Question Number 108175 by bemath last updated on 15/Aug/20
Commented by bemath last updated on 15/Aug/20
$${a}\:{man}\:\mathrm{6}\:{feet}\:{tall}\:{walks}\:{at}\:{a}\:{rate}\:\mathrm{6}\:{feet}\: \\ $$$${per}\:{second}\:{away}\:{from}\:{a}\:{light}\:{that}\: \\ $$$${is}\:\mathrm{15}\:{feet}\:{above}\:{the}\:{ground}. \\ $$$${When}\:{he}\:{is}\:\mathrm{10}\:{feet}\:{from}\:{the}\:{base}\:{of}\:{the}\:{light} \\ $$$${at}\:{what}\:{rate}\:{is}\:{the}\:{tip}\:{of}\:{his}\:{shadow} \\ $$$${moving}? \\ $$
Commented by john santu last updated on 15/Aug/20
Commented by john santu last updated on 15/Aug/20
$$\:\:\frac{{s}_{\mathrm{1}} }{{s}_{\mathrm{1}} +{s}_{\mathrm{2}} }\:=\:\frac{\mathrm{6}}{\mathrm{15}}\:\Rightarrow\frac{{s}_{\mathrm{1}} }{\mathrm{10}+{s}_{\mathrm{1}} }=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{5}{s}_{\mathrm{1}} =\:\mathrm{20}+\:\mathrm{2}{s}_{\mathrm{1}} \Rightarrow{s}_{\mathrm{1}} =\:\frac{\mathrm{20}}{\mathrm{3}}\:{feet} \\ $$$${v}\:=\:\frac{{s}_{\mathrm{1}} +{s}_{\mathrm{2}} }{{t}}\:=\:\frac{\mathrm{10}+\frac{\mathrm{20}}{\mathrm{3}}}{\left(\frac{\mathrm{5}}{\mathrm{3}}\right)}=\frac{\mathrm{50}}{\mathrm{5}}=\mathrm{10}\:{feet}/{second} \\ $$
Answered by mr W last updated on 15/Aug/20
$${x}={position}\:{of}\:{the}\:{man} \\ $$$${s}={posistion}\:{of}\:{the}\:{tip}\:{of}\:{the}\:{shadow} \\ $$$${l}={length}\:{of}\:{the}\:{shadow} \\ $$$$\frac{{s}−{x}}{\mathrm{6}}=\frac{{s}}{\mathrm{15}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{5}}{\mathrm{3}}{x} \\ $$$${l}={s}−{x}=\frac{\mathrm{2}}{\mathrm{3}}{x} \\ $$$$\frac{{ds}}{{dt}}=\frac{\mathrm{5}}{\mathrm{3}}×\frac{{dx}}{{dt}}=\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{6}=\mathrm{10}\:{ft}/{sec} \\ $$$$\frac{{dl}}{{dt}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{{dx}}{{dt}}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{6}=\mathrm{4}\:{ft}/{sec} \\ $$
Commented by bemath last updated on 15/Aug/20
$${thank}\:{you}\:{mr}\:\mathcal{W} \\ $$