Question

2. Boat Race (3 pt) Let's assume Galilean relativity. Two equally matched rowers race each other over courses shown in the figure below. Each oarsman rows at speed v and the current in the river moves at speed uv). Boat 1 goes from A to B, a distance L, and back. Boat 2 goes from A to C, also distance L, and back Ground River Ground (a) Calculate each round-trip time tABA and tACA ( b) Give the Maclaurin series for (1- x 2)-2 and (1-x2)-1 to the first order in x2, respectively (c) Using the previous results, calculate the time difference between the two round-trip times to the leading order in u/v, and answer which rower wins

Answer 1

a) The rower tries to row boat 1 away from perpendicular to river flow, so that resultant velocity of boat 1 is perpendicular to river. Resultant speed of boat 1 is v cos theta where sin theta = u/v v cos theta = v squareroot 1 - u^2/v^2 = squareroot v^2 - u^2 Time taken by boat 1 for the journey A to B and back is t_ABA = L/v cos theta + L/v cos theta = 2L/squareroot v^2 - u^2 For boat 2: Downstream speed of boat 2 is v + u and upstream speed is v + u and upstream speed is v - u Time taken by boat 2 for the journey A to C and back is t_ACA = L/v + u + L/v - u = 2Lv/v^2 - u^2 b) (1 + x)^n = 1 + nx + n (n - 1)/2! x^2 + Ellipsis (1 - x^2)^-1 = 1

a)The rower tries to row Boat 1 away from perpendicular to river flow, so that resultant velocity of boat 1 is perpendicular to river. Resultant speed of boat 1 is $v\cos\theta$, where $\sin\theta=\frac{u}{v}$

$v\cos\theta=v\sqrt{1-\frac{u^2}{v^2}}=\sqrt{v^2-u^2}$

Time taken by boat 1 for the journey A to B and back is  $t_{ABA}=\frac{L}{v\cos\theta}+\frac{L}{v\cos\theta}=\frac{2L}{\sqrt{v^2-u^2}}$

For boat 2: Downstream speed of boat 2 is  $v+u$ and upstream speed is $v-u$

Time taken by boat 2 for the journey A to C and back is $t_{ACA}=\frac{L}{v+u}+\frac{L}{v-u}=\frac{2Lv}{{v^2-u^2}}$

b)

$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\ldots$

$(1-x^2)^{-1}=1+(-1)(-x^2)+\frac{(-1)(-1-1)}{2!}(-x^2)^2+\ldots=1+x^2+x^4+\ldots$

${(1-x^2)^{-1/2}=1+(-1/2)(-x^2)+\frac{(-1/2)(-1/2-1)}{2!}(-x^2)^2+\ldots=1+\frac{x^2}{2}+\frac{3x^4}{8}+\ldots}$

c)

Using only first order in $x^2$

$(1-x^2)^{-1}\approx 1+x^2$

$(1-x^2)^{-1/2}\approx 1+\frac{x^2}{2}$

$\left ( 1-\frac{u^2}{v^2} \right )^{-1}\approx 1+\frac{u^2}{v^2}$

$\left ( 1-\frac{u^2}{v^2} \right )^{-1/2}\approx 1+\frac{u^2}{2v^2}$

$t_{ABA}=\frac{2L}{v}\frac{1}{\left ( 1-\frac{u^2}{v^2} \right )^{1/2}}=\frac{2L}{v}\left ( 1-\frac{u^2}{v^2} \right )^{-1/2}=\frac{2L}{v}\left ( 1+\frac{u^2}{2v^2} \right )$

$t_{ACA}=\frac{2L}{v}\frac{1}{\left ( 1-\frac{u^2}{v^2} \right )}=\frac{2L}{v}\left ( 1-\frac{u^2}{v^2} \right )^{-1}=\frac{2L}{v}\left ( 1+\frac{u^2}{v^2} \right )$

$t_{ACA}>t_{ABA}$

Difference between the two round-trip times is $t_{ACA}-t_{ABA}=\frac{Lu^2}{v^3}$

Boat 1 takes less time for the round trip, hence boat 1 wins.