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Chapter 5: Problem 4

A river is flowing towards with a velocity of \(5 \mathrm{~m} \mathrm{~s}^{-1}\). The boat velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The boat crosses the river by shortest path. Hence, (1) The direction of boat's velocity is \(30^{\circ}\) west of north. (2) The direction of boat's velocity is north-west. (3) Resultant velocity is \(5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\). (4) Resultant velocity of boat is \(5 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\).

Short Answer

Expert verified
The correct answers are (2) and (3).

Step by step solution

01

Understand the Problem

We have a river flow velocity of \(5 \mathrm{~m/s}\) towards the east and a boat velocity of \(10 \mathrm{~m/s}\) which needs to offset this velocity to cross by the shortest path. The options provide different directional angles and resultant velocities to verify.
02

Analyze the Boat's Trajectory for Shortest Path

For the boat to cross by the shortest path, its direction should be such that the eastward component of its velocity exactly cancels the river's flow velocity. Assume the boat moves at an angle \( \theta \) north west, then \(10 \cos \theta = 5\) since it needs to negate the eastward river flow.
03

Calculate Direction Angle

Rearrange \(10 \cos \theta = 5\) to find \(\theta\). Thus, \( \cos \theta = \frac{1}{2} \), giving \( \theta = 60^{\circ} \). Therefore, the boat's direction is \(60^{\circ}\) measured north of west.
04

Calculate Resultant Velocity

The northward component of the boat's velocity, using \( \sin 60^{\circ}\), is \(10 \sin 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\). Thus, the resultant velocity in terms of absolute velocity of crossing is \(5\sqrt{3} \mathrm{~m/s}\).
05

Verify Options

Option (1) is incorrect because direction is \(60^{\circ}\) not \(30^{\circ}\). Option (2) specifies a general direction and can match the calculated \(60^{\circ}\). Option (3) is correct for the resultant velocity \(5\sqrt{3} \mathrm{~m/s}\). Option (4) gives an incorrect resultant velocity.

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Key Concepts

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

River Crossing Problems
River crossing problems are classic examples in physics that involve understanding relative motion. When dealing with such problems, the aim is often to determine the path and time taken for an object, such as a boat, to cross a river that is flowing. The flow of the river introduces an additional velocity component that must be accounted for.
The art of solving these problems lies in accurately determining the path that results in the shortest crossing time or distance.
  • The shortest path is usually not a straight line across the river due to the river's flow.
  • To achieve the shortest path, the boat's velocity often includes an angle to counteract the river current.
Understanding river dynamics and boat velocity allows the calculation of the most efficient path, whether it be the shortest distance or the quickest time across.
Vector Components
In physics, vectors are used to represent quantities that have both magnitude and direction, such as velocity. When dealing with river crossing problems, it's essential to break down velocities into vector components.
The boat's velocity and the river's velocity can be represented as vectors, and these vectors can be resolved into perpendicular components:
  • A horizontal component that runs parallel to the river's flow.
  • A vertical component that runs perpendicular to the river's flow.
In our example, the boat's velocity of 10 m/s needs to be resolved into:
鈥� An east-west component intended to counteract the river's flow.
鈥� A north-south component that takes the boat directly across the river.
Resultant Velocity
The resultant velocity is the effective velocity of the boat as it crosses the river, taking into account both the boat's own velocity and the velocity of the river.
For a successful river crossing by the shortest path, this resultant velocity needs to be calculated precisely.
  • The northward component is essential for making progress across, as calculated using the formula for resultant velocity.
  • In our scenario, this resulted in a northward component of 5鈭�3 m/s.
This resultant velocity represents the diagonal path the boat actually takes across the river, counteracting the current and ensuring the shortest crossing path.
Angle of Velocity
The angle of velocity is crucial when determining the direction in which the boat should be steered. This angle often needs precise calculation to ensure that the boat compensates correctly for the river's flow.
  • The boat's direction is typically described relative to a cardinal direction, such as 'north of west.'
  • By calculating and knowing this angle, the boat can adjust its path accordingly to maintain a straight crossing path across the river.
In our exercise, the boat had to steer at a calculated angle of 60掳 north of west to effectively counter the eastward river flow of 5 m/s. This precise angle ensures that the eastward component negates the river's flow, allowing a direct north-south crossing.

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Most popular questions from this chapter

Two inclined planes \(O A\) and \(O B\) having inclination (with borizontal) \(30^{\circ}\) and \(60^{\circ}\), respectively, intersect each other at \(O\) as shown in figure. A particle is projected from point \(P\) with velocity \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\) along a direction perpendicular to plane \(O A\). If the particle strikes plane \(O B\) perpendicularly at \(Q\), calculate The velocity with which particle strikes the plane \(O B\), (1) \(15 \mathrm{~ms}^{-1}\) (2) \(30 \mathrm{~ms}^{\prime}\) (3) \(20 \mathrm{~ms}^{-1}\) (4) \(10 \mathrm{~ms}^{-1}\)

Projectile motion is a combination of twe enslonal motions: one in horizontal and other in vertical direction. Motion in \(2 D\) means in a plane. Necessary condition for \(2 D\) motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be \(0^{\circ}<\theta<180^{\circ} .\) During the projectile motion, the horizontal component of velocity remains unchanged but vertical component of velocity is time dependent. Now answer the following questions: A particle is projected from the origin in the \(x-y\) plane. The acceleration of particle in negative \(y\)-direction is \(\alpha\). If equation of path of the particle is \(y=a x-b x^{2}\), then initial velocity of the particle is (1) \(\sqrt{\frac{\alpha}{2 b}}\) (2) \(\sqrt{\frac{\alpha\left(1+a^{2}\right)}{2 b}}\) (3) \(\sqrt{\frac{\alpha}{a^{2}}}\) (4) \(\sqrt{\frac{\alpha b}{a^{2}}}\)

A train of \(150 \mathrm{~m}\) length is going toward north direction at a speed of \(10 \mathrm{~m} \mathrm{~s}^{-1}\). A parrot flies at a speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\) toward south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to (1) \(12 \mathrm{~s}\) (2) \(8 \mathrm{~s}\) (3) \(15 \mathrm{~s}\) (4) \(10 \mathrm{~s}\)

A river flows with a speed more than the maximum speed with which a person can swim in still water. He intends to cross the river by the shortest possible path (i.e., he wants to reach the point on the opposite bank which directly opposite to the starting point). Which of the following is correct? (1) He should start normal to the river bank. (2) He should start in such a way that he moves normal to the bank, relative to the bank. (3) He should start in a particular (calculated) direction making an obtuse angle with the direction of water current. (4) The man cannot cross the river in that way.

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 velocity of bullet at \(A\) : (1) \(50 \mathrm{~m} / \mathrm{s}\) (2) \(50 \sqrt{7} \mathrm{~m} / \mathrm{s}\) (3) \(60 \sqrt{7} \mathrm{~m} / \mathrm{s}\) (4) \(60 \mathrm{~m} / \mathrm{s}\)
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