Boat River Problem - Practice Questions & MCQ

 Important terms

$d=$ width of river
$U=$ speed of river
$V=$ Speed of Boat w.r.t. River
and $V_b=$ Speed of boat w.r.t. Ground
So, the relation between $\mathrm{u}, \mathrm{v}$ and $V_b$ is

$$
V_b=U+V
$$

Let's try to find out $V_b$ in some important cases

   I)  When boat travels downstream (u and v have the same direction)

              Then, $V_b=(U+V) \hat{i}$

II) When boat travels upstream (u and v has opposite direction)

      Then, $V_b=(U-V) \hat{i}$

III) If boat travels at some angle with river flow (u)

        Now resolve v in two component

Component of v along $U=v_x=v \cos \theta \hat{i}$
Component of v perpendicular to $U=v_y=v \sin \theta \hat{j}$
$\mathrm{So}_{\mathrm{o}}, V_b=(v \cos \theta+u) \hat{i}+v \sin \theta \hat{j}$
and, $\left|V_b\right|=\sqrt{u^2+v^2+2 u v \cos \theta}$
Now if time taken to cross the river is $t$
Then, $t=\frac{d}{v \sin \theta}$
Here $x=$ drift
And, $x=(u+v \cos \theta) t=\frac{(u+v \cos \theta) d}{v \sin \theta}$

2. Important cases

    I) To cross the river in the shortest time

Means $v$ is perpendicular to u
Or $\operatorname{Sin} \theta=1 \Rightarrow \theta=90^{\circ}$
So, $\left|V_b\right|=\sqrt{u^2+v^2}$

Time taken

$$
t_{\min }=\frac{d}{v}
$$

Drift along river flow, $\quad x=d\left(\frac{u}{v}\right)$

II) To cross river in the shortest path

Means drift $=0$

$$
\begin{aligned}
& x=(u+v \cos \theta) t=0 \Rightarrow \cos \theta=\frac{-u}{v} \\
& \left|V_b\right|=\sqrt{v^2-u^2}
\end{aligned}
$$

Time taken to cross the river is

$$
\begin{aligned}
t & =\frac{d}{v \sin \theta} \\
t & =\frac{d}{\sqrt{v^2-u^2}}
\end{aligned}
$$