Derivation of all formula of motion …

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Sia ? 6 years, 4 months ago
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 = at
v = u + at .....................(i)
Now, from the definition of velocity, we have
{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 to 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 = ut + {tex}\frac 12{/tex}at2
x = x0 + ut + {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} or vdv = adx
Integrating 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}; {tex}\frac { v ^ { 2 } - u ^ { 2 } } { 2 } = a \left( x - x _ { 0 } \right){/tex}
v2 = u2 + 2a(x - x0) ........................(iii)
Equations (i), (ii) and (iii) are the required equations of motion.
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