Question

A \[600\] meter wide river flows directly south at \[4.0m/s\]. A small motor boat travels at \[5.0m/s\] in still water and points in such a direction so that it will travel directly east relative to the land. The time taken it takes to cross the river is closest to
A) 67s
B) 120s
C) 150s
D) 200s

Answer 1

Hint:
Using the relative motion in 2D, the speed of the river governs the speed of the boat in the river. Let us imagine the boat is moving horizontally and the river flows vertically to the boat, the river's speed influence produces two direction components of the boat one is that of \[sin\text{ }x\] and another that of \[cos\text{ }x\]. Using the direction component of the boat speed we equate the boat’s speed with the vertical and horizontal components of the river speed and find the time taken.

Complete step by step solution:
Let us draw a diagram showing the components of the boat when flowing against the river side by side.

The vertical velocity component of the boat is: \[5\sin \theta \].
The horizontal velocity component of the boat is: \[5\cos \theta \].
The vertical velocity component of the river is: \[4\text{ }m/\sec \].
The time taken to cross the river is taken as: \[t\text{ }\sec \].
The horizon at the velocity component of the river is: \[\dfrac{600}{t}\text{ }m/\sec \].
Let us take the vertical components and then find the time taken. The velocity of the boat and river are equated together when checking vertically:
\[5\sin \theta =4\text{ }m/\sec \]
\[\theta =\sin \left( \dfrac{4}{5} \right)\]
Let us take the horizontal components together and then find the time taken. The velocity of the boat and river are equated together when checking horizontally:
\[5\cos \theta =\dfrac{600}{t}\text{ }m/\sec \]
Place the value of \[\theta \], in the above formula we get the value of time \[t\] as:
\[5\cos \left( \sin \left( \dfrac{4}{5} \right) \right)=\dfrac{600}{t}\text{ }m/\sec \]
The value of \[\cos \left( \sin \left( \dfrac{4}{5} \right) \right)=\dfrac{3}{5}\].
\[5\times \dfrac{3}{5}=\dfrac{600}{t}\text{ }m/\sec \]
\[5\times \dfrac{3}{5}=\dfrac{600}{t}\text{ }m/\sec \]
\[t=200\sec \]

Therefore, the time taken to cross the river in the shortest time is \[200\sec \]

So, correct option is D

Note: When checking the shortest distance, we need to check the components of the velocity, although by diagram, the shortest distance is vertically-straight line and speed of boat given along with the distance but that doesn't mean we can move straight as we need to keep the speed of the river in mind that is why we need to incorporate the river velocity with the boat velocity so both the velocity components are in same line for both horizontal and vertical value.