Question Number 182613 by sciencestudent last updated on 11/Dec/22
Answered by Acem last updated on 12/Dec/22
$$\:{Dis}.\:=\:{v}_{\mathrm{1}} \:\left({t}+\mathrm{4}\right)=\:{v}_{\mathrm{2}} \:\left({t}−\mathrm{2}\right) \\ $$$$\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{3}}\:\left({t}+\mathrm{4}\right)=\:\frac{\mathrm{5}}{\mathrm{12}}\:\left({t}−\mathrm{2}\right)\:,\:{t}=\:\mathrm{26}\:{min}\:_{\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{12}}\:\:{the}\:{same}} ^{{Exp}\:\frac{\mathrm{1}}{\mathrm{3}}=\:\frac{{v}_{\mathrm{1}} }{\mathrm{60}}\:{conv}.\:{to}\:{min}} \\ $$$$\:{Dis}=\:{v}_{\mathrm{1}} \:{T}=\:\mathrm{20}×\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{10}\:{km} \\ $$
Answered by a.lgnaoui last updated on 12/Dec/22
$${soit}\:{x}\::{distsnce}\:{to}\:{school} \\ $$$${t}_{\mathrm{1}} ={temps}\:{mis}\:\:{a}\:{la}\:{vitesse}\:{v}_{\mathrm{1}} =\mathrm{20}{km}/{h} \\ $$$${t}_{\mathrm{1}} ={t}+\mathrm{4}/\mathrm{60}\:\:\:\left({t}:{en}\:{heures}\right) \\ $$$${t}_{\mathrm{2}} ={temps}\:{mis}\:{a}\:{la}\:{vitesse}\:{v}_{\mathrm{2}} =\mathrm{25}{km}/{h} \\ $$$$\:\:{t}_{\mathrm{2}} ={t}−\frac{\mathrm{2}}{\mathrm{60}}\:{v} \\ $$$$\:\:{t}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{15}}={t}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{30}}\:\:\:\Rightarrow{t}_{\mathrm{1}} −{t}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:\:{t}_{\mathrm{1}} ={t}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:\:{x}={v}_{\mathrm{1}} ×{t}_{\mathrm{1}} ={v}_{\mathrm{2}} ×{t}_{\mathrm{2}} \Rightarrow\frac{{t}_{\mathrm{2}} }{{t}_{\mathrm{1}} }=\frac{\mathrm{4}}{\mathrm{5}}\:\:{t}_{\mathrm{1}} =\frac{\mathrm{5}{t}_{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\frac{\mathrm{5}{t}_{\mathrm{2}} }{\mathrm{4}}={t}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{10}}\:\:\:\:{t}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{5}}{h} \\ $$$${x}={v}_{\mathrm{2}} ×{t}_{\mathrm{2}} =\mathrm{25}×\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:{x}=\mathrm{10}{km} \\ $$