Question Number 181232 by mr W last updated on 23/Nov/22
Commented by mr W last updated on 23/Nov/22
$${a}\:{train}\:{drives}\:{from}\:{A}\:{in}\:{direction}\:{to} \\ $$$${D}\:{with}\:{speed}\:\mathrm{160}\:{km}/{h}.\:{at}\:{the}\:{same} \\ $$$${time}\:{a}\:{car}\:{drives}\:{from}\:{B}\:{in}\:{direction} \\ $$$${to}\:{E}\:{with}\:{speed}\:\mathrm{100}\:{km}/{h}. \\ $$$${after}\:{what}\:{time}\:{is}\:{the}\:{distance}\: \\ $$$${between}\:{them}\:{mininum}\:{and}\:{what}\:{is} \\ $$$${this}\:{minimum}\:{distance}? \\ $$$$\left({we}\:{treat}\:{both}\:{train}\:{and}\:{car}\:{as}\:{points}\right) \\ $$
Answered by mahdipoor last updated on 23/Nov/22
$$ \\ $$$${befor}\:{h}=\frac{\mathrm{1}}{\mathrm{10}}\:\left({at}\:{h}=\frac{\mathrm{1}}{\mathrm{10}}\:{trainB}\:{in}\:{C}\:\right), \\ $$$${d}\left(={distance}\:{among}\:{trains}\right)\:{was}\:{increas}\: \\ $$$${after}\:{h}=\frac{\mathrm{1}}{\mathrm{8}}\:\left({trainA}\:{in}\:{C}\right),\:{d}\:{will}\:{increas}\: \\ $$$${among}\:\mathrm{1}/\mathrm{8}\geqslant{h}\geqslant\mathrm{1}/\mathrm{10}\:\Rightarrow \\ $$$${d}^{\mathrm{2}} =\left(\mathrm{100}{h}−\mathrm{10}\right)^{\mathrm{2}} +\left(\mathrm{20}−\mathrm{160}{h}\right)^{\mathrm{2}} \\ $$$$−\mathrm{2}×\left(\mathrm{100}{h}−\mathrm{10}\right)\left(\mathrm{20}−\mathrm{160}{h}\right){cos}\mathrm{60} \\ $$$${d}\left({d}^{\mathrm{2}} \right)/{dh}=\mathrm{2}\left(\mathrm{100}{h}−\mathrm{10}\right)\mathrm{100}−\mathrm{2}\left(\mathrm{20}−\mathrm{160}{h}\right)\mathrm{160} \\ $$$$−\mathrm{100}\left(\mathrm{20}−\mathrm{160}{h}\right)+\mathrm{160}\left(\mathrm{100}{h}−\mathrm{10}\right)= \\ $$$$\mathrm{360}\left(\mathrm{100}{h}−\mathrm{10}\right)−\mathrm{420}\left(\mathrm{20}−\mathrm{160}{h}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1032}{h}−\mathrm{120}=\mathrm{0}\Rightarrow{h}=\frac{\mathrm{5}}{\mathrm{43}}\approx.\mathrm{12} \\ $$$$\Rightarrow\mathrm{0}.\mathrm{125}>\mathrm{0}.\mathrm{12}>\mathrm{0}.\mathrm{1}\Rightarrow \\ $$$${min}\left({d}^{\mathrm{2}} \right)={d}^{\mathrm{2}} \left({h}=\frac{\mathrm{5}}{\mathrm{43}}\right)=\frac{\mathrm{100}}{\mathrm{43}}\:{km}^{\mathrm{2}} \\ $$$$\Rightarrow{min}\:{d}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\approx\mathrm{1}.\mathrm{525}\:{km} \\ $$
Commented by mr W last updated on 23/Nov/22
$${thanks}\:{sir}! \\ $$
Commented by mahdipoor last updated on 23/Nov/22
$${tnx}\:,\:{i}\:{edited} \\ $$
Answered by mr W last updated on 23/Nov/22
Commented by mr W last updated on 23/Nov/22
$$\underline{{Using}\:{method}\:{of}\:{relative}\:{motion}} \\ $$$${assume}\:{the}\:{car}\:{doesn}'{t}\:{move}.\:{then} \\ $$$${the}\:{train}\:{moves}\:{with}\:{the}\:{relative} \\ $$$${velocity}\:{v}\:{in}\:{direction}\:{F},\:{see}\:{picture}. \\ $$$${the}\:{shortest}\:{distance}\:{between}\:{car}\: \\ $$$${and}\:{train}\:{is}\:{the}\:{distance}\:{from}\:{B}\:{to}\: \\ $$$${the}\:{line}\:{AF}. \\ $$$${v}=\sqrt{\mathrm{160}^{\mathrm{2}} +\mathrm{100}^{\mathrm{2}} +\mathrm{160}×\mathrm{100}}=\mathrm{20}\sqrt{\mathrm{129}}\:{km}/{h} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{100}}{\mathrm{20}\sqrt{\mathrm{129}}}×\mathrm{sin}\:\mathrm{120}=\frac{\mathrm{5}}{\:\mathrm{2}\sqrt{\mathrm{43}}} \\ $$$$\beta=\mathrm{90}°−\alpha \\ $$$$\mathrm{sin}\:\beta=\mathrm{cos}\:\alpha=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{43}}} \\ $$$$\frac{{BQ}}{\mathrm{sin}\:\mathrm{60}°}=\frac{{CQ}}{\mathrm{sin}\:\left(\mathrm{60}°+\beta\right)}=\frac{{CB}}{\mathrm{sin}\:\beta} \\ $$$${BQ}=\frac{\mathrm{10}×\mathrm{2}\sqrt{\mathrm{43}}}{\mathrm{7}\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{10}\sqrt{\mathrm{43}}}{\mathrm{7}} \\ $$$${CQ}=\frac{\mathrm{10}×\mathrm{2}\sqrt{\mathrm{43}}}{\mathrm{7}\sqrt{\mathrm{3}}}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{43}}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{43}}}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{60}}{\mathrm{7}} \\ $$$${PB}=\left(\mathrm{20}+\frac{\mathrm{60}}{\mathrm{7}}\right)\:\mathrm{sin}\:\alpha−\frac{\mathrm{10}\sqrt{\mathrm{43}}}{\mathrm{7}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\approx\mathrm{1}.\mathrm{525}\:{km} \\ $$$${t}=\frac{{AP}}{{v}}=\frac{\mathrm{1}}{\mathrm{20}\sqrt{\mathrm{129}}}\left(\mathrm{20}+\frac{\mathrm{60}}{\mathrm{7}}\right)\mathrm{cos}\:\alpha \\ $$$$\:\:=\frac{\mathrm{5}}{\mathrm{43}}\approx\mathrm{0}.\mathrm{116}\:{h}\:=\mathrm{6}\:{min}\:\mathrm{58}.\mathrm{6}\:{sec} \\ $$
Commented by Acem last updated on 23/Nov/22
Commented by Acem last updated on 23/Nov/22
$${The}\:{qusetion}\:{is}\:{very}\:{good}\:,\:{thank}\:{you}\:{for}\:{it}\:{Sir}. \\ $$$${Well},\:{please}\:{check}\:{the}\:{difference}\:{between} \\ $$$$\:\mathrm{2}.\mathrm{32}\:{and}\:\mathrm{1}.\mathrm{538}\:{km} \\ $$$$ \\ $$$$\mathrm{2}{nd}:\:{am}\:{confused}\:{that}\:{why}\:{i}\:{got} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\:{min}\::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2330}\:{meter} \\ $$$$\:{but}\:{time}\:\:\mathrm{6}\:{min}\:\mathrm{58}.\mathrm{6}\:{sec}\::\:\mathrm{2320}\:{meter} \\ $$$$\:{there}'{s}\:{a}\:{difference}\:{about}\:\mathrm{10}\:{meter} \\ $$$$ \\ $$$${Note}:\:{i}\:{will}\:{write}\:{mine}\:{when}\:{finish}\:{my}\:{work} \\ $$
Commented by mr W last updated on 23/Nov/22
$${correct}\:{answer}: \\ $$$${after}\:\frac{\mathrm{5}}{\mathrm{43}}\:{h},\:{i}.{e}.\:\mathrm{6}\:{min}\:\mathrm{58}.\mathrm{6}\:{sec}\:{the} \\ $$$${shortest}\:{distance}\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\approx\mathrm{1}.\mathrm{525}\:\:{km}. \\ $$
Answered by mr W last updated on 23/Nov/22
Commented by mr W last updated on 23/Nov/22
$$\underline{{An}\:{other}\:{geometrical}\:{way}} \\ $$$${say}\:{when}\:{train}\:{is}\:{at}\:{point}\:{A}_{\mathrm{1}} \:{and} \\ $$$${car}\:{is}\:{at}\:{point}\:{B}_{\mathrm{1}} \:{their}\:{distance}\:{is} \\ $$$${minimum},\:{at}\:{this}\:{moment}\:{the}\:{rate} \\ $$$${in}\:{which}\:{their}\:{distance}\:{changes}\:{is}\: \\ $$$${zero}: \\ $$$$\mathrm{160}\:\mathrm{cos}\:\theta_{\mathrm{1}} −\mathrm{100}\:\mathrm{cos}\:\theta_{\mathrm{2}} =\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{5}}{\mathrm{8}}\:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$$\theta_{\mathrm{1}} +\theta_{\mathrm{2}} =\mathrm{180}°−\mathrm{60}°=\mathrm{120}° \\ $$$$\mathrm{cos}\:\theta_{\mathrm{1}} =\mathrm{cos}\:\left(\mathrm{120}°−\theta_{\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\theta_{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\mathrm{8}}\:\mathrm{cos}\:\theta_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\theta_{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}\sqrt{\mathrm{3}}}\: \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{43}}}\:\Rightarrow\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{5}}{\:\mathrm{2}\sqrt{\mathrm{43}}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta_{\mathrm{1}} =\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{43}}},\:\mathrm{sin}\:\theta_{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{43}}} \\ $$$${A}_{\mathrm{1}} {C}=\mathrm{20}−\mathrm{160}{t} \\ $$$${B}_{\mathrm{1}} {C}=\mathrm{100}{t}−\mathrm{10} \\ $$$$\frac{{A}_{\mathrm{1}} {C}}{\mathrm{sin}\:\theta_{\mathrm{2}} }=\frac{{B}_{\mathrm{1}} {C}}{\mathrm{sin}\:\theta_{\mathrm{1}} }=\frac{{A}_{\mathrm{1}} {B}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{60}°} \\ $$$$\frac{\sqrt{\mathrm{43}}\left(\mathrm{20}−\mathrm{160}{t}\right)}{\:\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\sqrt{\mathrm{43}}\left(\mathrm{100}{t}−\mathrm{10}\right)}{\mathrm{7}\sqrt{\mathrm{3}}}=\frac{{A}_{\mathrm{1}} {B}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{60}°} \\ $$$$\frac{\sqrt{\mathrm{43}}\left(\mathrm{20}−\mathrm{160}{t}\right)}{\:\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\sqrt{\mathrm{43}}\left(\mathrm{100}{t}−\mathrm{10}\right)}{\mathrm{7}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{5}}{\mathrm{43}}\:\checkmark \\ $$$${A}_{\mathrm{1}} {B}_{\mathrm{1}} =\frac{\sqrt{\mathrm{43}}\left(\mathrm{20}−\mathrm{160}×\frac{\mathrm{5}}{\mathrm{43}}\right)}{\:\mathrm{3}\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\:\checkmark \\ $$
Commented by Acem last updated on 23/Nov/22
Commented by mr W last updated on 23/Nov/22
$${you}\:{miscalculated}\:{L}. \\ $$
Commented by Acem last updated on 23/Nov/22
$$\sqrt{{how}\:{am}\:{i}?}\: \\ $$$$ \\ $$$${We}\:{both}\:{have}\:{the}\:{same}\:{results}!\:{thank}\:{you} \\ $$$$\:{after}\:{a}\:{little}\:{will}\:{write}\:{the}\:{method} \\ $$