Question Number 13570 by Tinkutara last updated on 21/May/17
$$\mathrm{A}\:\mathrm{motor}\:\mathrm{boat}\:\mathrm{going}\:\mathrm{downstream} \\ $$$$\mathrm{overcomes}\:\mathrm{a}\:\mathrm{float}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:{A}.\:\mathrm{60}\:\mathrm{minutes} \\ $$$$\mathrm{later}\:\mathrm{it}\:\mathrm{turns}\:\mathrm{and}\:\mathrm{after}\:\mathrm{some}\:\mathrm{time}\:\mathrm{passes} \\ $$$$\mathrm{the}\:\mathrm{float}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{12}\:\mathrm{km}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{point}\:{A}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{stream} \\ $$$$\left(\mathrm{assuming}\:\mathrm{constant}\:\mathrm{velocity}\:\mathrm{for}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{boat}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}\right) \\ $$
Answered by ajfour last updated on 21/May/17
![boat moves from A for t_1 =1h with velocity u+v. [u=stream velocity] it turns back, moving with speed v−u , for time =t_2 , whereupon it meets the float again. the float has come down a distance d=12km . u(t_1 +t_2 )=d .....(i) (u+v)t_1 −(v−u)t_2 =d ....(ii) from eq. (ii) u(t_1 +t_2 )+v(t_1 −t_2 )=d using (i) in (ii), t_2 =t_1 =1h so, u=(d/((t_1 +t_2 )))=((12km)/(2h)) u=6km/h .](https://www.tinkutara.com/question/Q13574.png)
$$\mathrm{boat}\:\mathrm{moves}\:\mathrm{from}\:\mathrm{A}\:\mathrm{for}\:\boldsymbol{\mathrm{t}}_{\mathrm{1}} =\mathrm{1}\boldsymbol{\mathrm{h}} \\ $$$$\mathrm{with}\:\mathrm{velocity}\:\mathrm{u}+\mathrm{v}.\:\:\:\left[\mathrm{u}=\mathrm{stream}\:\mathrm{velocity}\right] \\ $$$$\mathrm{it}\:\mathrm{turns}\:\mathrm{back},\:\mathrm{moving}\:\mathrm{with}\:\mathrm{speed} \\ $$$$\mathrm{v}−\mathrm{u}\:,\:\mathrm{for}\:\mathrm{time}\:=\boldsymbol{\mathrm{t}}_{\mathrm{2}} \:,\:\mathrm{whereupon} \\ $$$$\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{float}\:\mathrm{again}. \\ $$$$\mathrm{the}\:\mathrm{float}\:\mathrm{has}\:\mathrm{come}\:\mathrm{down}\:\mathrm{a} \\ $$$$\mathrm{distance}\:\boldsymbol{\mathrm{d}}=\mathrm{12}\boldsymbol{\mathrm{km}}\:. \\ $$$$\boldsymbol{\mathrm{u}}\left(\boldsymbol{\mathrm{t}}_{\mathrm{1}} +\boldsymbol{\mathrm{t}}_{\mathrm{2}} \right)=\boldsymbol{\mathrm{d}}\:\:\:\:\:\:\:\:\:\:\:\:…..\left(\mathrm{i}\right) \\ $$$$\left(\boldsymbol{\mathrm{u}}+\boldsymbol{\mathrm{v}}\right)\boldsymbol{\mathrm{t}}_{\mathrm{1}} −\left(\boldsymbol{\mathrm{v}}−\boldsymbol{\mathrm{u}}\right)\boldsymbol{\mathrm{t}}_{\mathrm{2}} =\boldsymbol{\mathrm{d}}\:\:\:\:\:\:\:….\left(\mathrm{ii}\right) \\ $$$$\mathrm{from}\:\mathrm{eq}.\:\left(\mathrm{ii}\right) \\ $$$$\boldsymbol{\mathrm{u}}\left(\boldsymbol{\mathrm{t}}_{\mathrm{1}} +\boldsymbol{\mathrm{t}}_{\mathrm{2}} \right)+\boldsymbol{\mathrm{v}}\left(\boldsymbol{\mathrm{t}}_{\mathrm{1}} −\boldsymbol{\mathrm{t}}_{\mathrm{2}} \right)=\boldsymbol{\mathrm{d}} \\ $$$$\mathrm{using}\:\left(\mathrm{i}\right)\:\mathrm{in}\:\left(\mathrm{ii}\right), \\ $$$$\boldsymbol{\mathrm{t}}_{\mathrm{2}} =\boldsymbol{\mathrm{t}}_{\mathrm{1}} =\mathrm{1h} \\ $$$$\mathrm{so},\:\:\:\boldsymbol{\mathrm{u}}=\frac{\boldsymbol{\mathrm{d}}}{\left(\boldsymbol{\mathrm{t}}_{\mathrm{1}} +\boldsymbol{\mathrm{t}}_{\mathrm{2}} \right)}=\frac{\mathrm{12km}}{\mathrm{2h}} \\ $$$$\boldsymbol{\mathrm{u}}=\mathrm{6}\boldsymbol{\mathrm{km}}/\boldsymbol{\mathrm{h}}\:. \\ $$
Commented by Tinkutara last updated on 21/May/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$