Question Number 178782 by peter frank last updated on 21/Oct/22 | ||
Answered by mr W last updated on 22/Oct/22 | ||
$${m}\frac{{dv}}{{dt}}={mg}\:\mathrm{sin}\:\theta−{kmv}^{\mathrm{2}} \\ $$$${dt}=\frac{{dv}}{{g}\:\mathrm{sin}\:\theta−{kv}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{{t}} {dt}=\int_{\mathrm{0}} ^{{v}} \frac{{dv}}{{g}\:\mathrm{sin}\:\theta−{kv}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{{t}} {dt}=\int_{\mathrm{0}} ^{{v}} \frac{{d}\left({kv}\right)}{\left(\sqrt{{kg}\:\mathrm{sin}\:\theta}\right)^{\mathrm{2}} −\left({kv}\right)^{\mathrm{2}} } \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{kg}\:\mathrm{sin}\:\theta}}\left[\mathrm{ln}\:\frac{\sqrt{{kg}\:\mathrm{sin}\:\theta}+{kv}}{\:\sqrt{{kg}\:\mathrm{sin}\:\theta}−{kv}}\right]_{\mathrm{0}} ^{{v}} \\ $$$$\mathrm{2}\sqrt{{kg}\:\mathrm{sin}\:\theta}\:{t}=\mathrm{ln}\:\frac{\sqrt{{kg}\:\mathrm{sin}\:\theta}+{kv}}{\:\sqrt{{kg}\:\mathrm{sin}\:\theta}−{kv}} \\ $$$${e}^{\mathrm{2}\sqrt{{kg}\:\mathrm{sin}\:\theta}\:{t}} =\frac{\sqrt{{kg}\:\mathrm{sin}\:\theta}+{kv}}{\:\sqrt{{kg}\:\mathrm{sin}\:\theta}−{kv}} \\ $$$${let}\:\lambda=\sqrt{{kg}\:\mathrm{sin}\:\theta} \\ $$$${e}^{\mathrm{2}\lambda\:{t}} =\frac{\mathrm{2}\lambda}{\:\lambda−{kv}}−\mathrm{1} \\ $$$$\frac{\lambda}{{k}}\left(\mathrm{1}−\frac{\mathrm{2}}{\:\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }\right)={v}=\frac{{ds}}{{dt}} \\ $$$$\frac{\lambda}{{k}}\left(\mathrm{1}−\frac{\mathrm{2}}{\:\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }\right){dt}={ds} \\ $$$$\frac{\lambda}{{k}}\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}−\frac{\mathrm{2}}{\:\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }\right){dt}=\int_{\mathrm{0}} ^{{d}} {ds} \\ $$$${t}−\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{2}}{\:\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }{dt}=\frac{{kd}}{\lambda} \\ $$$${t}−\frac{\mathrm{1}}{\lambda}\int_{\mathrm{0}} ^{{t}} \frac{{d}\left(\mathrm{2}\lambda{t}\right)}{\:\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }=\frac{{kd}}{\lambda} \\ $$$${t}−\frac{\mathrm{1}}{\lambda}\left[\mathrm{2}\lambda{t}−\mathrm{ln}\:\left(\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} \right)\right]_{\mathrm{0}} ^{{t}} =\frac{{kd}}{\lambda} \\ $$$${t}−\frac{\mathrm{1}}{\lambda}\left[\mathrm{2}\lambda{t}−\mathrm{ln}\:\left(\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} \right)+\mathrm{ln}\:\mathrm{2}\right]=\frac{{kd}}{\lambda} \\ $$$${t}−\frac{\mathrm{1}}{\lambda}\mathrm{ln}\:\frac{\mathrm{2}{e}^{\mathrm{2}\lambda{t}} }{\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }=\frac{{kd}}{\lambda} \\ $$$$\mathrm{ln}\:{e}^{\lambda{t}} −\mathrm{ln}\:\frac{\mathrm{2}{e}^{\mathrm{2}\lambda{t}} }{\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }={kd} \\ $$$$\mathrm{ln}\:\frac{\left(\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} \right)}{\mathrm{2}{e}^{\lambda{t}} }={kd} \\ $$$$\:\frac{\mathrm{1}+{e}^{\mathrm{2}\lambda{t}} }{\mathrm{2}{e}^{\lambda{t}} }={e}^{{kd}} \\ $$$$\left({e}^{\lambda{t}} \right)^{\mathrm{2}} −\mathrm{2}{e}^{{kd}} {e}^{\lambda{t}} +\mathrm{1}=\mathrm{0} \\ $$$${e}^{\lambda{t}} ={e}^{{kd}} +\sqrt{{e}^{\mathrm{2}{kd}} −\mathrm{1}} \\ $$$$\lambda{t}=\mathrm{ln}\:\left({e}^{{kd}} +\sqrt{{e}^{\mathrm{2}{kd}} −\mathrm{1}}\right)=\mathrm{cosh}^{−\mathrm{1}} \:\left({e}^{{kd}} \right) \\ $$$$\Rightarrow{t}=\frac{\mathrm{cosh}^{−\mathrm{1}} \:\left({e}^{{kd}} \right)}{\lambda} \\ $$$$\Rightarrow{t}=\frac{\mathrm{cosh}^{−\mathrm{1}} \:\left({e}^{{kd}} \right)}{\:\sqrt{{kg}\:\mathrm{sin}\:\theta}}\:\checkmark \\ $$ | ||
Commented by mr W last updated on 22/Oct/22 | ||
Commented by Tawa11 last updated on 22/Oct/22 | ||
$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||
Commented by Spillover last updated on 22/Oct/22 | ||
$$\mathrm{Excellent}\:\mathrm{solution}\:\mathrm{Mr}\:\mathrm{W} \\ $$ | ||
Commented by peter frank last updated on 22/Oct/22 | ||
$$\mathrm{thanks} \\ $$ | ||
Answered by Spillover last updated on 22/Oct/22 | ||
$$\mathrm{F}=\mathrm{ma} \\ $$$$−\mathrm{kmv}^{\mathrm{2}} +\mathrm{mgsin}\:\theta=\mathrm{ma} \\ $$$$\mathrm{gsin}\:\theta−\mathrm{kv}^{\mathrm{2}} =\mathrm{a} \\ $$$$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{gsin}\:\theta−\mathrm{kv}^{\mathrm{2}} =\mathrm{k}\left[\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta−\mathrm{v}^{\mathrm{2}} \right] \\ $$$$\mathrm{dt}=\int\frac{\mathrm{dv}}{\mathrm{k}\left[\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta−\mathrm{v}^{\mathrm{2}} \right]}=\frac{\mathrm{1}}{\mathrm{k}}\int\frac{\mathrm{dv}}{\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta−\mathrm{v}^{\mathrm{2}} } \\ $$$$\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{k}}\int\frac{\mathrm{dv}}{\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta−\mathrm{v}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{k}}\int\frac{\mathrm{dv}}{\:\sqrt{\left(\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }−\mathrm{v}^{\mathrm{2}} } \\ $$$$\mathrm{let}\:\beta=\:\sqrt{\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta}\: \\ $$$$\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{k}}\int\frac{\mathrm{dv}}{\:\sqrt{\left(\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }−\mathrm{v}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{k}}\int\frac{\mathrm{dv}}{\beta^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} } \\ $$$$\mathrm{t}=\frac{\mathrm{1}}{\mathrm{k}}.\frac{\mathrm{1}}{\beta}\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{v}}{\beta}\right)+\mathrm{A} \\ $$$$\mathrm{t}=\mathrm{0}\:\:\:\mathrm{x}=\mathrm{0}\:\:\:\:\mathrm{v}=\mathrm{0} \\ $$$$\frac{\mathrm{v}}{\beta}=\mathrm{tanh}\:\left(\mathrm{kt}\beta\right)\:\:\:\mathrm{v}=\beta\mathrm{tanh}\:\left(\mathrm{kt}\beta\right) \\ $$$$\:\:\mathrm{v}=\beta\mathrm{tanh}\:\left(\mathrm{kt}\beta\right)\:\:\:\:\:\frac{\mathrm{dx}}{\mathrm{dt}}=\beta\mathrm{tanh}\:\left(\mathrm{kt}\beta\right)\:\:\:\:\:\:\mathrm{dx}=\beta\mathrm{tanh}\:\left(\mathrm{kt}\beta\right)\:\mathrm{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{d}} \mathrm{dx}=\int\beta\mathrm{tanh}\:\left(\mathrm{kt}\beta\right)\:\mathrm{dt}\:\: \\ $$$$\mathrm{d}=\beta\mathrm{ln}\:\frac{\left[\mathrm{cosh}\:\left(\mathrm{kt}\beta\right)\right.}{\mathrm{k}\beta} \\ $$$$\mathrm{d}=\frac{\mathrm{1}}{\mathrm{k}}\mathrm{ln}\:\mathrm{cosh}\:\left(\mathrm{kt}\beta\right) \\ $$$$\mathrm{kd}=\mathrm{ln}\:\mathrm{cosh}\:\left(\mathrm{kt}\beta\right) \\ $$$$\mathrm{e}^{\mathrm{kd}} =\mathrm{cosh}\:\left(\mathrm{kt}\beta\right) \\ $$$$\mathrm{cosh}\:^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{kd}} \right)=\mathrm{kt}\beta \\ $$$$\mathrm{t}=\frac{\mathrm{1}}{\mathrm{k}\beta}\mathrm{cosh}\:^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{kd}} \right) \\ $$$$\beta=\:\sqrt{\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta}\: \\ $$$$\mathrm{t}=\frac{\mathrm{1}}{\mathrm{k}.\:\sqrt{\frac{\mathrm{g}}{\mathrm{k}}\mathrm{sin}\:\theta}\:}.\mathrm{cosh}\:^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{kd}} \right) \\ $$$$\mathrm{t}=\frac{\mathrm{cosh}\:^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{kd}} \right)}{\:\sqrt{\mathrm{gksin}\:\theta}} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by peter frank last updated on 22/Oct/22 | ||
$$\mathrm{great}\:\mathrm{sir} \\ $$ | ||
Commented by mr W last updated on 22/Oct/22 | ||
$${we}\:{can}\:{see}\:{you}\:{are}\:{a}\:{fan}\:{of}\:{hyperbolic} \\ $$$${functions}.\:{that}'{s}\:{great}! \\ $$ | ||