/*! 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 21 An object moving with uniform ac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An object moving with uniform acceleration has a velocity of 12.0 \(\mathrm{cm} / \mathrm{s}\) in the positive \(x\) direction when its \(x\) coordinate is \(3.00 \mathrm{cm} .\) If its \(x\) coordinate 2.00 \(\mathrm{s}\) later is \(-5.00 \mathrm{cm},\) what is its acceleration?

Short Answer

Expert verified
The object's acceleration is \( -16.00 \frac{cm}{s^2} \) in the negative x direction.

Step by step solution

01

Identify Given Information

The problem provides the initial velocity (\(v_i = 12.0 \frac{cm}{s}\) in the positive x direction), the initial position (\(x_i = 3.00 cm\)), the final position (\(x_f = -5.00 cm\)), and the time interval (\(t = 2.00 s\) later). Note that the final position is to the left of the initial position, indicating that the object has moved in the negative x direction.
02

Apply the Kinematic Equation

To find acceleration, use the kinematic equation for position with uniform acceleration: \(x_f = x_i + v_i t + \frac{1}{2} a t^2\).Where \(x_f\) is the final position, \(x_i\) is the initial position, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time elapsed.
03

Solve for Acceleration

Rearrange the equation to solve for acceleration \(a\):\[a = \frac{2(x_f - x_i - v_i t)}{t^2}\].Substitute the given values into the equation and solve for \(a\):\[a = \frac{2((-5.00 \text{ cm}) - (3.00 \text{ cm}) - (12.0 \frac{cm}{s})(2.00 s))}{(2.00 s)^2}\].\[a = \frac{2(-5.00 \text{ cm} - 3.00 \text{ cm} - 24.00 \text{ cm})}{4.00 s^2}\].\[a = \frac{2(-32.00 \text{ cm})}{4.00 s^2}\].\[a = \frac{-64.00 \text{ cm}}{4.00 s^2}\].\[a = -16.00 \frac{cm}{s^2}\].This means the acceleration is \(16.00 \frac{cm}{s^2}\) in the negative x direction, indicating that it is decelerating towards the negative x.

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.

Kinematic Equations
Understanding kinematic equations is critical when studying how objects move. These equations allow us to predict future motion based on past and current states. In our exercise, we have dealt with uniform acceleration, which means the acceleration is constant over time.

In the realm of kinematics, there are a few equations that are key to solving such problems. One of these is the position equation for uniform acceleration, which is represented as \( x_f = x_i + v_i t + \frac{1}{2} a t^2 \). Here, \( x_f \) is the final position, \( x_i \) is the initial position, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. This equation becomes incredibly useful when you're trying to find unknown variables like displacement, velocity, or acceleration when the other parameters are known.

To grasp the practical application, let's break down the steps to calculate acceleration using kinematic equations. In our given example, we have the initial and final positions, the initial velocity, and the time interval. By substituting these values into the equation and rearranging for acceleration, we can deduce the rate at which the object's velocity is changing over time.
Initial Velocity
The concept of initial velocity is another key topic in understanding motion. Initial velocity, commonly denoted as \( v_i \), is the speed and direction an object has at the start of a time interval. In kinematics, it is crucial because it serves as the starting point for calculating changes in motion.

In our exercise, the initial velocity of the object is given as 12.0 cm/s in the positive x direction. This fact is a piece of the puzzle required to figure out the puzzle of the object's subsequent motion. With the initial velocity, you can anticipate how the position of an object will change over time when combined with information about its acceleration.

Moreover, initial velocity is instrumental in the kinematic equations. It serves as the baseline measure from which we observe acceleration's effect on an object’s motion. The sign of the initial velocity also helps us understand the direction of motion, providing us with a complete picture of how an object is moving.
Acceleration Calculation
Acceleration calculation involves determining the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. For an object with uniform acceleration, such as the one in our exercise, the change in velocity stays consistent during each time period.

To find the acceleration from the kinematic equations, we rearrange the equation to isolate \( a \). With the given values from our problem, we implement the following calculation: \( a = \frac{2(x_f - x_i - v_i t)}{t^2} \). After plugging in the appropriate values for final and initial positions, initial velocity, and time, we can find the acceleration.

This process highlights the intertwined nature of position, velocity, and time within kinematic problems. Moreover, it’s essential to note the sign of the acceleration value, as it also indicates direction. In the above exercise, an acceleration of \(-16.00 \frac{cm}{s^2}\) signifies a deceleration in the positive x direction or acceleration in the negative direction, hence the object is slowing down in the way it initially moved.

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 particle moves along the \(x\) axis. Its position is given by the equation \(x=2+3 t-4 t^{2}\) with \(x\) in meters and \(t\) in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at \(t=0\) .

A speedboat moving at 30.0 \(\mathrm{m} / \mathrm{s}\) approaches a no-wake buoy marker 100 \(\mathrm{m}\) ahead. The pilot slows the boat with a constant acceleration of \(-3.50 \mathrm{m} / \mathrm{s}^{2}\) by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?

Draw motion diagrams for (a) an object moving to the right at constant speed, (b) an object moving to the right and speeding up at a constant rate, (c) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a changes in speed were not uniform; that is, if the speed were not changing at a constant rate?

Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk \(J\) is constant, (a) determine expressions for its acceleration \(a_{x}(t),\) velocity \(v_{x}(t),\) and position \(x(t),\) given that its initial acceleration, velocity, and position are \(a_{x i}, v_{x i},\) and \(x_{i},\) respectively. (b) Show that \(a_{x}^{2}=\) \(a_{x i}^{2}+2 J\left(v_{x}-v_{x i}\right) .\)

A 50.0 -g superball traveling at 25.0 \(\mathrm{m} / \mathrm{s}\) bounces off a brick wall and rebounds at 22.0 \(\mathrm{m} / \mathrm{s}\) . A high-speed camera records this event. If the ball is in contact with the wall for 3.50 \(\mathrm{ms}\) , what is the magnitude of the average acceleration the ball during this time interval? (Note: \(1 \mathrm{ms}=10^{-3} \mathrm{s} .\)

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.