question_answer

A steam boat goes across a lake and comes back : (a) on a quiet day when the water is still and (b) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey backward. If the speed of launch on both days was same, in which case will it complete the journey in lesser time?

Answer:

                Let l be the width of lake and v be the velocity of steam boat. On a quiet day, time taken by steam boat in going and coming back is \[{{\text{t}}_{\text{Q}}}=\frac{l}{\upsilon }+\frac{l}{\upsilon }=\frac{2l}{\upsilon }\]                                                                                                                        ....(i) On a rough day, let \[\upsilon '\] be the velocity of air current. As in going across the lake, the air current helps the motion, so time taken is \[{{\text{t}}_{\text{1}}}=\frac{l}{\upsilon +\upsilon '}\] In coming back, as the air current opposes the motion, so time taken is \[{{\text{t}}_{2}}=\frac{l}{\upsilon -\upsilon '}\] Total time in going and coming back on a rough day \[{{\text{t}}_{\text{r}}}={{\text{t}}_{\text{1}}}+\text{ }{{\text{t}}_{\text{2}}}=\frac{2l\upsilon }{\left( {{\upsilon }^{2}}-\upsilon {{'}^{2}} \right)}=\frac{2l}{\upsilon [1-{{\left( \upsilon '/\upsilon  \right)}^{2}}]}\]                                                                             ....(ii) From (i) and (ii), we have                                 \[\frac{{{t}_{R}}}{{{t}_{Q}}}=\frac{1}{[1-{{(\upsilon '/\upsilon )}^{2}}]}>1\] or \[{{\text{t}}_{\text{R}}}>\text{ }{{\text{t}}_{\text{Q}}}\] Therefore the time taken to complete the journey on quiet day is less than on a rough day.