Boat River Problem

The boat and river problem is a classic example of relative motion, illustrating how the movement of a boat in a river is influenced by both the speed of the boat and the current of the river. This problem is highly applicable in real-life scenarios, such as navigating a boat across a river or planning the most efficient route in water bodies with strong currents. Understanding the dynamics between the boat's velocity and the river's current helps in determining the shortest path, the time taken to cross the river, or the boat's actual trajectory relative to the ground. This concept is crucial not only for sailors and river pilots but also in designing and planning transportation and logistics in riverine environments. By solving the boat and river problem, one can gain insights into optimizing travel time and fuel consumption, making it an essential topic in both practical navigation and theoretical physics.

Boat River Problem

To solve any riverboat problem, two things are to be kept in mind. A boat's speed with respect to the water is the same as its speed in still water. The velocity of the boat relative to water is equal to the difference in the velocities of the boat relative to the ground and the velocity of the water with respect to the ground.

Now, start with the important term related to relative velocity.

Important Terms

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

$\text { Let's try to find out } V_b \text { in some important cases }$

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

Then,

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

• When the boat travels upstream (u and v have opposite directions)

Then,

$V_b=(U-V) \hat{i}$

• $\text { If the boat travels at some angle } \theta \text { with river flow (u) }$

Now resolve v in two-component

Component of v along $U=v_x=v \cos \theta \hat{i}$

Component of v perpendicular to $U=v_y=v \sin \theta \hat{j}$

So,

$V_b=(v \cos \theta+u) \hat{i}+v \sin \theta \hat{j}$

and, $\left|V_b\right|=\sqrt{u^2+v^2+2 u v \cos \theta}$

Now if the time taken to cross the river is t

Then,

$t=\frac{d}{v \sin \theta}$

Here = drift

And,

$x=(u+v \cos \theta) t=\frac{(u+v \cos \theta) d}{v \sin \theta}$

Important Cases

Now, we will study some of the important cases which are mentioned below:

• To cross the river in the shortest time

This means v is perpendicular to u

$\begin{aligned}
& \text { Or } \operatorname{Sin} \theta=1 \Rightarrow \theta=90^{\circ} \\
& \text { So. }\left|V_b\right|=\sqrt{u^2+v^2}
\end{aligned}$

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

• To cross the river in the shortest path

Means drift = 0

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

The time taken to cross the river is: $t=\frac{d}{v \sin \theta}$

$ t=\frac{d}{\sqrt{v^2-u^2}}$

Recommended Topic Video

Solved Example Based On Boat River Problem

Example 1: A man wishes to cross a river in a boat. If he crosses the river in a minimum time he takes 10 minutes with a drift of 120m. If he crosses the river taking the shortest route it takes 12.5 min. The velocity of the boat with respect to water is :

1) $\frac{1}{3} m / s$
2) $\frac{1}{4} \mathrm{~m} / \mathrm{s}$
3) $\frac{1}{5} \mathrm{~m} / \mathrm{s}$
4) $\frac{1}{6} m / s$

Solution:

For minimum time

$t_1=\frac{d}{v}=10 \mathrm{~min}-(1)$

$\begin{aligned}
& \text { drift }=u \cdot t_1=120 \mathrm{~m} \\
& \text { or } u \cdot \frac{d}{v}=120 \Rightarrow u=\frac{120}{600} \mathrm{~m} / \mathrm{s} \Rightarrow u=\frac{1}{5} \mathrm{~m} / \mathrm{s}
\end{aligned}$

For Shortest route

$\text { Let it make angle } \theta \text { with the river flow. }$

$\begin{aligned}
& \Rightarrow v \cos \theta=u \\
& \& t_2=\frac{d}{v \sin \theta}=\frac{d}{v \cdot \sqrt{1-\frac{u^2}{v^2}}}=\frac{d}{\sqrt{v^2-u^2}} \\
& \text { or } \frac{d}{\sqrt{v^2-u^2}}=12.5 \mathrm{~min}-(2)
\end{aligned}$

$\begin{aligned}
& \text { Dividing Equation (2) to (1) } \\
& \frac{d}{v \cdot \frac{d}{\sqrt{v^2-u^2}}}=\frac{10}{12.5} \\
& \Rightarrow \sqrt{1-\frac{u^2}{v^2}}=\frac{4}{5} \\
& \Rightarrow 1-\frac{u^2}{v^2}=\frac{16}{25} \\
& \Rightarrow \frac{u^2}{v^2}=\frac{9}{25} \\
& \Rightarrow \frac{u}{v}=\frac{3}{5} \\
& \Rightarrow v=\frac{5}{3} \cdot u \\
& =\frac{5}{3} \times \frac{1}{5}=\frac{1}{3} \mathrm{~m} / \mathrm{s}
\end{aligned}$

Hence, the answer is option (1).

Example 2: A man can swim with a speed of 4km/hr in still water. He crosses a river 1 km wide that flows steadily at 3 kmph. If he makes his stroke normal to the river current, how far (in meters) down the river does he go when he reaches the other bank?

1) 750

2) 500

3) 700

4) 850

Solution:

Given- v= 4 km/hr, u=3 km/hr, D= 1km

$\text { Drift }=\left(\frac{d}{v}\right) \cdot u=1 \mathrm{~km} \times \frac{3}{4}=750 \mathrm{~m}$

Hence, the answer is option (1).

Example 3: A person is swimming at a speed of 10 m/s at an angle of $120^{\circ}$ with the flow and reaches a point directly opposite on the other side of the river. The speed of the flow is 'x' m/s. The value of 'x' to the nearest integer is ________.

1) 50

2) 5

3) 20

4) 20

Solution:

To reach a point directly opposite on the other side of the river
$
\begin{aligned}
& V_{M / R} \sin 30^{\circ}=V_R \\
& 10 \sin 30^{\circ}=V_R \\
& V_R=x=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

Hence, the answer is option (5).

Example 4: The swimmer crosses the river along the line making an angle of 45º with the direction of flow. The velocity of the river is 5 m/s. A swimmer takes 6 seconds to cross the river of width 60 m. The velocity of the swimmer with respect to water will be:

1) $10 \mathrm{~m} / \mathrm{s}$
2) $12 \mathrm{~m} / \mathrm{s}$
3) $5 \sqrt{5} \mathrm{~m} / \mathrm{s}$
4) $100 \mathrm{~m} / \mathrm{s}$

Solution:

River flow along y direction with a velocity of 5 j m/s.
The swimmer crossover along the vector: i + j since the angle is 45 degree
displacement of swimmer just after crossing = 60 i + 60 j meters
speed of the swimmer with respect to a stationary observer
= (60 i + 60 j) / 6 m/s = 10 i + 10 j m/s
velocity of swimmer relative to the river = 10 i + 10 j - 5 j
magnitude of velocity relative to river = 5√5 m/s

Hence, the answer is the Option (3).

Example 5: A man who has a speed of 5km/h in still water crosses a river of width 1km along the shortest possible path in 15 minutes. The velocity of river water in km/h is :

1) 3

2) 4

3) 8

4) 10

Solution:

$
\begin{aligned}
& \cos \theta=\frac{4}{5} \\
& \sin \theta=\frac{3}{5}
\end{aligned}
$

Now,
$
V_R=5 \times \frac{3}{5}=3 \mathrm{Km} / \mathrm{h}
$

Summary

In short, we have studied in depth about boat river problem along with different cases.. This is one of the most significant concepts in the kinematics chapter. In general, we encounter questions based on this principle in competitive exams like as NEET and JEE.