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Chapter 2: Problem 2

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 \(\mathrm{m} \mathrm{m} / \mathrm{s}(\mathrm{a})\) 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) 0 m/s.

Step by step solution

01

Convert Distance to Meters

First, convert the distance of 5150 km to meters. Distance in meters \( = 5150 \times 1000 = 5,150,000 \text{ meters} \).
02

Determine the Time in Seconds

Convert the time of 13.5 days to seconds. We know there are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute: \( 13.5 \times 24 \times 60 \times 60 = 1,166,400 \text{ seconds} \).
03

Calculate Average Velocity for Return Flight

The average velocity for the return flight is the total distance divided by the total time taken: \( v = \frac{5,150,000}{1,166,400} \approx 4.42 \text{ m/s} \).
04

Calculate Average Velocity for Whole Episode

For the entire episode, the displacement is zero because the bird returns to its starting point, the origin. Therefore, the average velocity \( v = \frac{0}{1,166,400 + t} = 0 \text{ m/s}\), where \( t \) is the initial travel time, which is not needed as displacement is zero.

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Key Concepts

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

Understanding Displacement
In physics, displacement refers to how far out of place an object is; it is the object's overall change in position. Unlike distance, which considers the total path traveled, displacement focuses only on the initial and final positions. This is why displacement can be zero even if the distance is significant.
  • Displacement is a vector quantity, meaning it has both magnitude and direction.
  • If an object returns to its starting point, the displacement is zero, despite the distance covered.
This concept was highlighted in the shearwater example, where the bird returned to its nest. Despite traveling a great distance, the displacement for the entire journey was zero because the starting and ending point were the same. This is why the average velocity for the whole episode was calculated as zero.
The Art of Unit Conversion
Converting units is essential in physics to ensure that calculations are accurate and meaningful. In the shearwater exercise, distance was initially given in kilometers, but needed to be expressed in meters to calculate average velocity.
  • To convert kilometers to meters, multiply by 1,000 because there are 1,000 meters in a kilometer.
  • For time, converting days to seconds involves several steps:
    • Days are converted to hours by multiplying by 24 (hours per day).
    • Hours are converted to minutes by multiplying by 60 (minutes per hour).
    • Finally, minutes are converted to seconds by multiplying by 60 (seconds per minute).
In this problem, we convert 5150 km to 5,150,000 meters and 13.5 days to 1,166,400 seconds. These are essential conversions for maintaining consistent units and ensuring precise calculations.
Cracking Distance-Time Calculations
Distance-time calculations are vital for deriving the average velocity of moving objects. Average velocity is given by the formula\[ v = \frac{\text{displacement}}{\text{time}}. \]This formula highlights the relationship between displacement and time to determine how fast an object is moving.
  • To find average velocity for the return flight, divide the total distance (in meters) by the total time (in seconds).
  • When the object returns to its starting point, such as in the complete journey of the bird, the displacement is zero, hence average velocity is zero.
Understanding these calculations allows us to differentiate between distance traveled and displacement, emphasizing that average velocity is contingent on initial and final positions rather than the total path covered.

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Most popular questions from this chapter

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

The position of a particle between \(t=0\) and \(t=2.00 \mathrm{s}\) is given by \(x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+(9.00 \mathrm{m} / \mathrm{s}) t\) (a) Draw the \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs of this particle. (b) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle instantaneously at rest? Does your numerical result agree with the \(v_{x}-t\) graph in part \((a) ?\) At each time calculated in part \((b)\) is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from \(a_{x}(t)\) and from the \(v_{x}-t\) graph. (d) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the velocity of the particle instantancously not changing? Locate this point on the \(v_{x} t\) and \(a_{x}-t\) graphs of part (a). (e) What is the particle's greatest distance from the origin \((x=0)\) between \(t=0\) and \(t=2.00 \mathrm{s} ?(\mathrm{f})\) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle speeding \(u p\) at the greatest rate? At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle slowing down at the greatest rate? Locate these points on the \(v_{x}-t\) and \(a_{x}-t\) graphs of part (a).

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 \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.

An antelope moving with constant acceleration covers the distance between two points 70.0 \(\mathrm{m}\) apart in 7.00 \(\mathrm{s}\) . Its speed as it passes the second point is 15.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is its speed at the first point? \((\mathrm{b})\) What is its acceleration?

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 \(\mathrm{m}\) away at a constant speed of \(2.50 \mathrm{m} / \mathrm{s},\) returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?
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