Derive the speed time equation by …
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Sia ? 6 years, 3 months ago
We know
a dt = dv
Integrating
{tex}\int\limits_0^tadt =\int\limits_u^v dv{/tex}
at = v - u
v = u+at
{tex}a = \frac { d v } { d t }{/tex}
Multiply and Divide by dx
{tex}a = \frac { d v } { d t } \times \frac { d x } { d x }{/tex}
{tex}a = \frac { d v } { d x } \times v{/tex}
{tex}adx = vdv{/tex} {tex}\left( \because \frac { d x } { d t } = v \right){/tex}
{tex}a \int \limits _{0}^{s}dx= \int \limits _{u}^{v}vdv{/tex}
{tex}as = \frac { \upsilon ^ { 2 } } { 2 } - \frac { \nu ^ { 2 } } { 2 }{/tex}
{tex}a s = \frac { \upsilon ^ { 2 } - \nu ^ { 2 } } { 2 }{/tex}
{tex}\upsilon ^ { 2 } - \nu ^ { 2 } = 2 a s{/tex}
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