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InterviewSolution
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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 same, in which case will it complete the journey in lesser time? |
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Answer» If the length of the lake is L and the velocity of boat is V, time taken in going and coming back on a quiet day: `t_(Q) = (L)/(V) + (L)/(V) = (2L)/(V)` ....(i) Now if `v` is the velocity of air-current, then in going across the lake `t_(1) = (L)/((V + v))` [as current helps the motion] and time taken in coming back `t_(2) = (L)/((V - v))` [as current opposes the motion] So, `t_(R) = t_(1) + t_(2) = (2LV)/((V^(2) - v^(2))) = (2L)/(V[1 - (v//V)^(2)])`....(ii) So, from Eqns. (i) and (ii), `(t_(R))/(t_(Q)) = (1)/([1 - (v//V)^(2)]) gt 1` [as `1 - (v//V)^(2) lt 1`] i.e., `t_(R) gt t_(Q)` i.e., time taken to complete the hourney on a quiet day is less than on a rough day. |
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