Question Number 59323 by Tawa1 last updated on 08/May/19 | ||
Answered by tanmay last updated on 08/May/19 | ||
$${when}\:{cab}\:{catch}\:{bus}\:{that}\:{both}\:{cover}\:{same}\:{distance} \\ $$$${d}_{{cab}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×{t}^{\mathrm{2}} \\ $$$${d}_{{bus}} =\mathrm{30}\left({t}+\mathrm{5}\right) \\ $$$${t}^{\mathrm{2}} =\mathrm{30}{t}+\mathrm{150} \\ $$$${t}^{\mathrm{2}} −\mathrm{30}{t}−\mathrm{150}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}×{t}×\mathrm{15}+\mathrm{225}=\mathrm{150}+\mathrm{225} \\ $$$$\left({t}−\mathrm{15}\right)^{\mathrm{2}} =\mathrm{375} \\ $$$${t}=\mathrm{15}+\mathrm{5}\sqrt{\mathrm{15}}\: \\ $$$${t}\approx\mathrm{34}.\mathrm{36}{sec} \\ $$$${d}_{{cab}} ={t}^{\mathrm{2}} \\ $$$${d}_{{cab}} \approx\mathrm{1180}.\mathrm{61}\:{meter} \\ $$ | ||
Commented by Tawa1 last updated on 08/May/19 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||