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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 160 km/h. at the same time a car drives from B in direction to E with speed 100 km/h. after what time is the distance between them mininum and what is this minimum distance? (we treat both train and car as points)

$${a}\:{train}\:{drives}\:{from}\:{A}\:{in}\:{direction}\:{to} \\ $$$${D}\:{with}\:{speed}\:\mathrm{160}\:{km}/{h}.\:{at}\:{the}\:{same} \\ $$$${time}\:{a}\:{car}\:{drives}\:{from}\:{B}\:{in}\:{direction} \\ $$$${to}\:{E}\:{with}\:{speed}\:\mathrm{100}\:{km}/{h}. \\ $$$${after}\:{what}\:{time}\:{is}\:{the}\:{distance}\: \\ $$$${between}\:{them}\:{mininum}\:{and}\:{what}\:{is} \\ $$$${this}\:{minimum}\:{distance}? \\ $$$$\left({we}\:{treat}\:{both}\:{train}\:{and}\:{car}\:{as}\:{points}\right) \\ $$

Answered by mahdipoor last updated on 23/Nov/22

befor h=(1/(10)) (at h=(1/(10)) trainB in C ), d(=distance among trains) was increas after h=(1/8) (trainA in C), d will increas among 1/8≥h≥1/10 ⇒ d^2 =(100h−10)^2 +(20−160h)^2 −2×(100h−10)(20−160h)cos60 d(d^2 )/dh=2(100h−10)100−2(20−160h)160 −100(20−160h)+160(100h−10)= 360(100h−10)−420(20−160h)=0 ⇒1032h−120=0⇒h=(5/(43))≈.12 ⇒0.125>0.12>0.1⇒ min(d^2 )=d^2 (h=(5/(43)))=((100)/(43)) km^2 ⇒min d=((10)/( (√(43))))≈1.525 km

$$ \\ $$$${befor}\:{h}=\frac{\mathrm{1}}{\mathrm{10}}\:\left({at}\:{h}=\frac{\mathrm{1}}{\mathrm{10}}\:{trainB}\:{in}\:{C}\:\right), \\ $$$${d}\left(={distance}\:{among}\:{trains}\right)\:{was}\:{increas}\: \\ $$$${after}\:{h}=\frac{\mathrm{1}}{\mathrm{8}}\:\left({trainA}\:{in}\:{C}\right),\:{d}\:{will}\:{increas}\: \\ $$$${among}\:\mathrm{1}/\mathrm{8}\geqslant{h}\geqslant\mathrm{1}/\mathrm{10}\:\Rightarrow \\ $$$${d}^{\mathrm{2}} =\left(\mathrm{100}{h}−\mathrm{10}\right)^{\mathrm{2}} +\left(\mathrm{20}−\mathrm{160}{h}\right)^{\mathrm{2}} \\ $$$$−\mathrm{2}×\left(\mathrm{100}{h}−\mathrm{10}\right)\left(\mathrm{20}−\mathrm{160}{h}\right){cos}\mathrm{60} \\ $$$${d}\left({d}^{\mathrm{2}} \right)/{dh}=\mathrm{2}\left(\mathrm{100}{h}−\mathrm{10}\right)\mathrm{100}−\mathrm{2}\left(\mathrm{20}−\mathrm{160}{h}\right)\mathrm{160} \\ $$$$−\mathrm{100}\left(\mathrm{20}−\mathrm{160}{h}\right)+\mathrm{160}\left(\mathrm{100}{h}−\mathrm{10}\right)= \\ $$$$\mathrm{360}\left(\mathrm{100}{h}−\mathrm{10}\right)−\mathrm{420}\left(\mathrm{20}−\mathrm{160}{h}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1032}{h}−\mathrm{120}=\mathrm{0}\Rightarrow{h}=\frac{\mathrm{5}}{\mathrm{43}}\approx.\mathrm{12} \\ $$$$\Rightarrow\mathrm{0}.\mathrm{125}>\mathrm{0}.\mathrm{12}>\mathrm{0}.\mathrm{1}\Rightarrow \\ $$$${min}\left({d}^{\mathrm{2}} \right)={d}^{\mathrm{2}} \left({h}=\frac{\mathrm{5}}{\mathrm{43}}\right)=\frac{\mathrm{100}}{\mathrm{43}}\:{km}^{\mathrm{2}} \\ $$$$\Rightarrow{min}\:{d}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\approx\mathrm{1}.\mathrm{525}\:{km} \\ $$

Commented by mr W last updated on 23/Nov/22

thanks sir!

$${thanks}\:{sir}! \\ $$

Commented by mahdipoor last updated on 23/Nov/22

tnx , i edited

$${tnx}\:,\:{i}\:{edited} \\ $$

Answered by mr W last updated on 23/Nov/22

Commented by mr W last updated on 23/Nov/22

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=(√(160^2 +100^2 +160×100))=20(√(129)) km/h sin α=((100)/(20(√(129))))×sin 120=(5/( 2(√(43)))) β=90°−α sin β=cos α=((7(√3))/(2(√(43)))) ((BQ)/(sin 60°))=((CQ)/(sin (60°+β)))=((CB)/(sin β)) BQ=((10×2(√(43)))/(7(√3)))×((√3)/2)=((10(√(43)))/7) CQ=((10×2(√(43)))/(7(√3)))×(((√3)/2)×(5/(2(√(43))))+(1/2)×((7(√3))/(2(√(43))))) =((60)/7) PB=(20+((60)/7)) sin α−((10(√(43)))/7) =((10)/( (√(43))))≈1.525 km t=((AP)/v)=(1/(20(√(129))))(20+((60)/7))cos α =(5/(43))≈0.116 h =6 min 58.6 sec

$$\underline{{Using}\:{method}\:{of}\:{relative}\:{motion}} \\ $$$${assume}\:{the}\:{car}\:{doesn}'{t}\:{move}.\:{then} \\ $$$${the}\:{train}\:{moves}\:{with}\:{the}\:{relative} \\ $$$${velocity}\:{v}\:{in}\:{direction}\:{F},\:{see}\:{picture}. \\ $$$${the}\:{shortest}\:{distance}\:{between}\:{car}\: \\ $$$${and}\:{train}\:{is}\:{the}\:{distance}\:{from}\:{B}\:{to}\: \\ $$$${the}\:{line}\:{AF}. \\ $$$${v}=\sqrt{\mathrm{160}^{\mathrm{2}} +\mathrm{100}^{\mathrm{2}} +\mathrm{160}×\mathrm{100}}=\mathrm{20}\sqrt{\mathrm{129}}\:{km}/{h} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{100}}{\mathrm{20}\sqrt{\mathrm{129}}}×\mathrm{sin}\:\mathrm{120}=\frac{\mathrm{5}}{\:\mathrm{2}\sqrt{\mathrm{43}}} \\ $$$$\beta=\mathrm{90}°−\alpha \\ $$$$\mathrm{sin}\:\beta=\mathrm{cos}\:\alpha=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{43}}} \\ $$$$\frac{{BQ}}{\mathrm{sin}\:\mathrm{60}°}=\frac{{CQ}}{\mathrm{sin}\:\left(\mathrm{60}°+\beta\right)}=\frac{{CB}}{\mathrm{sin}\:\beta} \\ $$$${BQ}=\frac{\mathrm{10}×\mathrm{2}\sqrt{\mathrm{43}}}{\mathrm{7}\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{10}\sqrt{\mathrm{43}}}{\mathrm{7}} \\ $$$${CQ}=\frac{\mathrm{10}×\mathrm{2}\sqrt{\mathrm{43}}}{\mathrm{7}\sqrt{\mathrm{3}}}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{43}}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{43}}}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{60}}{\mathrm{7}} \\ $$$${PB}=\left(\mathrm{20}+\frac{\mathrm{60}}{\mathrm{7}}\right)\:\mathrm{sin}\:\alpha−\frac{\mathrm{10}\sqrt{\mathrm{43}}}{\mathrm{7}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\approx\mathrm{1}.\mathrm{525}\:{km} \\ $$$${t}=\frac{{AP}}{{v}}=\frac{\mathrm{1}}{\mathrm{20}\sqrt{\mathrm{129}}}\left(\mathrm{20}+\frac{\mathrm{60}}{\mathrm{7}}\right)\mathrm{cos}\:\alpha \\ $$$$\:\:=\frac{\mathrm{5}}{\mathrm{43}}\approx\mathrm{0}.\mathrm{116}\:{h}\:=\mathrm{6}\:{min}\:\mathrm{58}.\mathrm{6}\:{sec} \\ $$

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 2.32 and 1.538 km 2nd: am confused that why i got 7 min : 2330 meter but time 6 min 58.6 sec : 2320 meter there′s a difference about 10 meter Note: i will write mine when finish my work

$${The}\:{qusetion}\:{is}\:{very}\:{good}\:,\:{thank}\:{you}\:{for}\:{it}\:{Sir}. \\ $$$${Well},\:{please}\:{check}\:{the}\:{difference}\:{between} \\ $$$$\:\mathrm{2}.\mathrm{32}\:{and}\:\mathrm{1}.\mathrm{538}\:{km} \\ $$$$ \\ $$$$\mathrm{2}{nd}:\:{am}\:{confused}\:{that}\:{why}\:{i}\:{got} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}\:{min}\::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2330}\:{meter} \\ $$$$\:{but}\:{time}\:\:\mathrm{6}\:{min}\:\mathrm{58}.\mathrm{6}\:{sec}\::\:\mathrm{2320}\:{meter} \\ $$$$\:{there}'{s}\:{a}\:{difference}\:{about}\:\mathrm{10}\:{meter} \\ $$$$ \\ $$$${Note}:\:{i}\:{will}\:{write}\:{mine}\:{when}\:{finish}\:{my}\:{work} \\ $$

Commented by mr W last updated on 23/Nov/22

correct answer: after (5/(43)) h, i.e. 6 min 58.6 sec the shortest distance ((10)/( (√(43))))≈1.525 km.

$${correct}\:{answer}: \\ $$$${after}\:\frac{\mathrm{5}}{\mathrm{43}}\:{h},\:{i}.{e}.\:\mathrm{6}\:{min}\:\mathrm{58}.\mathrm{6}\:{sec}\:{the} \\ $$$${shortest}\:{distance}\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\approx\mathrm{1}.\mathrm{525}\:\:{km}. \\ $$

Answered by mr W last updated on 23/Nov/22

Commented by mr W last updated on 23/Nov/22

An other geometrical way say when train is at point A_1 and car is at point B_1 their distance is minimum, at this moment the rate in which their distance changes is zero: 160 cos θ_1 −100 cos θ_2 =0 ⇒cos θ_1 =(5/8) cos θ_2 θ_1 +θ_2 =180°−60°=120° cos θ_1 =cos (120°−θ_2 )=−(1/2) cos θ_2 +((√3)/2) sin θ_2 (5/8) cos θ_2 =−(1/2) cos θ_2 +((√3)/2) sin θ_2 ⇒tan θ_2 =(9/(4(√3))) ⇒cos θ_2 =(4/( (√(43)))) ⇒cos θ_1 =(5/( 2(√(43)))) ⇒sin θ_1 =((7(√3))/(2(√(43)))), sin θ_2 =((3(√3))/( (√(43)))) A_1 C=20−160t B_1 C=100t−10 ((A_1 C)/(sin θ_2 ))=((B_1 C)/(sin θ_1 ))=((A_1 B_1 )/(sin 60°)) (((√(43))(20−160t))/( 3(√3)))=((2(√(43))(100t−10))/(7(√3)))=((A_1 B_1 )/(sin 60°)) (((√(43))(20−160t))/( 3(√3)))=((2(√(43))(100t−10))/(7(√3))) ⇒t=(5/(43)) ✓ A_1 B_1 =(((√(43))(20−160×(5/(43))))/( 3(√3)))×((√3)/2)=((10)/( (√(43)))) ✓

$$\underline{{An}\:{other}\:{geometrical}\:{way}} \\ $$$${say}\:{when}\:{train}\:{is}\:{at}\:{point}\:{A}_{\mathrm{1}} \:{and} \\ $$$${car}\:{is}\:{at}\:{point}\:{B}_{\mathrm{1}} \:{their}\:{distance}\:{is} \\ $$$${minimum},\:{at}\:{this}\:{moment}\:{the}\:{rate} \\ $$$${in}\:{which}\:{their}\:{distance}\:{changes}\:{is}\: \\ $$$${zero}: \\ $$$$\mathrm{160}\:\mathrm{cos}\:\theta_{\mathrm{1}} −\mathrm{100}\:\mathrm{cos}\:\theta_{\mathrm{2}} =\mathrm{0}\: \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{5}}{\mathrm{8}}\:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$$\theta_{\mathrm{1}} +\theta_{\mathrm{2}} =\mathrm{180}°−\mathrm{60}°=\mathrm{120}° \\ $$$$\mathrm{cos}\:\theta_{\mathrm{1}} =\mathrm{cos}\:\left(\mathrm{120}°−\theta_{\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\theta_{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\mathrm{8}}\:\mathrm{cos}\:\theta_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\theta_{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\theta_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}\sqrt{\mathrm{3}}}\: \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{43}}}\:\Rightarrow\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\mathrm{5}}{\:\mathrm{2}\sqrt{\mathrm{43}}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta_{\mathrm{1}} =\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{43}}},\:\mathrm{sin}\:\theta_{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{43}}} \\ $$$${A}_{\mathrm{1}} {C}=\mathrm{20}−\mathrm{160}{t} \\ $$$${B}_{\mathrm{1}} {C}=\mathrm{100}{t}−\mathrm{10} \\ $$$$\frac{{A}_{\mathrm{1}} {C}}{\mathrm{sin}\:\theta_{\mathrm{2}} }=\frac{{B}_{\mathrm{1}} {C}}{\mathrm{sin}\:\theta_{\mathrm{1}} }=\frac{{A}_{\mathrm{1}} {B}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{60}°} \\ $$$$\frac{\sqrt{\mathrm{43}}\left(\mathrm{20}−\mathrm{160}{t}\right)}{\:\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\sqrt{\mathrm{43}}\left(\mathrm{100}{t}−\mathrm{10}\right)}{\mathrm{7}\sqrt{\mathrm{3}}}=\frac{{A}_{\mathrm{1}} {B}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{60}°} \\ $$$$\frac{\sqrt{\mathrm{43}}\left(\mathrm{20}−\mathrm{160}{t}\right)}{\:\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\sqrt{\mathrm{43}}\left(\mathrm{100}{t}−\mathrm{10}\right)}{\mathrm{7}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{5}}{\mathrm{43}}\:\checkmark \\ $$$${A}_{\mathrm{1}} {B}_{\mathrm{1}} =\frac{\sqrt{\mathrm{43}}\left(\mathrm{20}−\mathrm{160}×\frac{\mathrm{5}}{\mathrm{43}}\right)}{\:\mathrm{3}\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{10}}{\:\sqrt{\mathrm{43}}}\:\checkmark \\ $$

Commented by Acem last updated on 23/Nov/22

Commented by mr W last updated on 23/Nov/22

you miscalculated L.

$${you}\:{miscalculated}\:{L}. \\ $$

Commented by Acem last updated on 23/Nov/22

(√(how am i?)) We both have the same results! thank you after a little will write the method

$$\sqrt{{how}\:{am}\:{i}?}\: \\ $$$$ \\ $$$${We}\:{both}\:{have}\:{the}\:{same}\:{results}!\:{thank}\:{you} \\ $$$$\:{after}\:{a}\:{little}\:{will}\:{write}\:{the}\:{method} \\ $$

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