Menu Close

Question-182582




Question Number 182582 by sciencestudent last updated on 11/Dec/22
Answered by Acem last updated on 11/Dec/22
a•Mathematical method: ; Distance= v.t   80= 2×2×4×5 ; diff. of time= 1 hr we take 4, 5   80= 4×20= 5×16   daily: v= 16 km/hr   this day: v= 20 km/hr
$${a}\bullet{Mathematical}\:{method}:\:;\:{Distance}=\:{v}.{t} \\ $$$$\:\mathrm{80}=\:\mathrm{2}×\mathrm{2}×\mathrm{4}×\mathrm{5}\:;\:{diff}.\:{of}\:{time}=\:\mathrm{1}\:{hr}\:{we}\:{take}\:\mathrm{4},\:\mathrm{5} \\ $$$$\:\mathrm{80}=\:\mathrm{4}×\mathrm{20}=\:\mathrm{5}×\mathrm{16} \\ $$$$\:{daily}:\:{v}=\:\mathrm{16}\:{km}/{hr} \\ $$$$\:{this}\:{day}:\:{v}=\:\mathrm{20}\:{km}/{hr} \\ $$
Answered by Acem last updated on 11/Dec/22
b• Phisical method   Distance= v.t= v′.t′= (v+4).(t−1)    v.t= v.t − v +4(t−1)   v= 4(t−1)   ((dis.)/t)= 4(t−1) ; dis.=80   20= t(t−1)   t_(s1) = −4  t_(s2) = 5 hr   v= 16 km/hr ′normal days   v′= 20 km/hr  ; t′= 4 hr ′this day
$${b}\bullet\:{Phisical}\:{method} \\ $$$$\:{Distance}=\:{v}.{t}=\:{v}'.{t}'=\:\left({v}+\mathrm{4}\right).\left({t}−\mathrm{1}\right) \\ $$$$\:\:{v}.{t}=\:{v}.{t}\:−\:{v}\:+\mathrm{4}\left({t}−\mathrm{1}\right) \\ $$$$\:{v}=\:\mathrm{4}\left({t}−\mathrm{1}\right) \\ $$$$\:\frac{{dis}.}{{t}}=\:\mathrm{4}\left({t}−\mathrm{1}\right)\:;\:{dis}.=\mathrm{80} \\ $$$$\:\mathrm{20}=\:{t}\left({t}−\mathrm{1}\right) \\ $$$$\:{t}_{{s}\mathrm{1}} =\:\cancel{−\mathrm{4}}\:\:{t}_{{s}\mathrm{2}} =\:\mathrm{5}\:{hr} \\ $$$$\:{v}=\:\mathrm{16}\:{km}/{hr}\:'{normal}\:{days} \\ $$$$\:{v}'=\:\mathrm{20}\:{km}/{hr}\:\:;\:{t}'=\:\mathrm{4}\:{hr}\:'{this}\:{day} \\ $$$$\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *