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A-boy-can-swim-with-a-speed-of-26m-s-in-still-water-He-wants-to-swim-across-a-150m-river-from-a-point-A-to-point-B-which-is-directly-opposite-the-other-side-of-the-river-The-river-flows-with-a-speed-




Question Number 30738 by NECx last updated on 25/Feb/18
A boy can swim with a speed of  26m/s in still water.He wants to  swim across a 150m river from  a point A to point B which is   directly opposite the other side  of the river.The river flows with  a speed of 10m/s.  i)if he always swim in the   direction parallel to AB,find how  far he lands downstream of B.  ii)In what direction relative to  the bank must he swim so as to  cross directly from A to B.
$${A}\:{boy}\:{can}\:{swim}\:{with}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{26}{m}/{s}\:{in}\:{still}\:{water}.{He}\:{wants}\:{to} \\ $$$${swim}\:{across}\:{a}\:\mathrm{150}{m}\:{river}\:{from} \\ $$$${a}\:{point}\:{A}\:{to}\:{point}\:{B}\:{which}\:{is}\: \\ $$$${directly}\:{opposite}\:{the}\:{other}\:{side} \\ $$$${of}\:{the}\:{river}.{The}\:{river}\:{flows}\:{with} \\ $$$${a}\:{speed}\:{of}\:\mathrm{10}{m}/{s}. \\ $$$$\left.{i}\right){if}\:{he}\:{always}\:{swim}\:{in}\:{the}\: \\ $$$${direction}\:{parallel}\:{to}\:{AB},{find}\:{how} \\ $$$${far}\:{he}\:{lands}\:{downstream}\:{of}\:{B}. \\ $$$$\left.{ii}\right){In}\:{what}\:{direction}\:{relative}\:{to} \\ $$$${the}\:{bank}\:{must}\:{he}\:{swim}\:{so}\:{as}\:{to} \\ $$$${cross}\:{directly}\:{from}\:{A}\:{to}\:{B}. \\ $$
Commented by NECx last updated on 25/Feb/18
please help
$${please}\:{help} \\ $$
Answered by mrW2 last updated on 26/Feb/18
assume the boy is a superman who  can really swim with a speed of  26 m/s (=93.6 km/h).  (i)  a=((150)/(26))×10=57.7 m  (ii)  cos θ=((10)/(26))  ⇒θ=cos^(−1) ((10)/(26))=67.4°
$${assume}\:{the}\:{boy}\:{is}\:{a}\:{superman}\:{who} \\ $$$${can}\:{really}\:{swim}\:{with}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{26}\:{m}/{s}\:\left(=\mathrm{93}.\mathrm{6}\:{km}/{h}\right). \\ $$$$\left({i}\right) \\ $$$${a}=\frac{\mathrm{150}}{\mathrm{26}}×\mathrm{10}=\mathrm{57}.\mathrm{7}\:{m} \\ $$$$\left({ii}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{10}}{\mathrm{26}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{10}}{\mathrm{26}}=\mathrm{67}.\mathrm{4}° \\ $$

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