/*! 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 17 A finch rides on the back of a G... [FREE SOLUTION] | 91Ó°ÊÓ

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A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of \(0.060 \mathrm{m} / \mathrm{s}\). After \(1.2 \mathrm{minutes}\) the finch tires of the tortoise"s slow pace, and takes flight in the same direction for another 1.2 minutes at \(12 \mathrm{m} / \mathrm{s}\). What was the average speed of the finch for this 2.4 -minute interval?

Short Answer

Expert verified
The average speed of the finch was approximately 6.03 m/s.

Step by step solution

01

Convert Time to Seconds

Convert the time intervals from minutes to seconds to use consistent units for speed calculations. The first time interval is 1.2 minutes, which is equivalent to \(1.2 \times 60 = 72\) seconds. The second time interval is also 1.2 minutes, which converts to another 72 seconds. Overall, the total time is \(72 + 72 = 144\) seconds.
02

Calculate Distance on Tortoise's Back

Calculate the distance the finch travels on the tortoise's back. Use the formula \(\text{distance} = \text{speed} \times \text{time}\). The speed of the tortoise is \(0.060\, \text{m/s}\) and the time is 72 seconds. Thus, the distance traveled is \(0.060 \times 72 = 4.32\, \text{m}\).
03

Calculate Distance While Flying

Calculate the distance the finch travels while flying. The finch flies at \(12 \text{ m/s}\) for 72 seconds. The distance covered while flying is \(12 \times 72 = 864\, \text{m}\).
04

Calculate Total Distance Traveled

Add the distances traveled on the tortoise's back and while flying to find the total distance. Thus, the total distance is \(4.32 + 864 = 868.32\, \text{m}\).
05

Calculate Average Speed

Use the formula for average speed, which is \(\text{average speed} = \frac{\text{total distance}}{\text{total time}}\). Substitute the total distance (868.32 m) and total time (144 seconds) to find the average speed: \(\frac{868.32}{144} \approx 6.03 \text{ m/s}\).

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Key Concepts

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

Distance Calculation
When we talk about distance calculation, we are determining how far an object travels while moving from one point to another. In this exercise, the finch moves in two phases: first on the tortoise's back and then by flying.

To calculate the distance in each phase, we use the formula: \[\text{distance} = \text{speed} \times \text{time}\]For example, while riding the tortoise, the finch's speed is a leisurely \(0.060 \text{ m/s}\). Over \(72\) seconds, the distance covered is \(0.060 \times 72 = 4.32 \text{ meters}\). Similarly, while flying at a swift \(12 \text{ m/s}\) for another \(72\) seconds, the distance is \(12 \times 72 = 864 \text{ meters}\).

This simple multiplication gives us the total distance traveled in each phase. Adding these values gives the total distance covered during the entire journey, which is \(868.32 \text{ meters}\). Understanding this step is crucial as it forms the basis of the speed and motion concepts in physics.
Unit Conversion
Unit conversion is an essential skill when dealing with problems in physics, as it ensures consistency across calculations. In the given problem, we encounter distances and speeds given in meters and meters per second, respectively.

However, time is initially provided in minutes. To make calculations seamless and compatible with these units, we convert minutes into seconds. Knowing there are \(60\) seconds in a minute, we multiply the given minutes by \(60\) to convert them.

For example, the finch spends \(1.2\) minutes in each phase of its journey. By converting this to seconds, we calculate \(1.2 \times 60 = 72\) seconds for each segment.

Performing this conversion ensures we are using the same unit of measure, which prevents errors and enhances the accuracy of our calculations.
Speed Formula
The speed formula is a fundamental concept that relates the distance traveled by an object to the time it takes. The formula is as follows:\[\text{speed} = \frac{\text{distance}}{\text{time}}\]In this exercise, we apply a specific version of this formula to find the average speed. When an object moves at different speeds during different intervals, its average speed gives us an understanding of the overall pace of the journey.

To find the average speed of the finch, we first calculate the total distance traveled, which is \(868.32\) meters, and divide it by the total time spent, \(144\) seconds. Thus, the average speed becomes:\[\frac{868.32}{144} \approx 6.03 \text{ m/s}\]

Remember, average speed is about the entire journey, irrespective of speed changes during different intervals. It's a simple yet powerful way to summarize motion.

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Most popular questions from this chapter

While riding on an elevator descending with a constant speed of \(3.0 \mathrm{m} / \mathrm{s}\), you accidentally drop a book from under your arm. (a) How long does it take for the book to reach the elevator floor, \(1.2 \mathrm{m}\) below your arm? (b) What is the book's speed relative to you when it hits the elevator floor?

You shoot an arrow into the air. Two seconds later ( 2.00 s) the arrow has gone straight upward to a height of \(30.0 \mathrm{m}\) above its launch point. (a) What was the arrow's initial speed? (b) How long did it take for the arrow to first reach a height of \(15.0 \mathrm{m}\) above its launch point?

You drive in a straight line at \(20.0 \mathrm{m} / \mathrm{s}\) for \(10.0 \mathrm{miles}\), then at \(30.0 \mathrm{m} / \mathrm{s}\) for another \(10.0 \mathrm{miles}\). (a) Is your average speed \(25.0 \mathrm{m} / \mathrm{s},\) more than \(25.0 \mathrm{m} / \mathrm{s},\) or less than \(25.0 \mathrm{m} / \mathrm{s}\) ? Explain. (b) Verify your answer to part (a) by calculating the average speed.

A hot-air balloon is descending at a rate of \(2.0 \mathrm{m} / \mathrm{s}\) when a passenger drops a camera. If the camera is \(45 \mathrm{m}\) above the ground when it is dropped, (a) how long does it take for the camera to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

Two bows shoot identical arrows with the same launch speed. To accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2 . (a) Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow \(2 ?\) (b) Choose the best explanation from among the following: The arrow in bow 2 accelerates for a greater time. II. Both arrows start from rest. III. The arrow in bow 1 accelerates for a greater time.

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