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InterviewSolution
| 1. |
Derive the speed time equation by calculas method |
| Answer» We know\t\u200b\u200b\u200b\u200b\u200b{tex}a = \\frac { d v } { d t }{/tex}\ta dt = dv\tIntegrating\t{tex}\\int\\limits_0^tadt =\\int\\limits_u^v dv{/tex}\tat = v - u\tv = u+at\tWe know\t{tex}a = \\frac { d v } { d t }{/tex}\tMultiply and Divide by dx\u200b\u200b\u200b\u200b\u200b\u200b\u200b\t{tex}a = \\frac { d v } { d t } \\times \\frac { d x } { d x }{/tex}\t{tex}a = \\frac { d v } { d x } \\times v{/tex}\t{tex}adx = vdv{/tex}\xa0{tex}\\left( \\because \\frac { d x } { d t } = v \\right){/tex}\tIntegrating within the limits\t{tex}a \\int \\limits _{0}^{s}dx= \\int \\limits _{u}^{v}vdv{/tex}\t{tex}as = \\frac { \\upsilon ^ { 2 } } { 2 } - \\frac { \\nu ^ { 2 } } { 2 }{/tex}\t{tex}a s = \\frac { \\upsilon ^ { 2 } - \\nu ^ { 2 } } { 2 }{/tex} \u200b\u200b\u200b\u200b\u200b\u200b\u200b{tex}\\upsilon ^ { 2 } - \\nu ^ { 2 } = 2 a s{/tex} | |