Question Number 37579 by ajfour last updated on 15/Jun/18
$${a}=\frac{−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:\:{with}\:{x}=\mathrm{1}\:{and}\:{v}=\mathrm{0}\:{at}\:{t}=\mathrm{0} \\ $$$${find}\:{time}\:{that}\:{particle}\:{takes}\:{to} \\ $$$${reach}\:{x}=\mathrm{0}.\mathrm{25}{m}\:. \\ $$
Answered by ajfour last updated on 15/Jun/18
$${since}\:{a}\:<\:\mathrm{0}\:{and}\:{v}_{\mathrm{0}} =\mathrm{0} \\ $$$${this}\:{means}\:{negative}\:{velocity} \\ $$$${starts}\:{developing}\:{as}\:{t}\:{increases} \\ $$$${from}\:{zero}. \\ $$$$\:\:\:\boldsymbol{{a}}=\frac{\boldsymbol{{vdv}}}{\boldsymbol{{dx}}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:\:{v}=−\sqrt{\frac{\mathrm{1}}{{x}}−\mathrm{1}} \\ $$$$\:\:\:\int_{\mathrm{1}} ^{\:\:{x}} \frac{{dx}}{\sqrt{\frac{\mathrm{1}}{{x}}−\mathrm{1}}}\:=\:−{t} \\ $$$${let}\:\:\:\:\:{x}=\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:{dx}\:=\:−\mathrm{sin}\:\mathrm{2}\theta{d}\theta \\ $$$$−{t}\:=\:−\int_{\mathrm{0}} ^{\:\:\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}} \:\:\frac{\mathrm{2sin}\:\theta\mathrm{cos}\:\theta{d}\theta}{\mathrm{tan}\:\theta} \\ $$$${t}=\int_{\mathrm{0}} ^{\:\:\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}} \:\:\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\ $$$$\:\:=\left(\theta+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}+{c}\right)\mid_{\mathrm{0}} ^{\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}} \\ $$$$\:\:{t}\:=\mathrm{cos}^{−\mathrm{1}} \sqrt{{x}}+\left(\sqrt{{x}}\right)\left(\sqrt{\mathrm{1}−{x}}\right) \\ $$$${And}\:{to}\:{reach}\:\:{x}=\mathrm{0}.\mathrm{25} \\ $$$$\:\:\:\:\:\boldsymbol{{t}}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:. \\ $$