/*! 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 78 A Coast Guard ship is traveling ... [FREE SOLUTION] | 91Ó°ÊÓ

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A Coast Guard ship is traveling at a constant velocity of \(4.20 \mathrm{m} / \mathrm{s}\), due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of \(2310 \mathrm{m}\) with respect to the ship, in a direction \(32.0^{\circ}\) south of east. Six minutes later, he notes that the object's position relative to the ship has changed to \(1120 \mathrm{m}, 57.0^{\circ}\) south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

Short Answer

Expert verified
The object's velocity relative to the water is approximately 1.47 m/s, 72.3° south of west.

Step by step solution

01

Convert Time to Seconds

First, convert the time of 6 minutes into seconds since velocities are given in meters per second. We have:\[ t = 6 \times 60 = 360 \text{ seconds} \]
02

Define Initial and Final Positions

The object's initial position vector relative to the ship is given in polar form as (2310 m, 32.0° south of east). Convert this to Cartesian coordinates. Assume due east is the positive x-axis.Initial position:\[ x_1 = 2310 \cos(32.0^{\circ}) \] \[ y_1 = -2310 \sin(32.0^{\circ}) \] The final position after 6 minutes is (1120 m, 57.0° south of west), which is:Final position:\[ x_2 = -1120 \cos(57.0^{\circ}) \] \[ y_2 = -1120 \sin(57.0^{\circ}) \]
03

Calculate Change in Position

Calculate the change in position (displacement) from the initial to final relative position of the object:\[ \Delta x = x_2 - x_1 \] \[ \Delta y = y_2 - y_1 \]
04

Determine Relative Velocity Components

Using the change in position and the time, calculate the velocity components relative to the ship:\[ v_{x, ext{relative}} = \frac{\Delta x}{t} \] \[ v_{y, ext{relative}} = \frac{\Delta y}{t} \]
05

Adjust Relative Velocity to Water

Since the object's velocity is relative to the ship which is moving at 4.20 m/s due east, add this velocity to get the object's velocity relative to the water:\[ v_{x, ext{object}} = v_{x, ext{relative}} + 4.20 \] \[ v_{y, ext{object}} = v_{y, ext{relative}} \]
06

Calculate Total Velocity Magnitude and Direction

Find the magnitude and direction of the object's velocity relative to the water. The magnitude is:\[ v = \sqrt{v_{x, ext{object}}^2 + v_{y, ext{object}}^2} \]The direction \(\theta\) is given by:\[ \theta = \arctan\left(\frac{v_{y, ext{object}}}{v_{x, ext{object}}}\right) \]Express the direction as an angle with respect to due west.

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Key Concepts

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

Velocity Components
When discussing relative velocities, especially in physics problems that involve moving objects, it's crucial to break down the velocities into components. This helps in understanding the movement in different directions. To do this, you often need to use terms like forward, sideways, east, west, etc.

Velocity can be split into two main components, usually along the x and y axes:
  • x-component: This gives the speed of the object along the horizontal direction (east-west in many exercises).
  • y-component: This provides the speed along the vertical direction (north-south or up-down).
For example, if you have a velocity given with directionals such as "south of east," you will need to determine how much of the velocity acts towards the east (\(x_1 = 2310 \, \cos(32.0^{\circ})\)) and how much acts towards the south (\(y_1 = -2310 \, \sin(32.0^{\circ})\)).

Converting these into Cartesian coordinates allows you to directly work with them in calculations, especially when calculating change over time.
Displacement Calculation
Displacement is a vector that helps you quantify how far an object has moved in a straight line from its initial position to its final position. In physics problems involving two positions given in polar forms, you must calculate this straight-line change.

Given initial and final positions, the displacement components can be determined as:
  • Change in x-component: Calculate the difference between the final x-position and the initial x-position, \(\Delta x = x_2 - x_1\).
  • Change in y-component: Similarly, determine the change in the y-direction, \(\Delta y = y_2 - y_1\).
This change in position is essential for determining the object's relative velocity, as it indicates both the distance and the direction over the specific time interval that has passed.

By subtracting these respective components, you form a new vector that represents the total displacement, which then can be divided by the duration to find the velocity components.
Polar to Cartesian Conversion
The polar to Cartesian conversion is a mathematical process used when handling vectors like displacement or velocity that are described using angles and magnitudes. Often, scenarios like navigation, weather patterns, or physics problems describe positions or directions using polar coordinates.

In polar coordinates, a point is defined by:
  • Magnitude: which provides how far from the origin it is.
  • Angle: which indicates its direction relative to a reference line, usually the positive x-axis (east direction).
To convert a point given in polar coordinates \((r, \theta)\) into Cartesian coordinates, you use the formulas:
  • x = r \(\cos(\theta)\): translates the radial distance in the direction of the horizontal (x) axis.
  • y = r \(\sin(\theta)\): computes the radial distance's projection onto the vertical (y) axis.
This conversion allows us to use straightforward algebraic methods for further analysis and calculations in exercises, where adding or subtracting vectors is necessary to determine things such as displacement or relative velocity.

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Most popular questions from this chapter

A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{m} / \mathrm{s}\), hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{m}\) below the jumping-off point. Find the maximum value for \(D\).

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