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How can we derive three equation of motion graphically

Answer» First Equation:Acceleration is defined as the rate of change of velocity.Let, V = final velocity;\xa0Vo\xa0= initial velocity, T= time, a =acceleration.By definition of acceleration:{tex}a = \\frac{{V - {V_o}}}{T}{/tex}{tex}at = V - {V_o}{/tex}{tex}V = {V_o} + {\\rm{ }}at{/tex}Since,\xa0Vo\xa0=u = initial velocitytherefore\xa0V = u +atSecond Equation:Let at time T=0, body moves with initial velocity u and at time ‘t’ body has final velocity ‘v’ and in time ‘t’ it covers a distance \'S\'.AC = v, AB = u, OA = t, DB = OA = t, BC = AC - AB = v - uArea under a v-t curve gives displacement so,S= Area of triangle DBC + Area of rectangle OABD .......(i)Area of {tex}\\Delta DBC = \\frac{1}{2} \\times Base \\times Height{/tex}=\xa0{tex}\\frac{1}{2} \\times DB \\times BC{/tex}=\xa0{tex}\\frac{1}{2} \\times t \\times \\left( {v - u} \\right){/tex} ......... (ii)Area of rectangle OABD\xa0{tex}= length \\times breadth{/tex}=\xa0{tex}OA \\times BA{/tex}=\xa0{tex}t \\times u{/tex} ......... (iii)From (i), (ii) and (iii)S= ut+\xa0{tex}\\frac{1}{2} \\times t \\times \\left( {v - u} \\right){/tex}S= ut +\xa0{tex}\\frac{1}{2} \\times t \\times at{/tex}S= ut +\xa0{tex}\\frac{1}{2}a{t^2}{/tex}\u200b\u200b\u200b\u200b\u200b\u200b\u200bThird Equation:\u200b\u200b\u200b\u200b\u200b\u200b\u200bLet at time t=0, the body moves with an initial velocity u and time at ‘t’ has final velocity ‘v’ and in time ‘t’ covers a distance ‘s’Area under v-t graph gives the displacementS = Area of {tex}\\Delta{/tex}DBC + Area of rectangle OABDS = {tex}\\frac{1}{2} \\times base \\times height + length \\times breadth{/tex}{tex}S = \\frac{1}{2} \\times DB \\times BC + OA \\times AB{/tex}{tex}S = \\frac{1}{2} \\times t \\times (v - u) + t \\times u{/tex}\xa0........(i)Now, v - u = at\xa0{tex}\\frac{{v - u}}{a} = t{/tex}put the value of \'t\' in equation (i){tex}S = \\frac{1}{2} \\times (v - u)\\frac{{(v - u)}}{a} + u \\times \\left( {\\frac{{(v - u)}}{a}} \\right){/tex}{tex}S = \\frac{{{{(v - u)}^2}2u(v - u)}}{{2a}}{/tex}{tex}S = \\frac{{{v^2} + {u^2} - 2uv + 2uv - 2{u^2}}}{{2a}}{/tex}{tex}S = \\frac{{{v^2} - {u^2}}}{{2a}}{/tex}{tex}2as = {v^2} - {u^2}{/tex}


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