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Chapter 5: Problem 28

Two ports, \(A\) and \(B\), on a North-South line are separated by a \(\pi_{\text {le }}\) of width \(D\). The river flows east with speed \(u\). A boat crosses the river starting from port \(A .\) The speed of the boat relative to the river is \(v\). Assume \(v=2 u\). The boat crosses the river from \(A\) to the other side in shortest possible time, then how far is the boat from the port \(B\) after crossing the river (1) \(D / \sqrt{2}\) (2) \(\sqrt{2} D\) (3) \(2 D\) (4) \(D / 2\)

Short Answer

Expert verified
The boat is \( \frac{D}{2} \) away from port \( B \).

Step by step solution

01

Determine the Boat's Actual Speed

The speed of the boat relative to the river is given as \( v = 2u \). Since the river flows east and the boat crosses in the shortest time, it means the boat moves perpendicular to the flow of the river. Therefore, its velocity component across the river is \( v = 2u \).
02

Calculate Time to Cross the River

The time it takes for the boat to cross the river is calculated by dividing the width of the river, \( D \), by the component of the boat's velocity that is perpendicular to the current. The velocity across the river is \( v = 2u \). Therefore, the time \( t \) to cross is given by:\[t = \frac{D}{v} = \frac{D}{2u}.\]
03

Determine Boat's Displacement Due to River's Flow

During the time taken to cross the river, the river will influence the boat's position parallel to its flow. The displacement \( x \) in the east direction can be calculated as the product of the river's speed \( u \) and the time \( t \):\[x = u \times \frac{D}{2u} = \frac{D}{2}.\]
04

Calculate Distance from Port B

Since the initial starting point \( A \) is directly across from port \( B \), the eastward displacement caused by the river's flow results in the boat being \( \frac{D}{2} \) units east of port \( B \) when it reaches the north bank.This means the boat is \( \frac{D}{2} \) away from port \( B \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Velocity Components
When analyzing the motion of the boat crossing the river, it's crucial to understand how velocity components come into play. Velocity components break down the boat's overall velocity into parts that act in specific directions. In this scenario, the boat has a velocity relative to the river of \( v = 2u \). Since the river flows east, the best way to achieve the shortest crossing time is for the boat to move perpendicular to the flow. This means it aims directly north or south, depending on direction.
  • The velocity of the boat across the river (north-south direction) is \( 2u \).
  • The river's velocity affects the boat's movement in the east-west direction.
Thus, the boat's perpendicular movement ensures it traverses the river quickest, minimizing the eastward drift caused by the river's own flow.
Calculating the Time of Crossing
The time it takes for the boat to cross the river entails dividing the river's width by the part of the boat's velocity that takes it directly across. Given the velocity across is \( 2u \), we find the time, \( t \), needed to cross the river as follows: \[ t = \frac{D}{v} = \frac{D}{2u}. \]This formula represents how long it will take the boat to move directly from one side to the other, discounting any influence from the river's current.
Displacement Calculation Due to River's Flow
While the boat crosses the river, it's important to acknowledge the influence of the river's flow on the boat, pushing it in an eastward direction. This resultant drift is an example of displacement due to relative motion. The displacement, \( x \), of the boat as the river flows can be calculated by multiplying the river's velocity \( u \) by the crossing time \( t \): \[ x = u \times t = u \times \frac{D}{2u} = \frac{D}{2}. \]This calculation shows that, while the boat moves across, the river's flow causes it to end up halfway across the distance in the east-west direction.
Determining the Shortest Crossing Path
The objective is to find out how far the boat is from port \( B \) after it crosses in the shortest time. Since the boat crosses directly north, yet is pushed east by the river's flow, it ends up displaced. The distance from port \( B \) is the same as the eastward displacement of \( \frac{D}{2} \).
  • Starting directly across from port \( B \),
  • Eastward drift makes the final distance \( \frac{D}{2} \) east of \( B \).
This highlights how considering shortest paths in relative motion also involves understanding how velocities in different directions affect a journey's endpoint.

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Most popular questions from this chapter

A particle has been projected with a speed of \(20 \mathrm{~ms}^{-1}\) at an angle of \(30^{\circ}\) with the horizontal. The time taken when the velocity vector becomes perpendicular to the initial 6 velocity vector is (1) \(4 \mathrm{~s}\) (2) \(2 \mathrm{~s}\) (3) \(3 \mathrm{~s}\) (4) Not possible in this case

A projectile has initially the same horizontal velocin as it would acquire if it had moved from rest with unifon acceleration of \(3 \mathrm{~m} \mathrm{~s}^{-2}\) for \(0.5 \mathrm{~min}\). If the maximum heigis reached by it is \(80 \mathrm{~m}\), then the angle of projection ip \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (1) \(\tan ^{-1} 3\) (2) \(\tan ^{-1}(3 / 2)\) (3) \(\tan ^{-1}(4 / 9)\) (4) \(\sin ^{-1}(4 / 9)\)

A truck is moving with a constant velocity of \(54 \mathrm{~km} \mathrm{~h}^{-1}\). In which direction (angle with the direction of motion of truck) should a stone be projected up with a velocity of \(20 \mathrm{~ms}^{-1}\), from the floor of the truck, so as to appear at right angles to the truck, for a person standing on earth? (1) \(\cos ^{-1}\left(-\frac{3}{4}\right)\) (2) \(\cos ^{-1}\left(-\frac{1}{4}\right)\) (3) \(\cos ^{-1}\left(\frac{2}{3}\right)\) (4) \(\cos ^{-1}\left(\frac{3}{4}\right)\)

During a projectile motion, if the maximum height equals the horizontal range, then the angle of projection with the horizontal is (1) \(\tan ^{-1}(1)\) (2) \(\tan ^{-1}(2)\) (3) \(\tan ^{-1}(3)\) (4) \(\tan ^{-1}(4)\)

Points \(A\) and \(C\) are on the horizontal ground and \(A\) and \(B\) are in same vertical plane at a distance of \(1500 \mathrm{~m}\). Simultaneously bullets are fired from \(A, B\) and \(C\) and they collide at \(O\). The bullet at \(B\) is fired at an angle of \(30^{\circ}\) with horizontal towards the ground at velocity \(100 \mathrm{~m} / \mathrm{s}\). The bullet at \(C\) is projected vertically upward at velocity of \(100 \mathrm{~m} / \mathrm{s}\). The bullet projected from \(A\) reaches its maximum height at \(O\). Find the time in which bullets will collide (seconds): (1) 10 (2) 15 (3) 20 (4) 25
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