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Two-swimmers-leave-point-A-on-one-bank-of-the-river-to-reach-point-B-lying-right-across-the-other-bank-One-of-them-crosses-the-river-along-the-straight-line-AB-while-the-other-swims-at-right-angle-to

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Question Number 19654 by Tinkutara last updated on 13/Aug/17
Two swimmers leave point A on one bank of the river to reach point B lying right across the other bank. One of them crosses the river along the straight line AB while the other swims at right angle to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity v of his walking if both swimmers reached the destination simultaneously? (The stream velocity v_0 = 2 km/h and the velocity v′ of each swimmer with respect to still water is 2.5 km/h).
$$\mathrm{Two}\:\mathrm{swimmers}\:\mathrm{leave}\:\mathrm{point}\:{A}\:\mathrm{on}\:\mathrm{one} \\ $$$$\mathrm{bank}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{point}\:{B}\:\mathrm{lying} \\ $$$$\mathrm{right}\:\mathrm{across}\:\mathrm{the}\:\mathrm{other}\:\mathrm{bank}.\:\mathrm{One}\:\mathrm{of} \\ $$$$\mathrm{them}\:\mathrm{crosses}\:\mathrm{the}\:\mathrm{river}\:\mathrm{along}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{line}\:{AB}\:\mathrm{while}\:\mathrm{the}\:\mathrm{other}\:\mathrm{swims}\:\mathrm{at}\:\mathrm{right} \\ $$$$\mathrm{angle}\:\mathrm{to}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{and}\:\mathrm{then}\:\mathrm{walks}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{that}\:\mathrm{he}\:\mathrm{has}\:\mathrm{been}\:\mathrm{carried}\:\mathrm{away} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{to}\:\mathrm{get}\:\mathrm{to}\:\mathrm{point}\:{B}.\:\mathrm{What} \\ $$$$\mathrm{was}\:\mathrm{the}\:\mathrm{velocity}\:{v}\:\mathrm{of}\:\mathrm{his}\:\mathrm{walking}\:\mathrm{if}\:\mathrm{both} \\ $$$$\mathrm{swimmers}\:\mathrm{reached}\:\mathrm{the}\:\mathrm{destination} \\ $$$$\mathrm{simultaneously}?\:\left(\mathrm{The}\:\mathrm{stream}\:\mathrm{velocity}\right. \\ $$$${v}_{\mathrm{0}} \:=\:\mathrm{2}\:\mathrm{km}/\mathrm{h}\:\mathrm{and}\:\mathrm{the}\:\mathrm{velocity}\:{v}'\:\mathrm{of}\:\mathrm{each} \\ $$$$\mathrm{swimmer}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{still}\:\mathrm{water}\:\mathrm{is} \\ $$$$\left.\mathrm{2}.\mathrm{5}\:\mathrm{km}/\mathrm{h}\right). \\ $$
Commented by ajfour last updated on 14/Aug/17
Commented by ajfour last updated on 14/Aug/17
T=(d/(vcos θ)) ; T=(d/v)+((u(d/v))/v_w ) ; vsin θ=u . ⇒ sin θ=(u/v)=(2/(2.5)) =(4/5) , so cos θ=(3/5) T=((5d)/(3v))=(d/v)+((ud)/(vv_w )) ⇒ (5/3)=1+(u/v_w ) ⇒ (u/v_w ) =(2/3) as u=2km/h , v_w =3km/h .
$$\mathrm{T}=\frac{\mathrm{d}}{\mathrm{vcos}\:\theta}\:\:;\:\:\mathrm{T}=\frac{\mathrm{d}}{\mathrm{v}}+\frac{\mathrm{u}\left(\mathrm{d}/\mathrm{v}\right)}{\mathrm{v}_{\mathrm{w}} }\:; \\ $$$$\:\:\:\:\:\:\:\mathrm{vsin}\:\theta=\mathrm{u}\:. \\ $$$$\Rightarrow\:\mathrm{sin}\:\theta=\frac{\mathrm{u}}{\mathrm{v}}=\frac{\mathrm{2}}{\mathrm{2}.\mathrm{5}}\:=\frac{\mathrm{4}}{\mathrm{5}}\:,\:\mathrm{so}\:\mathrm{cos}\:\theta=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{T}=\frac{\mathrm{5d}}{\mathrm{3v}}=\frac{\mathrm{d}}{\mathrm{v}}+\frac{\mathrm{ud}}{\mathrm{vv}_{\mathrm{w}} } \\ $$$$\Rightarrow\:\:\frac{\mathrm{5}}{\mathrm{3}}=\mathrm{1}+\frac{\mathrm{u}}{\mathrm{v}_{\mathrm{w}} }\:\:\:\:\Rightarrow\:\:\frac{\mathrm{u}}{\mathrm{v}_{\mathrm{w}} }\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{as}\:\mathrm{u}=\mathrm{2km}/\mathrm{h}\:,\:\:\:\:\:\mathrm{v}_{\mathrm{w}} =\mathrm{3km}/\mathrm{h}\:\:. \\ $$
Commented by Tinkutara last updated on 14/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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