{a}}\,\,\,\,\,\,\,\,\,\_\_\_\_\_\_\_\_\_\left( \text{a} \right)\end{array}

Again \displaystyle \text{S}=\,\,\left( \frac{\text{ }\!\!\upsilon\!\!\text{ }+\text{u}}{2} \right)\,\,\text{t} \displaystyle \text{S}\,\,\,=\,\,\,\left( \frac{\text{ }\!\!\upsilon\!\!\text{ }+\text{u}}{2} \right)\,\,\left( \frac{\text{ }\!\!\upsilon\!\!\text{ }-\text{u}}{\text{a}} \right) \displaystyle \begin{array}{l}\text{2}\,\,\text{a}\,\,\text{S}\,\,\,=\,\,{{\text{ }\!\!\upsilon\!\!\text{ }}^{\text{2}}}\,\,-\,\,{{\text{u}}^{\text{2}}}\\{{\text{ }\!\!\upsilon\!\!\text{ }}^{2}}\,\,\,=\,\,{{\text{u}}^{\text{2}}}\,\,\,+\,\,2\text{aS}\,\_\_\_\_\_\_\left( \text{III} \right)\,\text{equation}\end{array}

Let \displaystyle {{\text{S}}_{\text{n}}} be the displacement of the body in second

Let \displaystyle {{\text{S}}_{\text{n}-1}} be the displacement in \displaystyle \left( \text{n}-1 \right) second

we know that

\displaystyle \begin{array}{l}\text{S}=\,\,\text{ut}\,\,+\,\,\frac{1}{2}\,\,\text{a}\,{{\text{t}}^{\text{2}}}\\{{\text{S}}_{\text{n}}}\,\,\,=\,\,\,\text{un}\,\,\,+\,\,\frac{1}{2}\,\,\text{a}\,{{\text{n}}^{\text{2}}}\\{{\text{S}}_{\text{n}-1}}\,=\,\,\text{u}\left( \text{n}-1 \right)+\frac{1}{2}\,\,\text{a}\,{{\left( \text{n}-1 \right)}^{2}}\end{array}

Displacement during nth second

\displaystyle \begin{array}{l}{{\text{S}}_{\text{nth}}}\,\,\,\,=\,\,{{\text{S}}_{n}}-{{\text{S}}_{\text{n}-1}}\\{{\text{S}}_{\text{nth}}}\,\,=\,\,\left( \text{un}\,\,+\frac{1}{2}\,\,\text{a}{{\text{n}}^{\text{2}}} \right)-\left[ \text{u}\left( \text{n}-1 \right)+\frac{1}{2}\text{a}{{\left( \text{n}-1 \right)}^{2}} \right]\\{{\text{S}}_{\text{nth}}}=\,\,\text{u}\left[ \text{n}-\left( \text{n}-1 \right) \right]+\frac{a}{2}\left( {{\text{n}}^{2}}-{{\left( \text{n}-1 \right)}^{2}} \right)\\{{\text{S}}_{\text{nth}}}\,\,=\,\,\text{u}\,\,\,+\,\,\frac{\text{a}}{\text{2}}\,\,\left( 2\text{n}-1 \right)\end{array}

Derivation of equation for uniformly accelerated motion by calculus method.

Consider an object having mass m initial velocity u at time t = 0 and final velocity \displaystyle \text{ }\!\!\upsilon\!\!\text{ } at time due to uniform acceleration a.

If dv is small change in velocity in small time Interval dt

\displaystyle \begin{array}{l}\text{acceleration }\,\,\text{=}\,\,\frac{\text{Change in velocity}}{\text{time interval}}\\\text{a}=\,\,\frac{\text{dv}}{\text{dt}}\\\text{d }\!\!\upsilon\!\!\text{ }\,\,\text{=}\,\,\text{a}\,\,\text{dt}\end{array}

Integrate and take limit both side

\displaystyle \begin{array}{l}\int_{\text{u}}^{\text{ }\!\!\upsilon\!\!\text{ }}{\text{dv}}\,\,=\,\,\,\,\int_{\text{o}}^{\text{t}}{\text{a}}\,\,\text{dt}\\\left[ \text{ }\!\!\upsilon\!\!\text{ } \right]_{\text{u}}^{\text{ }\!\!\upsilon\!\!\text{ }}\,\,\,=\,\,\text{a}\left[ \text{t} \right]_{\text{o}}^{\text{t}}\\\text{ }\!\!\upsilon\!\!\text{ }-\text{u}\,\,=\,\,\text{a}\left( \text{t}-\text{o} \right)\\\text{ }\!\!\upsilon\!\!\text{ }\,\,\,\text{-}\,\,\text{u}\,\,\text{=}\,\,\text{at}\end{array}

Let ds be small displacement in small Interval dt

\displaystyle \text{ }\!\!\upsilon\!\!\text{ }\,\,=\,\,\frac{\text{ds}}{\text{dt}}\,\,;\,\,\,\,\,\,\text{ds}\,\,\,=\,\,\text{v}\,\,\text{dt}

But \displaystyle \text{ }\!\!\upsilon\!\!\text{ }\,\,\,=\,\,\text{u}\,\,\,+\,\,\text{at} \displaystyle \text{ds}\,\,=\,\,\left( \text{u}\,\,+\text{at} \right)\,\,\text{dt}

 

Taking limit both side

\displaystyle \begin{array}{l}\int_{\text{o}}^{\text{S}}{\text{ds}}\,\,\,\,=\,\,\,\,\int_{0}^{\text{t}}{\left( \text{u}+\text{at} \right)}\,\text{dt}\\\left[ \text{S} \right]_{\text{o}}^{\text{S}}\,\,=\,\,\,\int_{\text{o}}^{\text{t}}{\text{u}\,\,\text{dt}}\,\,\,+\,\,\,\,\int_{\text{o}}^{\text{t}}{\text{at}\,\,\text{dt}}\\\text{S}-\text{o}\,\,\text{=}\,\,\text{u}\left[ \text{t} \right]_{\text{0}}^{\text{t}}\,\,+\,\,\text{a}\left[ \frac{{{\text{t}}^{\text{2}}}}{\text{2}} \right]_{\text{o}}^{\text{t}}\\\text{S}=\,\,\text{ut}\,\,+\,\,\frac{1}{2}\text{a}\,{{\text{t}}^{\text{2}}}\end{array}

 

Velocity displacement equation

We know

\displaystyle \begin{array}{l}\text{a}=\frac{dv}{dt}\\\text{a}=\,\,\frac{dv}{ds}\cdot \,\frac{ds}{dt}\\\text{a}\,\,=\,\,\text{v}\frac{\text{dv}}{\text{ds}}\\\text{a}\,\,\text{ds}\,\,=\,\,\text{v}\,\,\text{dv}\end{array}

Integrating both side with limit

\displaystyle \begin{array}{l}\int_{u}^{v}{vdv=\int_{0}^{s}{ads}}\\\left| \frac{{{\upsilon }^{2}}}{2} \right|_{u}^{\upsilon }=a\left| S \right|_{0}^{S}\\\frac{{{\upsilon }^{2}}}{2}-\frac{{{u}^{2}}}{2}=a\left( S-0 \right)\\{{\upsilon }^{2}}-{{u}^{2}}=2aS\end{array}

Displacement in nth second

\displaystyle \begin{array}{l}\text{ }\!\!\upsilon\!\!\text{ }\,\,\,=\,\,\frac{\text{ds}}{\text{dt}}\,\,\,\text{or}\,\,\,\text{ds}\,\,\,=\,\text{ }\!\!\upsilon\!\!\text{ }\,\,\,\text{dt}\\\text{ds}\,\,\text{=}\,\,\left( \text{u}\,+\text{at} \right)\,\,\text{dt}\\\text{ds}\,\,=\,\,\text{u}\,\,\text{dt}\,\,+\,\,\text{at}\,\,\,\text{dt}\end{array}

If \displaystyle \text{t}\,=\,\text{n}-1 then \displaystyle \text{S}={{\text{S}}_{\text{n}-1}}

and if \displaystyle \text{t}=\text{n}, then \displaystyle \text{S}={{\text{S}}_{\text{n}}}

Integrating both side

\displaystyle \begin{array}{l}\int_{{{\text{S}}_{\text{n}-1}}}^{{{\text{S}}_{\text{n}}}}{\text{ds}}\,\,=\,\int_{\text{n}-1}^{\text{n}}{\text{u}\,\,\text{dt}}\,\,\,+\,\,\int_{\text{n}-1}^{\text{n}}{\text{at}\,\,\,\text{dt}}\\\left[ \text{S} \right]_{{{\text{S}}_{\text{n-1}}}}^{{{\text{S}}_{\text{n}}}}=\,\,\text{u}\left[ \text{t} \right]_{\text{n-1}}^{\text{n}}\,\,+\,\frac{\text{a}}{\text{2}}\left[ {{\text{t}}^{\text{2}}} \right]_{\text{n}-1}^{\text{n}}\\{{\text{S}}_{\text{n}}}-{{\text{S}}_{\text{n}-1}}\,\,=\,\,\text{u}\,\left[ \text{n}-\left( \text{n}-1 \right) \right]+\frac{\text{a}}{\text{2}}\left[ {{\text{n}}^{\text{2}}}-{{\left( \text{n}-1 \right)}^{2}} \right]\\{{\text{S}}_{\text{nth}}}\,\,=\,\,\text{u}\,\,+\,\frac{\text{a}}{\text{2}}\left( \text{2n}-1 \right)\end{array}

Post Author: E-Physics

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