Question

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

Short answer

(a) 4.42 m/s; (b) 0 m/s.

Step-by-step solution

Understand the Given Data

For the first part of the question, the shearwater travels 5150 km to return to its nest. This distance is the displacement for the return flight. The time taken for this return flight is 13.5 days.

Convert Units for Velocity Calculation

Convert the distance from kilometers to meters: 5150 km = 5150,000 m. Convert the time from days to seconds: 13.5 days = 13.5 \(\times \) 24 \(\times \) 60 \(\times \) 60 = 1166400 seconds.

Calculate Average Velocity for Return Flight

Average velocity is defined as total displacement divided by total time. Here, the displacement is 5150,000 m and the time is 1166400 s. So, average velocity \(v_{avg} = \frac{5150,000}{1166400} \text{ m/s}\).

Simplify for Part (a)

Solving for \(v_{avg}\), we get \(v_{avg} \approx 4.42 \text{ m/s}\). This is the average velocity for the return flight.

Understand Displacement for Whole Episode

For the entire episode from leaving the nest to coming back, the net displacement is 0 because the bird ends at its starting point (the nest).

Calculate Average Velocity for Entire Episode

The formula for average velocity over the entire trip is displacement divided by time. Since the displacement is 0, the average velocity for the entire episode is \(0 \text{ m/s}\).

Key concepts

Understanding Displacement

In the study of motion, displacement is a crucial concept. Displacement refers to the change in position of an object. It is a vector quantity, which means it has both magnitude and direction. In our exercise, the bird's displacement for the return flight is the straight-line distance back to its nest, not accounting for any detours or zigzagging. This distance is given as 5150 kilometers from the release point back to the nest.
The key aspect here is that for the entire episode, from the bird leaving the nest and returning, the displacement is zero. This is because displacement considers only the initial and final positions. Since the bird starts and ends at the same place, the net displacement is zero, regardless of the distance travelled.

Importance of Unit Conversion

Unit conversion is fundamental when dealing with physics problems, especially when calculating velocity. Measurements are often initially in different units. Here, we needed to convert distances from kilometers to meters and time from days to seconds.
This conversion is done because the standard unit of velocity in physics is meters per second (m/s). To convert 5150 km to meters, we multiply by 1000 (since 1 km = 1000 m), resulting in 5,150,000 meters.

• 5150 km = 5150 × 1000 = 5,150,000 meters.

Similarly, converting time from days to seconds involves converting days to hours, hours to minutes, and minutes to seconds. This step is crucial for accurate calculations.

Calculating Average Velocity

Average velocity is computed by dividing total displacement by the total time of travel. It provides insight into how fast something is moving along a straight path between two points. For the bird's return flight, the displacement was 5,150,000 meters, and the time taken was 1,166,400 seconds.
Using the formula for average velocity:

• \[v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{5,150,000 \, \text{m}}{1,166,400 \, \text{s}}\]

For this part of the journey, the average velocity was approximately 4.42 m/s.
For the entire episode, since the displacement is zero, the average velocity is zero. This illustrates that average velocity depends significantly on the net change in position, not just the distance traveled.

Understanding the Distance-Time Relationship

The relationship between distance and time is foundational in motion problems. Distance corresponds to how much ground an object covers, while time measures how long the movement takes. For uniform motion, velocity bridges these two concepts, showing how distance and time interact.
In this exercise, even though the bird traveled a significant path back to its nest, the time duration (13.5 days converted to seconds) influences the velocity calculation. This highlights an essential understanding:

• Long distance doesn't always translate to high velocity if the duration is extensive.
• Velocity is a vector and can differ considerably from speed, a scalar quantity that considers only distance covered irrespective of direction.

Hence, students should focus on both magnitude and direction to fully comprehend velocity and its related concepts.