A steamboat goes across a lake and comes back | vedclass

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(B) Let the breadth of the lake be $l$ and the velocity of the boat be $v_b$. On a quiet day,the time taken in going and coming back is:$t_Q = \frac{l

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Case $(a)$

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(B) Let the breadth of the lake be $l$ and the velocity of the boat be $v_b$. On a quiet day,the time taken in going and coming back is:$t_Q = \frac{l}{v_b} + \frac{l}{v_b} = \frac{2l}{v_b}$ .....$(i)$Now,if $v_a$ is the velocity of the air current,the time taken in going across the lake is:$t_1 = \frac{l}{v_b + v_a}$ [As the current helps the motion]And the time taken in coming back is:$t_2 = \frac{l}{v_b - v_a}$ [As the current opposes the motion]So,the total time on a rough day is:$t_R = t_1 + t_2 = \frac{l(v_b - v_a) + l(v_b + v_a)}{v_b^2 - v_a^2} = \frac{2lv_b}{v_b^2 - v_a^2} = \frac{2l}{v_b[1 - (v_a/v_b)^2]}$ .....(ii)Comparing equations $(i)$ and (ii):$\frac{t_R}{t_Q} = \frac{1}{1 - (v_a/v_b)^2}$. Since $1 - (v_a/v_b)^2  1$,which means $t_R > t_Q$.Therefore,the time taken to complete the journey on a quiet day is less than that on a rough day.

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