InterviewSolution
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InterviewSolution
| 1. |
Derivation of all formula of motion of constant acceleration |
| Answer» From the definition of average acceleration, we have{tex}a = \\frac { d v } { d t } \\Rightarrow d v = a d t{/tex}Integrating on both sides and taking the limit for velocity u to v and for time 0 to t, we get{tex}\\int _ { u } ^ { v } d v = \\int _ { 0 } ^ { t } a d t = a \\int _ { 0 } ^ { t } d t = a [ t ] _ { 0 } ^ { t }{/tex}[a is constant]v - u = atv = u + at .....................(i)Now, from the definition of velocity, we have\xa0{tex}v = \\frac { d x } { d t } \\Rightarrow d x = v d t{/tex}Integrating on both sides and taking the limit for displacement x0\xa0to x and for time 0 to t, we get{tex}\\int _ { x _ { 0 } } ^ { x } d x = \\int _ { 0 } ^ { t } v d t = \\int _ { 0 } ^ { t } ( u + a t ) d t = v _ { 0 } [ t ] _ { 0 } ^ { t } + a \\left[ \\frac { t ^ { 2 } } { 2 } \\right] _ { 0 } ^ { t }{/tex}x - x0\xa0= ut +\xa0{tex}\\frac 12{/tex}at2x = x0\xa0+ ut +\xa0{tex}\\frac 12{/tex}at2 ..........................(ii)Now, we can write{tex}a = \\frac { d v } { d t } = \\frac { d v } { d x } \\frac { d x } { d t } = v \\frac { d v } { d x }{/tex}\xa0or vdv = adxIntegrating on both sides and taking the limit for velocity u to v and for displacement x0 to x, we get{tex}\\int _ { u } ^ { v } v d v = \\int _ { x _ { 0 } } ^ { x } a d x{/tex};\xa0{tex}\\frac { v ^ { 2 } - u ^ { 2 } } { 2 } = a \\left( x - x _ { 0 } \\right){/tex}v2\xa0= u2\xa0+ 2a(x - x0) ........................(iii)Equations (i), (ii) and (iii) are the required equations of motion. | |