Question Number 103422 by aurpeyz last updated on 15/Jul/20
Commented by Worm_Tail last updated on 15/Jul/20
$${distance}\:{moved}\:{by}\:{car}=\mathrm{60}{t} \\ $$$${distance}\:{moved}\:{by}\:{A}=\mathrm{40}{t} \\ $$$${distance}\:{moved}\:{by}\:{B}=\mathrm{40}{t} \\ $$$${total}\:{distance}\:{moved}\:{by}\:{car}\:{andB}=\mathrm{50}=\mathrm{100}{t} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${d}_{{car}} =\mathrm{60}×\mathrm{0}.\mathrm{5}=\mathrm{30}{km}\:\:\:\:\:{d}_{{A}} =\mathrm{40}×\mathrm{0}.\mathrm{5}=\mathrm{20}{km} \\ $$$${dcar}−{d}_{{A}} =\mathrm{10}{km} \\ $$$$ \\ $$$$ \\ $$
Commented by aurpeyz last updated on 15/Jul/20
$$\mathrm{thanks}\:\mathrm{sir} \\ $$