Question Number 29907 by ajfour last updated on 13/Feb/18
Commented by ajfour last updated on 13/Feb/18
$${A}\:{boy}\:{swims}\:{at}\:{speed}\:\boldsymbol{{v}}\:{in}\:{still} \\ $$$${water}\:,\:{walks}\:{on}\:{ground}\:{with} \\ $$$${speed}\:\boldsymbol{\mathrm{w}}\:>\:\boldsymbol{{v}}\:.\:\:{If}\:{he}\:{needs}\:{to}\:{cross} \\ $$$${a}\:{river}\:{of}\:{width}\:\boldsymbol{{b}},\:{water}\:{flowing}\: \\ $$$${at}\:{speed}\:\boldsymbol{{u}},\:{find}\:{the}\:{minimum} \\ $$$${time}\:{in}\:{which}\:{he}\:{can}\:{reach}\:{B}, \\ $$$${starting}\:{from}\:{A}. \\ $$
Commented by 33 last updated on 13/Feb/18
$${sir},\:{isnt}\:{there}\:{any}\:{relation}\:{given} \\ $$$${between}\:{u}\:,\:{v}\:\:\&\:{w}\:\:{other}\:{than} \\ $$$${the}\:{inequality}?\:{like}\:{the}\:{ratio}\:{or}\: \\ $$$${something}? \\ $$
Commented by 33 last updated on 13/Feb/18
$${i}\:{think}\:{it}\:{might}\:{be}\:{required}. \\ $$
Commented by ajfour last updated on 13/Feb/18
$${Answer}\:{is}\:{to}\:{be}\:{obtained}\:{in} \\ $$$${terms}\:{of}\:{u},\:{v},\:{w},\:{b}\:. \\ $$
Commented by 33 last updated on 13/Feb/18
$${ok} \\ $$
Answered by mrW2 last updated on 13/Feb/18
![v_x =u−v sin θ v_y =v cos θ t_1 =(b/v_y )=(b/(v cos θ)) t_2 =((v_x t_1 )/w) t=t_1 +t_2 =(1+(v_x /w))t_1 =(1+((u−v sin θ)/w))(b/(v cos θ)) t=(b/w)×(((((w+u)/v)−sin θ))/(cos θ))=(b/w)(((β−sin θ)/(cos θ))) (dt/dθ)=(b/w)[−1+(((β−sin θ)sin θ)/(cos^2 θ))]=((βb)/w)×((sin θ−(1/β))/(cos^2 θ)) since v<w, (1/β)=(v/(w+u))<1 (dt/dθ)=0⇒sin β−(1/β)=0 ⇒sin θ=(1/β)=(v/(w+u)) ⇒θ=sin^(−1) ((v/(w+u))) cos θ=(√(1−((v/(w+u)))^2 )) t_(min) =((b(((w+u)/v)−(v/(w+u))))/(w(√(1−((v/(w+u)))^2 ))))=((b(√((w+u+v)(w+u−v))))/(wv))](https://www.tinkutara.com/question/Q29925.png)
$${v}_{{x}} ={u}−{v}\:\mathrm{sin}\:\theta \\ $$$${v}_{{y}} ={v}\:\mathrm{cos}\:\theta \\ $$$${t}_{\mathrm{1}} =\frac{{b}}{{v}_{{y}} }=\frac{{b}}{{v}\:\mathrm{cos}\:\theta} \\ $$$${t}_{\mathrm{2}} =\frac{{v}_{{x}} {t}_{\mathrm{1}} }{{w}} \\ $$$${t}={t}_{\mathrm{1}} +{t}_{\mathrm{2}} =\left(\mathrm{1}+\frac{{v}_{{x}} }{{w}}\right){t}_{\mathrm{1}} =\left(\mathrm{1}+\frac{{u}−{v}\:\mathrm{sin}\:\theta}{{w}}\right)\frac{{b}}{{v}\:\mathrm{cos}\:\theta} \\ $$$${t}=\frac{{b}}{{w}}×\frac{\left(\frac{{w}+{u}}{{v}}−\mathrm{sin}\:\theta\right)}{\mathrm{cos}\:\theta}=\frac{{b}}{{w}}\left(\frac{\beta−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right) \\ $$$$\frac{{dt}}{{d}\theta}=\frac{{b}}{{w}}\left[−\mathrm{1}+\frac{\left(\beta−\mathrm{sin}\:\theta\right)\mathrm{sin}\:\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta}\right]=\frac{\beta{b}}{{w}}×\frac{\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\beta}}{\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${since}\:{v}<{w},\:\frac{\mathrm{1}}{\beta}=\frac{{v}}{{w}+{u}}<\mathrm{1} \\ $$$$\frac{{dt}}{{d}\theta}=\mathrm{0}\Rightarrow\mathrm{sin}\:\beta−\frac{\mathrm{1}}{\beta}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\beta}=\frac{{v}}{{w}+{u}} \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{v}}{{w}+{u}}\right) \\ $$$$\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\left(\frac{{v}}{{w}+{u}}\right)^{\mathrm{2}} } \\ $$$${t}_{{min}} =\frac{{b}\left(\frac{{w}+{u}}{{v}}−\frac{{v}}{{w}+{u}}\right)}{{w}\sqrt{\mathrm{1}−\left(\frac{{v}}{{w}+{u}}\right)^{\mathrm{2}} }}=\frac{{b}\sqrt{\left({w}+{u}+{v}\right)\left({w}+{u}−{v}\right)}}{{wv}} \\ $$
Commented by ajfour last updated on 13/Feb/18
$${Excellent}\:{Sir},\:{and}\:{even}\:{quick}\:! \\ $$
Commented by 33 last updated on 13/Feb/18
$${nice}\:{one}\:{sir} \\ $$