/*! 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 16 A cheetah can accelerate from re... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A cheetah can accelerate from rest to a speed of \(30.0 \mathrm{m} / \mathrm{s}\) in 7.00 s. What is its acceleration?

Short Answer

Expert verified
The cheetah’s acceleration is approximately \(4.29 \mathrm{m/s^2}\).

Step by step solution

01

Define the initial and final velocities

Start by identifying the initial velocity (\(v_i\) and the final velocity (\(v_f\) of the cheetah. The cheetah starts from rest, so its initial velocity is 0 m/s. The final velocity is given as 30.0 m/s after 7.00 seconds.
02

Calculate the acceleration

Use the formula for acceleration, which is the change in velocity over time. The formula is \( a = \frac{\Delta v}{\Delta t} \) where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time. Here, \( \Delta v = v_f - v_i = 30.0 \mathrm{m/s} - 0 \mathrm{m/s} = 30.0 \mathrm{m/s} \) and \( \Delta t = 7.00 s \).
03

Substitute values into the acceleration formula

Substitute the given values into the acceleration formula to find the cheetah's acceleration: \( a = \frac{30.0 \mathrm{m/s}}{7.00 s} \).
04

Calculate the result

Calculate the acceleration by dividing the change in velocity by the time interval: \( a = \frac{30.0 \mathrm{m/s}}{7.00 s} = 4.29 \mathrm{m/s^2} \). Round the acceleration to two decimal places following the rules of significant figures.

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.

Initial Velocity
In the study of motion, 'initial velocity' refers to the speed of an object before it has undergone any change in speed or direction. It's denoted as \( v_i \). In our cheetah example, the initial velocity is particularly simple to identify because the cheetah starts from rest, which means its initial velocity is \( 0 \text{ m/s} \). This is a common starting point in physics problems dealing with acceleration.

Understanding the initial velocity is essential because it sets the stage for computing changes in motion. Whenever an object starts moving from a complete stop, its initial velocity will always be zero. However, initial velocity can also involve directional components, so in more complex problems, it might be necessary to consider the direction of motion as well.
Final Velocity
Moving on from the starting point of an object's motion, the 'final velocity' is the speed at which it moves after experiencing acceleration or deceleration, symbolized as \( v_f \). When analyzing our running cheetah, we're given that it reaches a final velocity of \( 30.0 \text{ m/s} \) after accelerating. This figure is a key piece of information because the difference between final and initial velocities aids in determining the rate of acceleration.

The concept of final velocity isn't just about the speed value; it is also vectored, which means it takes the direction into account. It's important for students to note that while speed is scalar, both velocity and acceleration are vectored quantities in physics.
Change in Velocity
The term 'change in velocity' (\( \Delta v \)), in physics, signifies the difference between the final velocity and the initial velocity of an object. It comes into play when you’re calculating acceleration. In our example, we calculate the change in velocity by subtracting the initial velocity from the final velocity. With the cheetah beginning at rest (\( 0 \text{ m/s} \)), and ending at \( 30.0 \text{ m/s} \), the change in velocity is simply \( 30.0 \text{ m/s} \).

This concept is integral to the equation \( a = \frac{\Delta v}{\Delta t} \), which defines acceleration. In practice, if an object decelerates, the change in velocity might be negative, indicating a decrease in speed. It's vital for students to understand that this change can be both positive or negative, depending on the context.
Time Interval
The 'time interval' (\( \Delta t \)) is a fundamental component necessary for calculating acceleration. It represents the duration over which the change in velocity occurs. In our cheetah's case, the time interval is the 7.00 seconds it takes for it to go from rest to \( 30.0 \text{ m/s} \).

The precision of the time measurement is critical and should reflect the degree of precision of the velocity measurements. It's also important to note that time is a scalar quantity. This means it doesn't have direction, unlike the velocity measurements we're working with, which makes computing with time intervals more straightforward.
Significant Figures
Significant figures play a vital role in presenting final results in science and engineering accurately. These are the digits in a number that carry meaning contributing to its precision. When we calculated the cheetah's acceleration to be \( 4.28571428571 \text{ m/s}^2 \), we rounded it to \( 4.29 \text{ m/s}^2 \) in order to reflect the precision of the data we were provided with, which had three significant figures.

The use of significant figures helps communicate the certainty of the measurements. In other words, it ensures that any figures reported are as accurate as the measurements allow. If the given values in a problem are to three significant figures, as they are in our cheetah problem, the answer should also be reported to three significant figures unless the calculation dictates otherwise.

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

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is \(2.10 \times 10^{4} \mathrm{m} / \mathrm{s}^{2},\) and \(1.85 \mathrm{ms}\) \(\left(1 \mathrm{ms}=10^{-3} \mathrm{s}\right)\) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Dragsters can actually reach a top speed of \(145 \mathrm{m} / \mathrm{s}\) in only \(4.45 \mathrm{s}-\) considerably less time than given in Example 2.10 and Example 2.11. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint. Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of \(6.20 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\) for \(8.10 \times 10^{-4}\) s. What is its muzzle velocity (that is, its final velocity)?

Calculate the average pressure exerted on the palm of a shot-putter's hand by the shot if the area of contact is \(50.0 \mathrm{~cm}^{2}\) and he exerts a force of \(800 \mathrm{~N}\) on it. Express the pressure in \(\mathrm{N} / \mathrm{m}^{2}\) and compare it with the \(1.00 \times 10^{6}\) Pa pressures sometimes encountered in the skeletal system.

A rock with a mass of \(540 \mathrm{~g}\) in air is found to have an apparent mass of \(342 \mathrm{~g}\) when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density? Is this consistent with the value for granite?

See all solutions

Recommended explanations on Physics 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.