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Chapter 2: Problem 9

In \(1954,\) Roger Bannister became the first human to run a mile in less than 4 minutes. Suppose that a runner on a straight track covers a distance of 1.00 mi in exactly 4.00 min. What is his average speed in (a) \(\mathrm{mi} / \mathrm{h},\) (b) \(\mathrm{ft} / \mathrm{s},\) and \((\mathrm{c}) \mathrm{m} / \mathrm{s}\) ?

Short Answer

Expert verified
(a) 15 mi/h, (b) 22 ft/s, (c) 6.71 m/s

Step by step solution

01

Convert Time to Hours

Since we want to find the average speed in miles per hour, we first convert the time from minutes to hours. There are 60 minutes in an hour, so for 4 minutes: \[\text{Time in hours} = \frac{4 \text{ minutes}}{60 \text{ minutes per hour}} = \frac{4}{60} = \frac{1}{15} \text{ hours}.\]
02

Calculate Average Speed in Miles per Hour

With the distance known as 1 mile and the time as \(\frac{1}{15}\) hours, the average speed (V) is given by the formula:\[V = \frac{\text{Distance}}{\text{Time}} = \frac{1 \text{ mile}}{\frac{1}{15} \text{ hour}} = 15 \text{ miles per hour}.\]
03

Convert Miles to Feet

To find the speed in feet per second, first convert the distance from miles to feet. There are 5280 feet in a mile, so:\[1 \text{ mile} = 1 \times 5280 \text{ feet} = 5280 \text{ feet}.\]
04

Convert Time to Seconds

Next, convert the time from minutes to seconds. There are 60 seconds in a minute, so for 4 minutes:\[4 \text{ minutes} = 4 \times 60 \text{ seconds} = 240 \text{ seconds}.\]
05

Calculate Average Speed in Feet per Second

Using the distance in feet and the time in seconds, the average speed (V) is:\[V = \frac{\text{Distance}}{\text{Time}} = \frac{5280 \text{ feet}}{240 \text{ seconds}} = 22 \text{ feet per second}.\]
06

Convert Feet to Meters

For speed in meters per second, convert feet to meters. One foot is 0.3048 meters, so:\[5280 \text{ feet} \times 0.3048 \text{ meters per foot} = 1609.344 \text{ meters}.\]
07

Calculate Average Speed in Meters per Second

Using the distance in meters and time in seconds, the average speed (V) is:\[V = \frac{\text{Distance}}{\text{Time}} = \frac{1609.344 \text{ meters}}{240 \text{ seconds}} \approx 6.7056 \text{ meters per second}.\]

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Key Concepts

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

Unit Conversion
Unit conversion involves changing a measurement from one unit to another. It is an essential step in solving many physics problems, as units must be consistent to accurately calculate values like speed, distance, or time.
For instance, if you are given time in minutes but need to find speed in miles per hour, you must first convert minutes to hours. This ensures all units align correctly, allowing precise and meaningful calculations.
  • Convert time: minutes to seconds or hours.
  • Convert distance: miles to feet or meters.
Using proper unit conversion makes it easier to understand the relationships between different measurements, which is crucial for overall accuracy in calculations.
Mile to Kilometer Conversion
Converting miles to kilometers is necessary when you need to perform calculations with different unit systems, or when international standards require metric units. The conversion factor between miles and kilometers is that 1 mile equals approximately 1.60934 kilometers.
To perform the conversion, multiply the number of miles by the conversion factor:
  • 1 mile = 1.60934 kilometers
  • 10 miles = 10 脳 1.60934 = 16.0934 kilometers
Knowing this conversion factor is incredibly useful, especially when dealing with calculations across the miles and kilometers divide, ensuring that all participants understand the data in their familiar measurement system.
Distance-Time Relationship
In physics, understanding the distance-time relationship is crucial for calculating speed. Speed is a measure of how fast an object moves, often represented as distance traveled over time. The standard formula for average speed is:\[ V = \frac{\text{Distance}}{\text{Time}} \]
This formula provides a clear understanding that the speed depends on how much distance is covered over a given time period. Whether the problem involves converting units of time or distance, the relationship remains constant. Grasping this concept can simplify many physics problems.
Physics Problem Solving
Physics problem solving often involves breaking down complex problems into smaller, more manageable steps. This approach increases accuracy and can make tackling challenging problems more intuitive.
Here's an effective strategy:
  • Identify the given information and what needs to be found.
  • Use appropriate formulas, making sure units are consistent.
  • Perform necessary unit conversions.
  • Substitute numbers into formulas and solve.
By systematically approaching a problem, students can navigate through even the most intricate physics questions with confidence, achieving solutions that are both correct and easy to understand.

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Most popular questions from this chapter

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is its speed after falling freely for 2.00 s? (b) How far does it fall in 2.00 s? (c) What is the magnitude of its velocity after falling 10.0 \(\mathrm{m} ?\)

Freeway ramps. In designing a freeway on-ramp, you must make it long enough for cars to be able to accelerate and merge safely with traffic traveling at speeds of 70 mph. For an off-ramp, the cars must be able to come to a stop from 70 mph over the length of the ramp. Your traffic engineers provide the following data: Powerful cars can accelerate from rest to 60 mph in 5.0 s, while less powerful cars take twice as long to do this. When slowing down from 60 mph, a car with good tire treads will take 12.0 s to stop, whereas one with bald tires takes 20.0 s. In order to safely accommodate all the types of vehicles, what should be the minimum length (in meters) of on-ramps and off- ramps? Assume constant acceleration.

A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no appreciable air drag. (a) How tall, in meters, is the building? (b) How fast is the brick moving just before it reaches the ground? (c) Sketch graphs of this falling brick's acceleration, velocity, and vertical position as functions of time.

Faster than a speeding bullet! The Beretta Model 92\(S\) (the standard-issue U.S. army pistol) has a barrel 127 \(\mathrm{mm}\) long. The bullets leave this barrel with a muzzle velocity of 335 \(\mathrm{m} / \mathrm{s}\) (a) What is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g^{\prime} \mathrm{s}\) . (b) For how long is the bullet in the barrel?

Egg drop. You are on the roof of the physics building of your school, 46.0 \(\mathrm{m}\) above the ground. (See Figure \(2.53 . )\) Your physics professor, who is 1.80 \(\mathrm{m}\) tall, is walking alongside the building at a constant speed of 1.20 \(\mathrm{m} / \mathrm{s} .\) If you wish to drop an egg on your professor's head, where should the professor be when you release the egg, assuming that the egg encounters no appreciable air drag.
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