Multiple choice

A boat can go across a lake and return in time $T_{0}$ at a speed $v$. On a rough day there is a uniform current at speed $v_{1}$ to help the onward journey and impede the return journey. If the time taken to go across and return on the same day be $T$, then $T/T_{0}$ will be

  1. $\dfrac {1}{(1 - v_{1}^{2}/ v^{2})}$
  2. $\dfrac {1}{(1 + v_{1}^{2}/ v^{2})}$
  3. $(1 - v_{1}^{2} / v^{2})$
  4. $\left (1 + \dfrac {v_{1}^{2}}{v_{2}}\right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

T0 = 2d/v. With current v1, time T = d/(v+v1) + d/(v-v1) = 2dv / (v^2 - v1^2). Then T/T0 = (2dv / (v^2 - v1^2)) / (2d/v) = v^2 / (v^2 - v1^2) = 1 / (1 - v1^2/v^2).

AI explanation

Let the distance across the lake be D, so the time in still water is T0 = 2D/v. On the rough day, the time for the onward journey is D divided by the sum of the speeds (v + v1), and the time for the return journey is D divided by the difference of the speeds (v - v1). Combining these terms gives a total time T of 2Dv divided by the difference of the squares of the speeds (v squared minus v1 squared). Dividing this total time T by the original time T0 gives the ratio v squared divided by the difference between v squared and v1 squared. This simplifies to 1 divided by the quantity (1 minus v1 squared over v squared).

Keep practicing — related questions