Question

A river is flowing with speed $v_0$ m/s. Width of the river is $d$. A boat starts from point $P$. Boat crosses the river such that its velocity w.r.t. river is always directed towards a fix point $Q$ which is directly opposite to point $P$. Find drift of the boat if speed of boat in still water is also $v_0$.

• A. $\frac{d}{3}$
• B. $\frac{d}{2}$
• C. $\frac{d}{4}$
• D. $\frac{d}{\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{d}{2}$

Answer 2

Let the river flow along the x-axis with velocity $\vec{v}_r = v_0 \hat{i}$. The width of the river is $d$ along the y-axis. Let the starting point $P$ be at the origin $(0,0)$ and the destination point $Q$ be at $(0,d)$.

The velocity of the boat relative to the river, $\vec{v}_{br}$, has a magnitude of $v_0$ and is always directed towards point $Q$. If the boat is at position $(x,y)$, the vector from the boat to $Q$ is $(-x, d-y)$. Thus, the velocity of the boat relative to the river is: $\vec{v}_{br} = v_0 \frac{(-x, d-y)}{\sqrt{(-x)^2 + (d-y)^2}} = v_0 \left( \frac{-x}{\sqrt{x^2 + (d-y)^2}} \hat{i} + \frac{d-y}{\sqrt{x^2 + (d-y)^2}} \hat{j} \right)$

The velocity of the boat relative to the ground is $\vec{v}_b = \vec{v}_{br} + \vec{v}_r$. $\vec{v}_b = v_0 \left( \frac{-x}{\sqrt{x^2 + (d-y)^2}} \hat{i} + \frac{d-y}{\sqrt{x^2 + (d-y)^2}} \hat{j} \right) + v_0 \hat{i}$ $\vec{v}_b = v_0 \left( 1 - \frac{x}{\sqrt{x^2 + (d-y)^2}} \right) \hat{i} + v_0 \frac{d-y}{\sqrt{x^2 + (d-y)^2}} \hat{j}$

Let $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$. We have: $\frac{dx}{dt} = v_0 \left( 1 - \frac{x}{\sqrt{x^2 + (d-y)^2}} \right)$ $\frac{dy}{dt} = v_0 \frac{d-y}{\sqrt{x^2 + (d-y)^2}}$

To find the path of the boat, we can find the ratio $\frac{dx}{dy}$: $\frac{dx}{dy} = \frac{v_x}{v_y} = \frac{v_0 \left( 1 - \frac{x}{\sqrt{x^2 + (d-y)^2}} \right)}{v_0 \frac{d-y}{\sqrt{x^2 + (d-y)^2}}} = \frac{\sqrt{x^2 + (d-y)^2} - x}{d-y}$ Let $u = d-y$. Then $y = d-u$, so $\frac{dy}{du} = -1$, which means $\frac{dx}{du} = \frac{dx}{dy} \frac{dy}{du} = -\frac{dx}{dy}$. The differential equation becomes: $-\frac{dx}{du} = \frac{\sqrt{x^2+u^2} - x}{u}$ $\frac{dx}{du} = \frac{x - \sqrt{x^2+u^2}}{u} = \frac{x}{u} - \frac{\sqrt{x^2+u^2}}{u}$ This is a homogeneous differential equation. Let $x = vu$, so $\frac{dx}{du} = v + u\frac{dv}{du}$. $v + u\frac{dv}{du} = \frac{vu}{u} - \frac{\sqrt{(vu)^2+u^2}}{u}$ $v + u\frac{dv}{du} = v - \frac{u\sqrt{v^2+1}}{u}$ $u\frac{dv}{du} = -\sqrt{v^2+1}$ $\frac{dv}{\sqrt{v^2+1}} = -\frac{du}{u}$ Integrating both sides: $\int \frac{dv}{\sqrt{v^2+1}} = -\int \frac{du}{u}$ $\ln|v+\sqrt{v^2+1}| = -\ln|u| + C_1$ $\ln|v+\sqrt{v^2+1}| = \ln\left(\frac{1}{|u|}\right) + C_1$ $v+\sqrt{v^2+1} = \frac{A}{u} \quad (\text{where } A = e^{C_1})$ Substitute back $v = x/u$: $\frac{x}{u} + \sqrt{\frac{x^2}{u^2}+1} = \frac{A}{u}$ $\frac{x}{u} + \frac{\sqrt{x^2+u^2}}{u} = \frac{A}{u}$ $x + \sqrt{x^2+u^2} = A$ The boat starts at $P(0,0)$, so $x=0$ when $y=0$. When $y=0$, $u = d-0 = d$. Substituting $x=0$ and $u=d$: $0 + \sqrt{0^2+d^2} = A \implies A = d$ So the equation of the boat's path is: $x + \sqrt{x^2+u^2} = d$ Substitute $u=d-y$: $x + \sqrt{x^2+(d-y)^2} = d$ The drift is the value of $x$ when the boat reaches the other bank, i.e., when $y=d$. When $y=d$, $u = d-d = 0$. $x + \sqrt{x^2+0^2} = d$ $x + \sqrt{x^2} = d$ Since the boat moves in the positive x direction, $x>0$, so $\sqrt{x^2}=x$. $x + x = d$ $2x = d$ $x = \frac{d}{2}$ The drift of the boat is $\frac{d}{2}$.