Question Number 182582 by sciencestudent last updated on 11/Dec/22
Answered by Acem last updated on 11/Dec/22
$${a}\bullet{Mathematical}\:{method}:\:;\:{Distance}=\:{v}.{t} \\ $$$$\:\mathrm{80}=\:\mathrm{2}×\mathrm{2}×\mathrm{4}×\mathrm{5}\:;\:{diff}.\:{of}\:{time}=\:\mathrm{1}\:{hr}\:{we}\:{take}\:\mathrm{4},\:\mathrm{5} \\ $$$$\:\mathrm{80}=\:\mathrm{4}×\mathrm{20}=\:\mathrm{5}×\mathrm{16} \\ $$$$\:{daily}:\:{v}=\:\mathrm{16}\:{km}/{hr} \\ $$$$\:{this}\:{day}:\:{v}=\:\mathrm{20}\:{km}/{hr} \\ $$
Answered by Acem last updated on 11/Dec/22
$${b}\bullet\:{Phisical}\:{method} \\ $$$$\:{Distance}=\:{v}.{t}=\:{v}'.{t}'=\:\left({v}+\mathrm{4}\right).\left({t}−\mathrm{1}\right) \\ $$$$\:\:{v}.{t}=\:{v}.{t}\:−\:{v}\:+\mathrm{4}\left({t}−\mathrm{1}\right) \\ $$$$\:{v}=\:\mathrm{4}\left({t}−\mathrm{1}\right) \\ $$$$\:\frac{{dis}.}{{t}}=\:\mathrm{4}\left({t}−\mathrm{1}\right)\:;\:{dis}.=\mathrm{80} \\ $$$$\:\mathrm{20}=\:{t}\left({t}−\mathrm{1}\right) \\ $$$$\:{t}_{{s}\mathrm{1}} =\:\cancel{−\mathrm{4}}\:\:{t}_{{s}\mathrm{2}} =\:\mathrm{5}\:{hr} \\ $$$$\:{v}=\:\mathrm{16}\:{km}/{hr}\:'{normal}\:{days} \\ $$$$\:{v}'=\:\mathrm{20}\:{km}/{hr}\:\:;\:{t}'=\:\mathrm{4}\:{hr}\:'{this}\:{day} \\ $$$$\: \\ $$