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The average person passes out at an acceleration of \(7 \mathrm{~g}\) (that is, seven times the gravitational acceleration on Earth). Suppose a car is designed to accelerate at this rate. How much time would be required for the car to accelerate from rest to \(60.0\) miles per hour? (The car would need rocket boosters!)

Short Answer

Expert verified
It would take approximately 0.391 seconds for the car to accelerate from rest to 60.0 miles per hour at an acceleration of 7g.

Step by step solution

01

Convert velocity to same units

First, convert the speed from miles per hour (mph) to meters per second (m/s) for use in the formula. We know that 1 mile = 1609 meters, and 1 hour = 3600 seconds. Thus, \(60.0 \ mph = 60.0*1609/3600 = 26.8 \ m/s\).
02

Calculate acceleration

Next, calculate the acceleration. The problem states that the car accelerates at a rate of 7g. We know that the gravitational acceleration on Earth, g, is approximately \(9.8 \ m/s^2\). So, the acceleration the car is designed for is \(7*9.8 = 68.6 \ m/s^2\).
03

Solve for time

Now we can use the formula for acceleration \(a = \Delta v/ \Delta t\) and solve for time. Rearrange the formula to solve for \(\Delta t\) gives \(\Delta t = \Delta v / a\). Inserting our values gives: \(\Delta t = 26.8 \ m/s/68.6 \ m/s^2 = 0.391 \ seconds\).

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Key Concepts

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

Acceleration
Acceleration is defined as the rate at which an object's velocity changes with time. In simple terms, it describes how fast or slow an object speeds up or slows down. If a car accelerates, it means that its velocity increases over time. The formula for acceleration, which is commonly used in physics, is given by:\[ a = \frac{\Delta v}{\Delta t} \]where \( a \) is the acceleration, \( \Delta v \) represents the change in velocity, and \( \Delta t \) is the time over which this change occurs. The units of acceleration are usually meters per second squared \( (m/s^2) \).
In the exercise, the car accelerates at a rate of 7 times the gravitational acceleration on Earth, which is denoted as \( 7g \). Often, we deal with acceleration when analyzing motion, especially in physics problems involving changes in speed or direction. Understanding acceleration is crucial when calculating how long it takes for vehicles or objects to reach certain speeds.
Gravitational Force
Gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another. The most familiar example of this is the force that gives weight to physical objects and causes them to fall to the ground when dropped.
In physics, the standard acceleration due to Earth's gravity is symbolized by \( g \). On the surface of the Earth, \( g \) has an approximate value of \( 9.8 \, \text{m/s}^2 \).
This particular value tells us how fast objects speed up as they fall when only gravity is acting on them. Despite these details, the gravitational force affects everything, large or small.
  • In the problem, acceleration is expressed in terms of gravitational force, \( 7g \).
  • To solve physics problems, it's common to convert this to a standard acceleration form, \( m/s^2 \).
This constant is immensely useful in calculations.Applying concepts of gravitational force can help predict the movement and the required force to change an object's motion.
Units Conversion
Units conversion is an essential step in many physics problems, especially when different parts of a problem are given in various units. Converting all terms to compatible units is crucial for ensuring consistent calculations and finding the correct answers.In the exercise given, it is necessary to convert the velocity from miles per hour to meters per second. Here’s a simplified way to look at such conversions:
  • 1 mile = 1609 meters
  • 1 hour = 3600 seconds
  • Therefore, 60 mph is converted to 26.8 m/s using \( 60 \times \frac{1609}{3600} = 26.8 \).
Converting units helps in utilizing formulae seamlessly, such as the acceleration equation. It's like speaking the same language across all parts of your physics problem.Keep in mind that in physics and engineering, staying consistent with SI units (meters, seconds, kilograms, etc.) is a generally followed standard, making calculations easier and reducing errors.

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Most popular questions from this chapter

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\), It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

In 1865 Jules Verne proposed sending men to the Moon by firing a space capsule from a 220 -m-long cannon with final speed of \(10.97 \mathrm{~km} / \mathrm{s}\). What would have been the unrealistically large acceleration experienced by the space travelers during their launch? (A human can stand an acceleration of \(15 g\) for a short time.) Compare your answer with the free- fall acceleration, \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

A ball is thrown upward from the ground with an initial speed of \(25 \mathrm{~m} / \mathrm{s}\); at the same instant, another ball is dropped from a building \(15 \mathrm{~m}\) high. After how long will the balls be at the same height?

A bullet is fired through a board \(10.0 \mathrm{~cm}\) thick in such a way that the bullet's line of motion is perpendicular to the face of the board. If the initial speed of the bullet is \(400 \mathrm{~m} / \mathrm{s}\) and it emerges from the other side of the board with a speed of \(300 \mathrm{~m} / \mathrm{s}\), find (a) the acceleration of the bullet as it passes through the board and (b) the total time the bullet is in contact with the board.

Two cars travel in the same direction along a straight highway, one at a constant speed of \(55 \mathrm{mi} / \mathrm{h}\) and the other at \(70 \mathrm{mi} / \mathrm{h}\). (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination \(10 \mathrm{mi}\) away? (b) How far must the faster car travel before it has a 15 -min lead on the slower car?

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