/*! 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 19 At a certain time a particle had... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At a certain time a particle had a speed of \(18 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction, and \(2.4 \mathrm{~s}\) later its speed was \(30 \mathrm{~m} / \mathrm{s}\) in the opposite direction. What is the average acceleration of the particle during this \(2.4 \mathrm{~s}\) interval?

Short Answer

Expert verified
The average acceleration is \(-20 \, m/s^2\).

Step by step solution

01

Understand the Situation

The problem involves a particle that changes its velocity from a positive direction to a negative direction over a given time period. We need to calculate the average acceleration over this interval.
02

Identify Initial and Final Velocities

The initial velocity (\(v_i\)) of the particle is \(18 \, m/s\) in the positive x direction. The final velocity (\(v_f\)) is \(30 \, m/s\) in the opposite direction, which we'll consider as \(-30 \, m/s\).Thus, we have:\[v_i = 18 \, \text{m/s} , \, v_f = -30 \, \text{m/s}\]
03

Determine the Time Interval

The time interval over which this change occurs is \(2.4 \, s\). We denote this as \(t\).\[t = 2.4 \, \text{s}\]
04

Use the Average Acceleration Formula

The average acceleration (\(a_{avg}\)) can be calculated using the formula:\[a_{avg} = \frac{v_f - v_i}{t}\]Substitute the known values into the formula:\[a_{avg} = \frac{-30 \, \text{m/s} - 18 \, \text{m/s}}{2.4 \, \text{s}}\]
05

Calculate the Average Acceleration

Solve the equation from Step 4:\[a_{avg} = \frac{-30 - 18}{2.4}\]\[a_{avg} = \frac{-48}{2.4}\]\[a_{avg} = -20 \, \text{m/s}^2\]
06

Interpret the Result

The average acceleration of the particle is \(-20 \, m/s^2\), indicating that it speeds up in the negative direction, as expected since it ended up moving in the opposite direction to its initial motion.

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.

velocity change
Velocity change is a crucial part of understanding how objects move. When you look at an object's velocity, you see both the speed and direction it's traveling. A change in velocity means either a change in speed, a change in direction, or both. For our exercise, the particle starts at a speed of 18 m/s in the positive x direction and changes to 30 m/s in the negative x direction. Let's break this down:
  • The initial velocity (\(v_i\)) is +18 m/s.
  • The final velocity (\(v_f\)) is -30 m/s.
This means the particle not only speeds up but also reverses its direction. The change in velocity (\(\Delta v\)) can be calculated by subtracting the initial velocity from the final velocity: \(\Delta v = v_f - v_i = -30 \text{ m/s} - 18 \text{ m/s} = -48 \text{ m/s}\).Understanding velocity change provides insight into how fast or slow an object is moving and the direction of its motion.
time interval
The time interval is the duration over which changes in motion occur. In physics, it's typically denoted by the letter \(t\). Knowing the time interval helps us understand how quickly or slowly a velocity change happens.In our problem, the time interval is given as 2.4 seconds. This is the duration from when the particle's initial velocity is measured to when the final velocity is taken. This crucial piece of information allows us to compute the average acceleration. When you calculate things like average acceleration, the time interval serves as the denominator in the equation: \(a_{avg} = \frac{\Delta v}{t}\).This means the longer the interval, assuming the same change in velocity, the smaller the average acceleration, and vice versa.
kinematics
Kinematics is a branch of physics dealing with motion without considering the forces that cause it. It focuses on parameters like velocity, acceleration, and time, helping us understand the movement patterns of objects.When solving our exercise, we are applying kinematic principles. We determined the average acceleration using the formula: \(a_{avg} = \frac{v_f - v_i}{t}\).This calculation hinges on understanding the change in velocity and the time over which this change occurs. In real-world applications, kinematics allows us to predict how an object will move under known conditions, aiding in fields like engineering, astronomy, and even sports science. So, by grasping the basic kinematic equations, we can predict and analyze motion with precision and clarity.

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

The head of a rattlesnake can accelerate at \(50 \mathrm{~m} / \mathrm{s}^{2}\) in striking a victim. If a car could do as well, how long would it take to reach a speed of \(100 \mathrm{~km} / \mathrm{h}\) from rest?

A car moves along an \(x\) axis through a distance of \(900 \mathrm{~m}\), starting at rest (at \(x=0\) ) and cnding at rest (at \(x=900 \mathrm{~m}\) ). Through the first \(\frac{1}{4}\) of that distance, its acceleration is \(+2.25 \mathrm{~m} / \mathrm{s}^{2}\). Through the rest of that distance, its acceleration is \(-0.750 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) its travel time through the \(900 \mathrm{~m}\) and \((\mathrm{b})\) its maximum speed? (c) Graph position \(x,\) velocity \(v,\) and acceleration \(a\) versus time \(t\) for the trip.

A hot rod can accelerate from 0 to \(60 \mathrm{~km} / \mathrm{h}\) in \(5.4 \mathrm{~s}\). (a) What is its average acceleration. in \(\mathrm{m} / \mathrm{s}^{2}\), during this time? (b) How far will it travel during the \(5.4 \mathrm{~s}\), assuming its acceleration is constant? (c) From rest, how much time would it require to go a distance of \(0.25 \mathrm{~km}\) if its acceleration could be maintained at the value in (a)?

A parachutist bails out and freely falls \(50 \mathrm{~m}\). Then the parachute opens, and thercafter she decclerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2} .\) She reaches the ground with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). (a) How long is the parachutist in the air? (b) At what height does the fall begin?

A rocket-driven sled running on a straight, levcl track is used to investigate the effects of large accelerations on humans. One such sled can attain a speed of \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\), starting from rest. Find (a) the acceleration (assumed constant) in terms of \(g\) and (b) the distance traveled.

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.