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Question Number 69207 by Rio Michael last updated on 21/Sep/19
help please.    A river is 5m wide and flows at 3.0ms^(−1) . A man can swim at 2.0ms^(−1)   in still water. if he sets off at an angle of 90° to the bank  calculate  a) the mans time and velocity  b) his distance downstream from the starting point till  when he reaches the other side of the river bank  c) the actual distance he swims through the water.
$${help}\:{please}. \\ $$$$ \\ $$$${A}\:{river}\:{is}\:\mathrm{5}{m}\:{wide}\:{and}\:{flows}\:{at}\:\mathrm{3}.\mathrm{0}{ms}^{−\mathrm{1}} .\:{A}\:{man}\:{can}\:{swim}\:{at}\:\mathrm{2}.\mathrm{0}{ms}^{−\mathrm{1}} \\ $$$${in}\:{still}\:{water}.\:{if}\:{he}\:{sets}\:{off}\:{at}\:{an}\:{angle}\:{of}\:\mathrm{90}°\:{to}\:{the}\:{bank} \\ $$$${calculate} \\ $$$$\left.{a}\right)\:{the}\:{mans}\:{time}\:{and}\:{velocity} \\ $$$$\left.{b}\right)\:{his}\:{distance}\:{downstream}\:{from}\:{the}\:{starting}\:{point}\:{till} \\ $$$${when}\:{he}\:{reaches}\:{the}\:{other}\:{side}\:{of}\:{the}\:{river}\:{bank} \\ $$$$\left.{c}\right)\:{the}\:{actual}\:{distance}\:{he}\:{swims}\:{through}\:{the}\:{water}. \\ $$
Commented by mr W last updated on 21/Sep/19
when the river is only 5 m wide, i′ll  try to jump to the other side directly  instead of swim.
$${when}\:{the}\:{river}\:{is}\:{only}\:\mathrm{5}\:{m}\:{wide},\:{i}'{ll} \\ $$$${try}\:{to}\:{jump}\:{to}\:{the}\:{other}\:{side}\:{directly} \\ $$$${instead}\:{of}\:{swim}. \\ $$
Commented by Rio Michael last updated on 21/Sep/19
lol man its just a question
$${lol}\:{man}\:{its}\:{just}\:{a}\:{question} \\ $$
Commented by Prithwish sen last updated on 21/Sep/19
ha ha ha ! Yes mr.W sir I agree with you.
$$\mathrm{ha}\:\mathrm{ha}\:\mathrm{ha}\:!\:\mathrm{Yes}\:\mathrm{mr}.\mathrm{W}\:\mathrm{sir}\:\mathrm{I}\:\mathrm{agree}\:\mathrm{with}\:\mathrm{you}. \\ $$
Answered by mr W last updated on 21/Sep/19
a)  t=(5/2)=2.5 sec  v=(√(2^2 +3^2 ))=(√(13)) m/s  b)  2.5×3=7.5 m  c)  (√(5^2 +7.5^2 ))=((5(√(13)))/2)≈9 m
$$\left.{a}\right) \\ $$$${t}=\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{2}.\mathrm{5}\:{sec} \\ $$$${v}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\mathrm{13}}\:{m}/{s} \\ $$$$\left.{b}\right) \\ $$$$\mathrm{2}.\mathrm{5}×\mathrm{3}=\mathrm{7}.\mathrm{5}\:{m} \\ $$$$\left.{c}\right) \\ $$$$\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{7}.\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{2}}\approx\mathrm{9}\:{m} \\ $$
Commented by Rio Michael last updated on 21/Sep/19
thanks sir
$${thanks}\:{sir} \\ $$

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