Question

A steam boat goes across a lake and comes back (1) On a quite day when the water is still and (2) On a rough day when there is uniform current so as to help the journey onward and to impede the journey back. If the speed of the launch on both days was same, in which case it will complete the journey in lesser time

• A. Case (1)
• B. Case (2)
• C. Same in both
• D. Nothing can be predicted

Answer 1

Answer: (B)

Case (2)

Answer 2

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

$t_{Q} = \frac{l}{v_{b}} + \frac{l}{v_{b}} = \frac{2l}{v_{b}}$ .....(i)

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

$t_{1} = \frac{l}{v_{b} + v_{a}}$ [as current helps the motion]

and time taken in coming back $t_{2} = \frac{l}{v_{b} - v_{a}}$ [as current opposes the motion]

So $t_{R} = t_{1} + t_{2} = \frac{2l}{v_{b}\lbrack 1 - (v_{a}/v_{b})^{2}\rbrack}$ .....(ii)

From equation (i) and (ii)

$\frac{t_{R}}{t_{Q}} = \frac{1}{\lbrack 1 - (v_{a}/v_{b})^{2}\rbrack} > 1\lbrack\text{as }1 - \frac{v_{a}^{2}}{v_{b}^{2}} < 1\rbrack$ i.e. $t_{R} > t_{Q}$

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