/*! 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 27 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 acceleration of the object is -16 cm/s^2.

Step by step solution

01

Identify Given Information

Identify the information given in the problem and convert all quantities to the same unit. The initial velocity \(v_0\) is 12.0 cm/s, the initial position \(x_0\) is 3.00 cm, the final position \(x\) is -5.00 cm, and the time \(t\) is 2.00 s.
02

Substitute Given Information into Formula

Substitute the given information into the formula for position under uniform acceleration: -5 cm = 3 cm + 12 cm/s * 2 s + 0.5 * a * (2s)^2. Multiply each term as necessary, which gives -5 cm = 3 cm + 24 cm + a cm/s^2 * 2 s.
03

Solve for Acceleration

Sanitize the equation by subtracting 24 cm and 3 cm from both sides to isolate the acceleration term, which results in -32 cm = a * 2 s. Finally, divide each side by 2 s to solve for acceleration \(a = -16 cm/s^2\).

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.

Understanding Initial Velocity
Initial velocity is a fundamental concept in the study of motion in physics. It refers to the speed and direction at which an object starts moving. In our case, the initial velocity of the object is specified as 12.0 cm/s in the positive x-direction. This information is crucial as it serves as the starting point for any calculations involving kinematic equations.

To fully grasp the importance of initial velocity, consider that any change in an object's motion is evaluated with respect to its starting conditions. The initial velocity not only influences an object's position over time but also affects its acceleration, especially when the acceleration is uniform as in our exercise. Understanding initial velocity is the first step towards solving problems related to motion under uniform acceleration.
Demystifying Kinematic Equations
Kinematic equations are the backbone of solving motion problems in physics. These equations allow us to relate an object's displacement, initial velocity, acceleration, and the time it takes for these changes to occur. In problems with uniform acceleration, one of the key kinematic equations is:\[ x = x_0 + v_0t + \frac{1}{2}at^2 \]where\( x \)is the final position,\( x_0 \)is the initial position,\( v_0 \)is the initial velocity, \( a \)is the acceleration, and\( t \)is the time.

Understanding this equation is critical because it integrates all the primary variables involved in uniformly accelerated motion. When an object is consistently speeding up or slowing down at the same rate, these kinematic equations become indispensable tools for predicting future motion or understanding past motion.
Solving for Acceleration
Solving for acceleration involves rearranging the kinematic equations to find the rate at which an object's velocity changes over time. When given certain information about an object's motion, as with our exercise, acceleration can be determined using the appropriate manipulation of the kinematic formula.

Following the solution steps, we start by organizing the given data and identifying the kinematic equation that includes the unknown variable we wish to solve for — in this case, acceleration (\( a \)). By substituting the known values into the equation and solving for \( a \), we find the object's acceleration. This process requires careful algebraic manipulation and an understanding of the units involved. The fact that acceleration was found to be \( -16 \mathrm{cm/s^2} \) signifies that the object is decelerating or slowing down in the positive x-direction. Solving for acceleration is a fundamental skill in physics that allows us to predict and articulate how an object will behave as it moves over time.

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 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?

The driver of a truck slams on the brakes when he sees a tree blocking the road. The truck slows down uniformly with an acceleration of \(-5.60 \mathrm{~m} / \mathrm{s}^{2}\) for \(4.20 \mathrm{~s}\), making skid marks \(62.4 \mathrm{~m}\) long that end at the tree. With what speed does the truck then strike the tree?

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 \(15 \mathrm{~g}\) for a short time.) Compare your answer with the free-fall acceleration, \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

A train is traveling down a straight track at \(20 \mathrm{~m} / \mathrm{s}\) when the engineer applies the brakes, resulting in an acceleration of \(-1.0 \mathrm{~m} / \mathrm{s}^{2}\) as long as the train is in motion. How far does the train move during a 40 -s time interval starting at the instant the brakes are applied?

A mountain climber stands at the top of a \(50.0-\mathrm{m}\) cliff that overhangs a calm pool of water. She throws two stones vertically downward \(1.00 \mathrm{~s}\) apart and observes that they cause a single splash. The first stone had an initial velocity of \(-2.00 \mathrm{~m} / \mathrm{s}\). (a) How long after release of the first stone did the two stones hit the water? (b) What initial velocity must the second stone have had, given that they hit the water simultaneously? (c) What was the velocity of each stone at the instant it hit the water?

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.