Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com