/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 On a pleasure cruise a boat is t... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 52

On a pleasure cruise a boat is traveling relative to the water at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) due south. Relative to the boat, a passenger walks toward the back of the boat at a speed of \(1.5 \mathrm{~m} / \mathrm{s} .\) (a) What is the magnitude and direction of the passenger's velocity relative to the water? (b) How long does it take for the passenger to walk a distance of \(27 \mathrm{~m}\) on the boat? (c) How long does it take for the passenger to cover a distance of \(27 \mathrm{~m}\) on the water?

Short Answer

Expert verified
(a) 3.5 m/s due south; (b) 18 s; (c) 7.71 s.

Step by step solution

01

Determine the Passenger's Velocity Relative to the Water

To find the passenger's velocity relative to the water, we need to consider both the velocity of the boat and the velocity of the passenger relative to the boat.- The boat travels due south at a speed of \(5.0 \text{ m/s}\).- The passenger walks toward the back of the boat at a speed of \(1.5 \text{ m/s}\) (which is opposite the direction of the boat's motion).Thus, the total velocity of the passenger relative to the water is calculated as:\[v_{pw} = v_{bw} - v_{pb} = 5.0 \text{ m/s} - 1.5 \text{ m/s} = 3.5 \text{ m/s}\]The passenger's velocity relative to the water is \(3.5 \text{ m/s}\) due south.
02

Calculate Time on the Boat

To calculate the time it takes for the passenger to walk \(27 \text{ m}\) on the boat, use the equation:\[time = \frac{distance}{speed}\]Here, the speed of the passenger relative to the boat is \(1.5 \text{ m/s}\). Therefore:\[time = \frac{27 \text{ m}}{1.5 \text{ m/s}} = 18 \text{ s}\]It takes the passenger 18 seconds to walk \(27 \text{ m}\) on the boat.
03

Calculate Time on the Water

To calculate the time it takes for the passenger to cover \(27 \text{ m}\) relative to the water, use the same time formula:\[time = \frac{distance}{speed}\]Here, the passenger's speed relative to the water is \(3.5 \text{ m/s}\) (from Step 1). Therefore:\[time = \frac{27 \text{ m}}{3.5 \text{ m/s}} \approx 7.71 \text{ s}\]It takes the passenger approximately 7.71 seconds to cover \(27 \text{ m}\) on the water.

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics that studies the motion of objects without considering the forces that cause the motion. It deals with quantities like displacement, velocity, and acceleration. In this exercise, kinematics helps us understand how the passenger's movement on the boat translates to the movement relative to the water.
Understanding kinematics involves considering concepts like:
  • Displacement: The change in position of an object. Measured in meters.
  • Velocity: The rate of change of displacement. Unlike speed, velocity includes a direction.
  • Acceleration: The rate of change of velocity over time. It is how quickly an object speeds up or slows down.
In our scenario, we look at how the passenger's walking speed affects their velocity relative to both the boat and the water. Understanding kinematics helps us calculate position changes over time and solve real-world problems like this one.
Velocity Addition
Velocity addition is a vital concept in kinematics when dealing with multiple moving objects. It allows us to find an object's velocity relative to a different frame of reference. In this problem, we have the boat moving relative to the water and the passenger moving on the boat.
The process involves:
  • Determining Reference Frames: Identify the different reference frames. In this problem, we have the boat as one frame and the water as another.
  • Relative Velocities: Calculate how fast an object is moving in one frame relative to another. Here, the passenger鈥檚 velocity relative to the water is found by subtracting their velocity on the boat from the boat鈥檚 velocity.
  • Equation: The formula for velocity addition is: \[ v_{PQ} = v_{PB} + v_{BQ} \] where \(v_{PB}\) is the velocity of object P relative to object B, and \(v_{BQ}\) is the velocity of object B relative to object Q.
By applying velocity addition, we found that the passenger moves at \(3.5 \text{ m/s}\) relative to the water.
Relative Motion
Relative motion is the calculation of the movement of an object with respect to another moving object. This exercise is a classic example of understanding how relative motion affects perceived velocity.
In situations where multiple entities are in motion, such as a boat and a passenger:
  • Frame of Reference: Choose an initial point or object from which you measure motion. In this case, the water serves as a stationary reference point.
  • Direction Influences Velocity: Movements in opposite directions need subtraction of velocities, as they oppose the larger motion.
  • Understanding Perceived Motion: Although the passenger walks towards the back of the boat, his movement relative to the water is a combination of his walking speed and the boat's motion.
Relative motion is essential for scenarios involving multiple moving objects, where each object's velocity affects how you see others moving.
Speed and Direction Calculations
Speed and direction are fundamental to understanding and calculating velocity. Velocity is speed with a directional component, which is crucial in vector calculations.
For calculations involving speed and direction:
  • Scalar vs. Vector: Speed is a scalar, meaning it only has magnitude. Velocity, being a vector, has both magnitude and direction.
  • Component Addition: When adding velocities, you must consider the direction. Opposite directions involve subtraction.
  • Conversion to Magnitude and Direction: The magnitude can be found using the Pythagorean theorem if vectors are not colinear. Direction can give context to motion, such as heading north or south.
In the solution provided, calculating the passenger's effective speed involves understanding that her walking in the opposite direction decreases her speed relative to the water. This exercise teaches how speed and direction combine to define an entity's velocity accurately.

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Most popular questions from this chapter

The ball is \(26.9 \mathrm{~m}\) from the goalpost. The ball is kicked with an initial velocity of \(19.8 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) above the ground. Between what two angles, \(\theta_{1}\) and \(\theta_{2}\), will the ball clear the \(2.74\) -m-high crossbar? (Hint: The following trigonometric identities may be useful: \(\sec \theta=1 /(\cos \theta)\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)

A major-league pitcher can throw a baseball in excess of \(41.0 \mathrm{~m} / \mathrm{s}\). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \(17.0 \mathrm{~m}\) away from the point of release?

A bullet is fired from a rifle that is held \(1.6 \mathrm{~m}\) above the ground in a horizontal position. The initial speed of the bullet is \(1100 \mathrm{~m} / \mathrm{s}\). Find (a) the time it takes for the bullet to strike the ground and (b) the horizontal distance traveled by the bullet.

The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you loft the ball with an initial speed of \(15.0 \mathrm{~m} / \mathrm{s},\) at an angle of \(50.0^{\circ}\) above the horizontal. At this instant your opponent is \(10.0 \mathrm{~m}\) away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is \(2.10 \mathrm{~m}\) above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by \(1.3 \mathrm{~m}\). A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is \(1.0 \mathrm{~m}\) above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is \(1.7 \mathrm{~m}\) above the ground. Finally, he leaps back to the other tree, now landing at a spot that is \(2.5 \mathrm{~m}\) above the ground. What is the magnitude of the squirrel's displacement?
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