/*! 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 4 In an experiment, a shearwater (... [FREE SOLUTION] | 91Ó°ÊÓ

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In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 \(\mathrm{km}\) away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin in the nest and extend the \(+x\) -axis to the release point, what was the bird's average velocity in \(\mathrm{m} / \mathrm{s}\) (a) for the return flight, and (b) for the whole episode, from leaving the nest to returning?

Short Answer

Expert verified
(a) 4.42 m/s, (b) 8.83 m/s.

Step by step solution

01

Understanding the Problem

We need to calculate the average velocity of the shearwater in two scenarios: (a) for the return journey from release to the nest, and (b) for the entire journey, which includes both the outward and return trips.
02

Convert Distance to Meters

The given distance the bird was flown is 5150 km. To convert this to meters, we use the conversion 1 km = 1000 meters. Therefore, the distance in meters is: \[ 5150 \text{ km} \times 1000 = 5,150,000 \text{ meters} \]
03

Duration in Seconds

The bird took 13.5 days to return. First, we convert the days into seconds. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, \[ 13.5 \text{ days} \times 24 \times 60 \times 60 = 1,166,400 \text{ seconds} \]
04

Calculate Average Velocity for Return Flight

The average velocity \( v \) can be calculated with the formula \( v = \frac{d}{t} \), where \( d \) is distance and \( t \) is time. For the return flight, \( d = 5,150,000 \text{ meters} \), and \( t = 1,166,400 \text{ seconds} \). Thus, \[ v = \frac{5,150,000}{1,166,400} \approx 4.42 \text{ m/s} \]
05

Calculate Average Velocity for the Whole Episode

The total distance for the whole episode is the sum of the onward and return journeys, which is \( 5,150 \text{ km} \times 2 \), so \( 10,300 \text{ km} \). Convert to meters: \[ 10,300 \times 1000 = 10,300,000 \text{ meters} \] The total time for the whole episode is also 13.5 days (since the outward journey time is not considered separately here), which is \( 1,166,400 \text{ seconds} \). Thus, the average velocity for the whole episode is \[ \frac{10,300,000}{1,166,400} \approx 8.83 \text{ m/s} \]

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Key Concepts

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

Kinematics
Let's dive into the world of kinematics to understand the behavior of moving objects, like our bird. Kinematics focuses on the motion of objects, without considering the forces that cause such movement. One of the primary variables in kinematics is velocity, which describes how fast an object is moving in a specific direction. Average velocity, in our case, involves calculating the overall change in position over the total time taken. This concept is key in both parts of our exercise because the bird's journey involves understanding how its position changes over time. In simple terms, if you know the starting point and how far and fast something travels over a period, you can figure out details about its entire motion.
Unit Conversion
Converting units is essential when solving physics problems, as it ensures all measurements are in the same system, allowing for accurate calculations. For our exercise, we initially have the distance in kilometers, but we need it in meters for calculating velocity in meters per second. To convert kilometers to meters, simply multiply by 1,000 because there are 1,000 meters in each kilometer. This conversion is straightforward: Just think of it as shifting the decimal point three places to the right. The same goes for converting days to seconds. Since time is given in days and we need it in seconds, remember that:
  • 1 day = 24 hours
  • 1 hour = 60 minutes
  • 1 minute = 60 seconds
Multiply each step to get the full time duration in seconds, enabling the calculation of velocity.
Distance and Displacement
In physics, distance and displacement might sound similar but are quite different. Distance is the total path length traveled by an object, without considering the direction, while displacement focuses on the overall change in position from start to finish and takes direction into account. For the shearwater's return journey, displacement is taken as the straight-line distance from the release point back to the nest, while the total distance for the entire trip accounts for both flying away and returning, doubling the single trip's distance. Understanding these differences helps clarify why calculations differ based on whether we're examining the return journey (displacement) or the entire round trip (total distance).
Time Conversion
Time conversion is often needed to ensure uniformity when computing other values like velocity. In this exercise, we start with the bird's return journey time in days and need it in seconds for calculating average velocity. To effectively handle time conversion:
  • Remember the basic time units: days, hours, minutes, and seconds.
  • Use step-by-step multiplication: 1 day equals 24 hours, each hour has 60 minutes, and each minute contains 60 seconds.
  • This breakdown helps convert days into seconds seamlessly, leading to an accurate calculation of time over which the velocity is averaged.
Doing so ensures that the computed average velocity conforms with the units needed, giving precise results in m/s.

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Most popular questions from this chapter

Airplane \(A,\) starting from rest with constant acceleration, requires a runway 500 \(\mathrm{m}\) long to become airborne. Airplane \(B\) requires a takeoff speed twice as great as that of airplane \(A\) but has the same acceleration, and both planes start from rest. (a) How long must the runway be for airplane \(B ?\) (b) If airplane \(A\) takes time \(T\) to travel the length of its runway, how long (in terms of \(T )\) will airplane \(B\) take to travel the length of its runway?

Blackout? A jet fighter pilot wishes to accelerate from rest at 5\(g\) to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5\(g\) before blacking out?

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s}\) . Air resistance may be ignored. (a) At what time after being ejected is the boulder moving at 20.0 \(\mathrm{m} / \mathrm{s}\) upward? (b) At what time is it moving at 20.0 \(\mathrm{m} / \mathrm{s}\) downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_{y}-t, v_{v}-t,\) and \(y-t\) graphs for the motion.

(. A tennis ball on Mars, where the acceleration due to gravity is 0.379\(g\) and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go? (b) How fast was it moving just after being hit? (c) Sketch clear graphs of the ball's vertical position, vertical velocity, and vertical acceleration as functions of time while it's in the Martian air.

A fast pitch. The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . If the pitcher was in contact with the ball over a distance of 1.50 \(\mathrm{m}\) and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

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