Question Number 215523 by Ismoiljon_008 last updated on 09/Jan/25
$$ \\ $$$$\:\:\:\mathcal{T}{wo}\:{friends}\:{set}\:{off}\:\:{by}\:{train}\:{at}\:{dawn}\:{to}\:{visit} \\ $$$$\:\:\:{each}\:{other}.\:{The}\:{two}\:{friends}\:{caught}\:{sight}\:{of} \\ $$$$\:\:\:{each}\:{other}\:{through}\:{the}\:{window}\:{as}\:{the}\:{trains}\: \\ $$$$\:\:\:{passed}\:{in}\:{opposite}\:{direction}\:{on}\:{adjacent}\:{tracks}− \\ $$$$\:\:\:{it}\:{was}\:\mathrm{12}^{{oo}} \:{hours}.\:{The}\:{friends}\:{helplessly}\:{reached}\: \\ $$$$\:\:\:{their}\:{destinations}.\:{If}\:{the}\:{first}\:{of}\:{them}\:{reached}\: \\ $$$$\:\:\:{their}\:{destination}\:{at}\:\mathrm{16}^{{oo}} \:{and}\:{the}\:{second}\:{at}\:\mathrm{21}^{{oo}} , \\ $$$$\:\:\:{what}\:{time}\:{did}\:{dawn}\:{break}\:{on}\:{that}\:{day}\:? \\ $$$$\:\:\:{Help}\:{me},\:{please} \\ $$
Commented by nikif99 last updated on 09/Jan/25
Commented by mr W last updated on 10/Jan/25
$${great}\:{translation}\:{for}\:{the}\:{unclear} \\ $$$${question}! \\ $$
Answered by nikif99 last updated on 09/Jan/25
![Let t the dawn. [12:00]⇒v_1 =(a/(12−t)), v_2 =((s−a)/(12−t)) [16:00]⇒ v_1 =(s/(16−t)), [16:00−12:00]⇒v_1 =((s−a)/4) [21:00]⇒ v_2 =(s/(21−t)), [21:00−12:00]⇒v_2 =(a/9) (v_1 /v_2 )=((9(s−a))/(4a))=(a/(s−a))=((21−t)/(16−t)) (1) (1)⇒9(s−a)^2 =4a^2 ⇒5a^2 −18sa+9s^2 =0⇒ a_(1,2) =((18s±(√(18^2 s^2 −4×5×9s^2 )))/(2×5))=3s or ((3s)/5) (1)⇒a(16−t)=(s−a)(21−t)⇒ 37a−2at−21s+st=0⇒t=((37a−21s)/(2a−s)) for a_1 =3s⇒t=18:00h rejected for a_2 =((3s)/5)⇒t=6:00 accepted](https://www.tinkutara.com/question/Q215530.png)
$$ \\ $$$${Let}\:{t}\:{the}\:{dawn}.\: \\ $$$$\left[\mathrm{12}:\mathrm{00}\right]\Rightarrow{v}_{\mathrm{1}} =\frac{{a}}{\mathrm{12}−{t}},\:{v}_{\mathrm{2}} =\frac{{s}−{a}}{\mathrm{12}−{t}} \\ $$$$\left[\mathrm{16}:\mathrm{00}\right]\Rightarrow\:{v}_{\mathrm{1}} =\frac{{s}}{\mathrm{16}−{t}},\:\left[\mathrm{16}:\mathrm{00}−\mathrm{12}:\mathrm{00}\right]\Rightarrow{v}_{\mathrm{1}} =\frac{{s}−{a}}{\mathrm{4}} \\ $$$$\left[\mathrm{21}:\mathrm{00}\right]\Rightarrow\:{v}_{\mathrm{2}} =\frac{{s}}{\mathrm{21}−{t}},\:\left[\mathrm{21}:\mathrm{00}−\mathrm{12}:\mathrm{00}\right]\Rightarrow{v}_{\mathrm{2}} =\frac{{a}}{\mathrm{9}} \\ $$$$\frac{{v}_{\mathrm{1}} }{{v}_{\mathrm{2}} }=\frac{\mathrm{9}\left({s}−{a}\right)}{\mathrm{4}{a}}=\frac{{a}}{{s}−{a}}=\frac{\mathrm{21}−{t}}{\mathrm{16}−{t}}\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\mathrm{9}\left({s}−{a}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} \Rightarrow\mathrm{5}{a}^{\mathrm{2}} −\mathrm{18}{sa}+\mathrm{9}{s}^{\mathrm{2}} =\mathrm{0}\Rightarrow \\ $$$${a}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{18}{s}\pm\sqrt{\mathrm{18}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{4}×\mathrm{5}×\mathrm{9}{s}^{\mathrm{2}} }}{\mathrm{2}×\mathrm{5}}=\mathrm{3}{s}\:{or}\:\frac{\mathrm{3}{s}}{\mathrm{5}} \\ $$$$\left(\mathrm{1}\right)\Rightarrow{a}\left(\mathrm{16}−{t}\right)=\left({s}−{a}\right)\left(\mathrm{21}−{t}\right)\Rightarrow \\ $$$$\mathrm{37}{a}−\mathrm{2}{at}−\mathrm{21}{s}+{st}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{37}{a}−\mathrm{21}{s}}{\mathrm{2}{a}−{s}} \\ $$$${for}\:{a}_{\mathrm{1}} =\mathrm{3}{s}\Rightarrow{t}=\mathrm{18}:\mathrm{00}{h}\:{rejected} \\ $$$${for}\:{a}_{\mathrm{2}} =\frac{\mathrm{3}{s}}{\mathrm{5}}\Rightarrow{t}=\mathrm{6}:\mathrm{00}\:{accepted} \\ $$
Commented by Ismoiljon_008 last updated on 10/Jan/25
$$\:\:{thank}\:{you}\:{very}\:{much} \\ $$
Answered by mr W last updated on 10/Jan/25
Commented by mr W last updated on 10/Jan/25
$${this}\:{is}\:{the}\:{time}\:{distance}\:{diagram}\: \\ $$$${for}\:{both}\:{trains}. \\ $$$$\frac{{x}}{\mathrm{4}}=\frac{\mathrm{4}+\mathrm{5}}{{x}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{4}×\mathrm{9}=\mathrm{36}\:\Rightarrow{x}=\mathrm{6}\:{h} \\ $$$$?=\mathrm{12}−\mathrm{6}=\mathrm{6} \\ $$$${i}.{e}.\:{the}\:{dawn}\:{breaks}\:{at}\:\mathrm{6}\:{o}'{clock}. \\ $$
Commented by mr W last updated on 10/Jan/25
$${additional}\:{question}: \\ $$$${say}\:{both}\:{trains}\:{take}\:{a}\:{pause}\:{of}\: \\ $$$$\mathrm{2}\:{hours}\:{and}\:{then}\:{go}\:{back}.\:{at}\:{what} \\ $$$${time}\:{can}\:{the}\:{friends}\:“{see}''\:{each}\: \\ $$$${other}\:{again}\:{through}\:{the}\:{windows}? \\ $$
Commented by mr W last updated on 10/Jan/25
Commented by mr W last updated on 10/Jan/25
$${say}\:{they}\:“{see}''\:{each}\:{other}\:{y}\:{hours}\: \\ $$$${after}\:\mathrm{23}^{\mathrm{00}} . \\ $$$$\frac{{y}}{\mathrm{15}−{y}}=\frac{\mathrm{5}−{y}}{\mathrm{5}+{y}} \\ $$$$\Rightarrow{y}=\mathrm{3} \\ $$$$?=\mathrm{23}^{\mathrm{00}} +\mathrm{3}=\mathrm{2}^{\mathrm{00}} \\ $$$${i}.{e}.\:{the}\:{friends}\:{see}\:{each}\:{other}\:{again}\: \\ $$$${at}\:\mathrm{2}\:{o}'{clock}\:{on}\:{next}\:{day}. \\ $$
Commented by mr W last updated on 10/Jan/25
$${following}\:{is}\:{an}\:{example}\:{how}\:{a}\:{time} \\ $$$${distance}\:{diagram}\:{for}\:{real}\:{train}\: \\ $$$${traffic}\:{may}\:{look}\:{like}. \\ $$
Commented by mr W last updated on 10/Jan/25
Commented by Ismoiljon_008 last updated on 10/Jan/25
$$\:\:{thank}\:{you}\:{very}\:{much} \\ $$