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Chapter 4: Problem 3

As you walk to class with a constant speed of \(1.75 \mathrm{m} / \mathrm{s},\) you are moving in a direction that is \(18.0^{\circ}\) north of east. How much time does it take to change your displacement by (a) \(20.0 \mathrm{m}\) east or (b) \(30.0 \mathrm{m}\) north?

Short Answer

Expert verified
(a) 12.05 seconds for 20.0 m east; (b) 55.56 seconds for 30.0 m north.

Step by step solution

01

Understanding the Components of Velocity

The problem provides a velocity of \(1.75 \text{ m/s}\) in a specified direction. We need to separate this velocity into components along east (x-axis) and north (y-axis). This is done using trigonometric functions. Let \(v_x\) be the eastward component and \(v_y\) be the northward component.
02

Calculate Eastward and Northward Velocity Components

Using trigonometry, we find \(v_x = 1.75 \times \cos(18^\circ)\) and \(v_y = 1.75 \times \sin(18^\circ)\). Calculating these values gives us: \[ v_x = 1.75 \times \cos(18^\circ) \approx 1.66 \text{ m/s} \] \[ v_y = 1.75 \times \sin(18^\circ) \approx 0.54 \text{ m/s} \]
03

Determine Time for 20.0 m East Displacement

To find time \(t\) taken for a displacement of \(20.0 \text{ m}\) east, use the formula \(t = \frac{d}{v_x}\), where \(d\) is the displacement. Substituting the values: \[ t = \frac{20.0}{1.66} \approx 12.05 \text{ seconds} \]
04

Determine Time for 30.0 m North Displacement

To find time \(t\) for a displacement of \(30.0 \text{ m}\) north, use \(t = \frac{d}{v_y}\), where \(d\) is the displacement. Substituting the values: \[ t = \frac{30.0}{0.54} \approx 55.56 \text{ seconds} \]

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Key Concepts

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

Velocity Components
When objects move in directions that are not aligned with standard axes (like the north or east), describing their motion can become tricky. This is where velocity components are crucial. Each velocity vector can be broken down into components that correspond to the directions of interest, typically horizontal and vertical. In this exercise, our initial velocity is directed at an angle of \(18^{\circ}\) north of east.
These directions form our axes:
  • Eastward Direction (x-axis)
  • Northward Direction (y-axis)
By using the initial velocity and some trigonometric functions, we can extract:
  • The eastward velocity component (\(v_x\))
  • The northward velocity component (\(v_y\))
Breaking down the velocity into components helps in analyzing and solving this exercise systematically.
Trigonometry in Physics
Trigonometry plays a pivotal role in physics, especially when dealing with forces and velocities that act at angles. In our exercise, we use trigonometric functions to decompose a velocity vector into its components.
The two main functions we use here are:
  • Cosine function (\(\cos\theta\)) for eastward or horizontal component
    Equation: \(v_x = v \times \cos(\theta)\)
  • Sine function (\(\sin\theta\)) for northward or vertical component
    Equation: \(v_y = v \times \sin(\theta)\)
This decomposition allows us to understand how much of the velocity contributes to each axis. Whether it's calculating travel routes or object motion in physics, trigonometry is essential for breaking down and understanding angles in vector quantities.
Displacement Calculation
Displacement tells us how far and in what direction an object has moved. It differs from distance; displacement focuses on the shortest path between the starting and ending points.
In this exercise, we need to find out how long it takes to change displacement by specific amounts in the east and north directions.
  • Eastward Displacement: Calculated using the formula \(t = \frac{d}{v_x}\) where \(d\) is the desired displacement and \(v_x\) is the eastward velocity component.
    For a \(20.0 \text{ m}\) change, it takes approximately \(12.05 \text{ seconds}\).
  • Northward Displacement: Similarly calculated with \(t = \frac{d}{v_y}\) where \(v_y\) is the northward velocity component.
    A \(30.0 \text{ m}\) northward change takes about \(55.56 \text{ seconds}\).
These calculations help correlate velocity components to specific directional displacements, showing how time is involved in the process.

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Most popular questions from this chapter

A ball rolls off a table and falls \(0.75 \mathrm{m}\) to the floor, landing with a speed of \(4.0 \mathrm{m} / \mathrm{s}\). (a) What is the acceleration of the ball just before it strikes the ground? (b) What was the initial speed of the ball? (c) What initial speed must the ball have if it is to land with a speed of \(5.0 \mathrm{m} / \mathrm{s} ?\)

Astronomers have discovered several volcanoes on Io, a moon of Jupiter. One of them, named Loki, ejects lava to a maximum height of \(2.00 \times 10^{5} \mathrm{m}\). (a) What is the initial speed of the lava? (The acceleration of gravity on lo is \(1.80 \mathrm{m} / \mathrm{s}^{2}\).) (b) If this volcano were on Earth, would the maximum height of the ejected lava be greater than, less than, or the same as on Io? Explain.

A soccer ball is kicked with an initial speed of \(10.2 \mathrm{m} / \mathrm{s}\) in a direction \(25.0^{\circ}\) above the horizontal. Find the magnitude and direction of its velocity (a) \(0.250 \mathrm{s}\) and (b) 0.500 s after being kicked. (c) Is the ball at its greatest height before or after 0.500 s? Explain.

When the twin Mars exploration rovers, Spirit and Opportunity, set down on the surface of the red planet in January of 2004 their method of landing was both unique and elaborate. After initial braking with retro rockets, the rovers began their long descent through the thin Martian atmosphere on a parachute until they reached an altitude of about \(16.7 \mathrm{m}\). At that point a system of four air bags with six lobes each were inflated, additional retro rocket blasts brought the craft to a virtual standstill, and the rovers detached from their parachutes. After a period of free fall to the surface, with an acceleration of \(3.72 \mathrm{m} / \mathrm{s}^{2}\), the rovers bounced about a dozen times before coming to rest. They then deflated their air bags, righted themselves, and began to explore the surface. Figure \(4-25\) shows a rover with its surrounding cushion of air bags making its first contact with the Martian surface. After a typical first bounce the upward velocity of a rover would be \(9.92 \mathrm{m} / \mathrm{s}\) at an angle of \(75.0^{\circ}\) above the horizontal. Assume this is the case for the problems that follow. How far does a rover travel in the horizontal direction between its first and second bounces? A. \(13.2 \mathrm{m}\) B. \(49.4 \mathrm{m}\) C. \(51.1 \mathrm{m}\) D. \(98.7 \mathrm{m}\)

Babe Didrikson holds the world record for the longest baseball throw \((296 \mathrm{ft})\) by a woman. For the following questions, assume that the ball was thrown at an angle of \(45.0^{\circ}\) above the horizontal, that it traveled a horizontal distance of \(296 \mathrm{ft},\) and that it was caught at the same level from which it was thrown. (a) What was the ball's initial speed? (b) How long was the ball in the air?
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