Boat River Problem

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.