If the breadth of the lake is l and velocity of boat is v _b . Time in going and coming back on a quite day

\(t_Q = \frac{l}{v_b} + \frac{l}{v_b} = \frac{2l}{v_b} \quad \ldots (i)\)

Now if v _a is the velocity of air- current then time taken in going across the lake,

\(t_1 = \frac{1}{v_a + v_b}\) [As current helps the motion]

and time taken in coming back \(t_2 = \frac{1}{v_b - v_a}\)

[As current opposes the motion]

So \(t_R = t_1 + t_2 = \frac{2l}{v_b [1 - (v_a / v_b)^2]} \quad \ldots \text{(ii)}\)

From equation (i) and (ii)

\(\frac{t_{R}}{t_{Q}} = \frac{1}{\left[ 1 - \left( v_{a} / v_{b} \right)^{2} \right]} > 1 \left[ \text{as } 1 - \frac{v_{a}^{2}}{v_{b}^{2}} < 1 \right]\) i.e. t_{R} > t_{Q}

i.e. time taken to complete the journey on quite day is lesser than that on rough day.