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Question-114615




Question Number 114615 by I want to learn more last updated on 19/Sep/20
Answered by Olaf last updated on 19/Sep/20
t is the scheduled time  x = v_1 t_1  = 45(t+(1/(20))) (1)  x = v_2 t_2  = 60(t−(3/(20))) (2)  ((1/(20)) h = 3 min and (3/(20)) h = 9 min)  (1)−(3/4)×(2) :  x−(3/4)x = 45(t+(1/(20)))−(3/4)×60(t−(3/(20)))  (1/4)x = ((45)/(20))+((45×3)/(20)) = 9 km  ⇒ x = 4×9 = 36 km  with (1) : t+(1/(20)) = (x/(45)) = ((36)/(45)) = (4/5)  t = (4/5)−(1/(20)) = (3/4) h = 45 min
$${t}\:\mathrm{is}\:\mathrm{the}\:\mathrm{scheduled}\:\mathrm{time} \\ $$$${x}\:=\:{v}_{\mathrm{1}} {t}_{\mathrm{1}} \:=\:\mathrm{45}\left({t}+\frac{\mathrm{1}}{\mathrm{20}}\right)\:\left(\mathrm{1}\right) \\ $$$${x}\:=\:{v}_{\mathrm{2}} {t}_{\mathrm{2}} \:=\:\mathrm{60}\left({t}−\frac{\mathrm{3}}{\mathrm{20}}\right)\:\left(\mathrm{2}\right) \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{20}}\:{h}\:=\:\mathrm{3}\:{min}\:\mathrm{and}\:\frac{\mathrm{3}}{\mathrm{20}}\:{h}\:=\:\mathrm{9}\:{min}\right) \\ $$$$\left(\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{4}}×\left(\mathrm{2}\right)\:: \\ $$$${x}−\frac{\mathrm{3}}{\mathrm{4}}{x}\:=\:\mathrm{45}\left({t}+\frac{\mathrm{1}}{\mathrm{20}}\right)−\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{60}\left({t}−\frac{\mathrm{3}}{\mathrm{20}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{x}\:=\:\frac{\mathrm{45}}{\mathrm{20}}+\frac{\mathrm{45}×\mathrm{3}}{\mathrm{20}}\:=\:\mathrm{9}\:{km} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{4}×\mathrm{9}\:=\:\mathrm{36}\:{km} \\ $$$$\mathrm{with}\:\left(\mathrm{1}\right)\::\:{t}+\frac{\mathrm{1}}{\mathrm{20}}\:=\:\frac{{x}}{\mathrm{45}}\:=\:\frac{\mathrm{36}}{\mathrm{45}}\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${t}\:=\:\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{20}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:{h}\:=\:\mathrm{45}\:{min} \\ $$$$ \\ $$
Commented by I want to learn more last updated on 20/Sep/20
Thanks sir, i appreciate
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate} \\ $$

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