Question Number 4624 by sanusihammed last updated on 14/Feb/16
Answered by Yozzii last updated on 15/Feb/16
$$\left(\mathrm{12}\right)\:{The}\:{equation}\:{of}\:{motion}\:{of}\:{the}\:{particle} \\ $$$${is}\:{given}\:{as} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}}{{ds}}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right)=−{g}−{kv}^{\mathrm{2}} . \\ $$$${v}\:{is}\:{a}\:{function}\:{of}\:{s}\:{as}\:{s}\:{and}\:{v}\:{are}\: \\ $$$${related}\:{to}\:{each}\:{other}\:{by}\:{time}\:{t}. \\ $$$$\therefore\:\frac{{d}}{{ds}}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{v}\frac{{dv}}{{ds}}={v}\frac{{dv}}{{ds}}. \\ $$$$\therefore\:{v}\frac{{dv}}{{ds}}=−{g}−{kv}^{\mathrm{2}} =−\left({g}+{kv}^{\mathrm{2}} \right)\:\:\:\:\:\:\left(\ast\right) \\ $$$${At}\:{the}\:{maximum}\:{height}\:{s}={h},{v}=\mathrm{0}\:{and} \\ $$$${at}\:{v}={u},{s}=\mathrm{0}. \\ $$$$\therefore\:{Separating}\:{the}\:{variables}\:{of}\:\left(\ast\right)\:{and} \\ $$$${integrating}\:{between}\:{the}\:{limits}\: \\ $$$${v}=\mathrm{0},{v}={u}\:{and}\:{s}={h},{s}=\mathrm{0},\:{we}\:{get} \\ $$$$\int_{{u}} ^{\mathrm{0}} \frac{{v}}{{g}+{kv}^{\mathrm{2}} }{dv}=\int_{\mathrm{0}} ^{{h}} −\mathrm{1}{ds} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{k}}\int_{{u}} ^{\mathrm{0}} \frac{\mathrm{2}{kv}}{{g}+{kv}^{\mathrm{2}} }{dv}=−{s}\mid_{\mathrm{0}} ^{{h}} \:\:\:\:\left({k}\neq\mathrm{0}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{k}}{log}_{{e}} \left({g}+{kv}^{\mathrm{2}} \right)\mid_{{u}} ^{\mathrm{0}} =−{h}=\mathrm{0} \\ $$$${h}=\frac{−\mathrm{1}}{\mathrm{2}{k}}\left({log}_{{e}} \left({g}\right)−{log}_{{e}} \left({g}+{ku}^{\mathrm{2}} \right)\right) \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}{k}}{log}_{{e}} \left(\frac{{g}+{ku}^{\mathrm{2}} }{{g}}\right)\:\:\left({k}\neq\mathrm{0},{g}+{ku}^{\mathrm{2}} >\mathrm{0}\Rightarrow{k}>−{g}/{u}^{\mathrm{2}} \right) \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}{k}}{log}_{{e}} \left(\mathrm{1}+\frac{{ku}^{\mathrm{2}} }{{g}}\right)\:\:\: \\ $$$$ \\ $$$$\left(\mathrm{13}\right)\left({i}\right)\:{Let}\:{a}\:{be}\:{the}\:{acceleration}\:{of}\:{the}\: \\ $$$${particle}\:{in}\:{ms}^{−\mathrm{2}} .\:{a}\:{is}\:{given}\:{as}\:{a}\:{function} \\ $$$${of}\:{time}\:{t}\:{in}\:{the}\:{form}\:{a}={t}+\mathrm{3}.\: \\ $$$${Since}\:{a}=\frac{{dv}}{{dt}}\Rightarrow\int_{{v}_{\mathrm{0}} } ^{{v}} {dv}=\int_{{t}_{\mathrm{0}} } ^{{t}} {adt} \\ $$$$\Rightarrow{v}\mid_{{v}_{\mathrm{0}} } ^{{v}} =\int_{{t}_{\mathrm{0}} } ^{{t}} \left({t}+\mathrm{3}\right){dt} \\ $$$${v}−{v}_{\mathrm{0}} =\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}{t}\mid_{{t}_{\mathrm{0}} } ^{{t}} \\ $$$${v}={v}_{\mathrm{0}} +\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{0}.\mathrm{5}{t}_{\mathrm{0}} ^{\mathrm{2}} −\mathrm{3}{t}_{\mathrm{0}} \\ $$$${At}\:{t}_{\mathrm{0}} =\mathrm{0},{v}_{\mathrm{0}} =\mathrm{2}. \\ $$$$\therefore{v}=\mathrm{2}+\mathrm{3}{t}+\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} −\mathrm{0}−\mathrm{0} \\ $$$${v}=\mathrm{2}+\mathrm{3}{t}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \:\:\:\:\left({t}\geqslant\mathrm{0}\right). \\ $$$${At}\:{t}=\mathrm{2},\:{v}=\mathrm{2}+\mathrm{3}×\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}=\mathrm{10}{ms}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{Since}\:{v}=\frac{{ds}}{{dt}}\Rightarrow\int_{{s}_{\mathrm{0}} } ^{{s}} {ds}=\int_{{t}_{\mathrm{0}} } ^{{t}} {vdt}. \\ $$$${At}\:{t}_{\mathrm{0}} =\mathrm{0},{s}_{\mathrm{0}} =\mathrm{0}. \\ $$$$\therefore\int_{\mathrm{0}} ^{{s}} {ds}=\int_{\mathrm{0}} ^{{t}} \left(\mathrm{2}+\mathrm{3}{t}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right){dt} \\ $$$${s}\mid_{\mathrm{0}} ^{{s}} =\mathrm{2}{t}+\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{3}} \mid_{\mathrm{0}} ^{{t}} \\ $$$${s}−\mathrm{0}=\mathrm{2}{t}+\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{3}} −\mathrm{0} \\ $$$${s}={t}\left(\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}{t}+\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{2}} \right)\:\:\left({t}\geqslant\mathrm{0}\right) \\ $$$${s}\:{is}\:{an}\:{increasing}\:{function}\:{for}\:{all}\:{t}\geqslant\mathrm{0} \\ $$$${since}\:\frac{{ds}}{{dt}}={v}=\mathrm{2}+\mathrm{3}{t}+\mathrm{0}.\mathrm{5}{t}^{\mathrm{2}} \geqslant\mathrm{2}+\mathrm{0}+\mathrm{0}=\mathrm{2}>\mathrm{0}. \\ $$$${The}\:{particle}\:{moves}\:{along}\:{a}\:{straight}\: \\ $$$${line}\:{so}\:{that}\:{since}\:{s}\geqslant\mathrm{0},\:{the}\:{distance}={displacement}. \\ $$$${At}\:{t}=\mathrm{2}\:{s},{s}\left(\mathrm{2}\right)=\mathrm{2}\left(\mathrm{2}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{34}}{\mathrm{3}}{m}. \\ $$$${At}\:{t}=\mathrm{4}\:{s},\:{s}\left(\mathrm{4}\right)=\mathrm{4}\left(\mathrm{2}+\mathrm{6}+\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\mathrm{128}}{\mathrm{3}}\:{m}\: \\ $$$$\therefore\:{required}\:{distance}={s}\left(\mathrm{4}\right)−{s}\left(\mathrm{2}\right)=\frac{\mathrm{128}−\mathrm{34}}{\mathrm{3}}=\frac{\mathrm{94}}{\mathrm{3}}{m} \\ $$$$ \\ $$