/*! 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 48 The world record for the hundred... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The world record for the hundred meter dash is \(9.74 \mathrm{~s}\). What is the corresponding average speed in units of \(\mathrm{m} / \mathrm{s}, \mathrm{km} / \mathrm{h}, \mathrm{ft} / \mathrm{s}\), and \(\mathrm{mi} / \mathrm{h}\) ? At this speed, how long would it take to run \(1.00 \times 10^{2}\) yards?

Short Answer

Expert verified
The corresponding average speeds for the world record 100-meter dash time of 9.74 seconds are as follows: 10.27 m/s, 36.97 km/h, 33.69 ft/s, and 22.97 mi/h. At this speed, it would take approximately 8.90 seconds to run 100 yards.

Step by step solution

01

Convert time to seconds

First, let's represent the given time in seconds: \(9.74 \mathrm{~s}\)
02

Calculate average speed in m/s

Now, we can find the average speed in m/s simply by dividing the distance by the time: Average speed = \(\frac{Distance}{Time}\) = \(\frac{100\mathrm{m}}{9.74\mathrm{s}}\) = \(10.27 \mathrm{~m/s}\)
03

Convert m/s to km/h

To convert the average speed from m/s to km/h, we can use the following conversion: 1 m/s = 3.6 km/h So, the average speed in km/h = \(10.27 \mathrm{~m/s} * 3.6 \frac{\mathrm{km}}{\mathrm{h}}\) = \(36.97 \mathrm{~km/h}\)
04

Convert m/s to ft/s

To convert the average speed in m/s to ft/s, we can use the following conversion: 1 meter = 3.281 feet So, the average speed in ft/s = \(10.27 * 3.281 \mathrm{~ft/s}\) = \(33.69 \mathrm{~ft/s}\)
05

Convert ft/s to mi/h

To convert the average speed from ft/s to mi/h, we can use the following conversion: 1 ft/s = 0.6818 mi/h So, the average speed in mi/h = \(33.69 * 0.6818 \mathrm{~mi/h}\) = \(22.97 \mathrm{~mi/h}\)
06

Convert the distance to yards

Next, we need to convert the given distance of 100 yards to meters: 1 yard = 0.9144 meters So, the distance in meters = \(100 * 0.9144 \mathrm{~m}\) = \(91.44 \mathrm{~m}\)
07

Calculate time to run 100 yards

Now, we can use the average speed in m/s and the distance in meters to calculate the time required to run 100 yards at the world record speed: Time = \(\frac{Distance}{Speed}\) = \(\frac{91.44\mathrm{~m}}{10.27\mathrm{~m/s}}\) = \(8.90 \mathrm{~s}\) Thus, it would take approximately 8.90 seconds to run 100 yards at the world record speed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Speed Conversion
Understanding speed conversion is crucial when dealing with exercises that require reporting the result in various units. This concept refers to the process of changing the value of speed from one unit of measurement to another without altering the actual pace. For example, converting the average speed of a runner from meters per second (m/s) to miles per hour (mi/h) involves multiplying or dividing by specific conversion factors, such as using 3.6 to go from m/s to kilometers per hour (km/h).

An important step is to know the most common conversion factors or to have access to a conversion chart. Keep in mind the unit congruence; units on both sides of an equation must be consistent. In some cases, multiple conversions might be necessary to get the final desired unit—each conversion being a simple multiplication or division operation. Practice with various speeds and units strengthens one's ability to quickly and accurately perform these conversions.
Unit Conversion
Unit conversion is absolutely necessary not only in physics but also in many real-world applications. It involves calculating the equivalent amount of one measurement unit in terms of another unit of the same dimension. For example, when converting feet to meters or yards to meters, as seen in the exercise, the key lies in knowing the precise conversion factor between the units. In the provided solution, using 0.9144 meters for each yard was essential to ascertain the distance in a unit compatible with the speed previously calculated in meters per second.

Since units such as distance and time are fundamentals in speed calculations, effective problem-solving requires students to master this skill. Utilizing the correct conversion factors lets us accurately translate one measurement into another, securing the integrity of computations and conclusions derived from them.
Problem-solving in Physics
Physics problems often involve a systematic approach to finding a solution. The process begins with understanding the question, identifying what is given and what needs to be found—a fundamental step to avoid confusion during calculations. Next involves formingulating a strategy, such as using the formula for average speed, which is the distance traveled divided by the time taken. In this example, the initial task was to determine the average speed in various units given the world record for a distance in meters and time in seconds.

In physics, problem-solving requires careful reasoning and logical steps, verified with unit consistency and conversion where necessary. Knowing to work sequentially, like how it was first necessary to calculate the speed in m/s before converting it to other units, is key. Applying these problem-solving techniques allows for a clear path to an accurate and comprehensive solution in physics exercises.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

During a very cold winter, the temperature may remain below freezing for extended periods. However, fallen snow can still disappear, even though it cannot melt. This is possible because a solid can vaporize directly, without passing through the liquid state. Is this process (sublimation) a physical or a chemical change?

You go to a convenience store to buy candy and find the owner to be rather odd. He allows you to buy pieces in multiples of four, and to buy four, you need \(\$ 0.23 .\) He only allows you to do this by using 3 pennies and 2 dimes. You have a bunch of pennies and dimes, and instead of counting them, you decide to weigh them. You have \(636.3 \mathrm{~g}\) of pennies, and each penny weighs \(3.03 \mathrm{~g}\). Each dime weighs \(2.29 \mathrm{~g}\). Each piece of candy weighs \(10.23 \mathrm{~g}\). a. How many pennies do you have? b. How many dimes do you need to buy as much candy as possible? c. How much should all these dimes weigh? d. How many pieces of candy could you buy? (number of dimes from part b) e. How much would this candy weigh? f. How many pieces of candy could you buy with twice as many dimes?

Consider multiplying \(26.2\) by \(16.43 .\) What would a mathematician say the answer is? What would a scientist say? Justify the scientist's answer, not merely citing the rule, but explaining it.

An experiment was performed in which an empty \(100-\mathrm{mL}\) graduated cylinder was weighed. It was weighed once again after it had been filled to the \(10.0-\mathrm{mL}\) mark with dry sand. A \(10-\mathrm{mL}\) pipet was used to transfer \(10.00 \mathrm{~mL}\) of methanol to the cylinder. The sand- methanol mixture was stirred until bubbles no longer emerged from the mixture and the sand looked uniformly wet. The cylinder was then weighed again. Use the data obtained from this experiment (and displayed at the end of this problem) to find the density of the dry sand, the density of methanol, and the density of sand particles. Does the bubbling that occurs when the methanol is added to the dry sand indicate that the sand and methanol are reacting? Mass of cylinder plus wet sand \(\quad 45.2613 \mathrm{~g}\) Mass of cylinder plus dry sand \(\quad 37.3488 \mathrm{~g}\) Mass of empty cylinder \(22.8317 \mathrm{~g}\) Volume of dry sand \(10.0 \mathrm{~mL}\) Volume of sand plus methanol \(\quad 17.6 \mathrm{~mL}\) Volume of methanol \(\quad 10.00 \mathrm{~mL}\)

What is the difference between homogeneous and heterogeneous matter? Classify each of the following as homogeneous or heterogeneous. a. a door b. the air you breathe c. a cup of coffee (black) d. the water you drink e. salsa f. your lab partner

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.