Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 30

The position of a particle moving along the \(x\) -axis is given by \(x=\left(11+14 t-2.0 t^{2}\right),\) where \(t\) is in seconds and \(x\) is in meters. What is the average velocity during the time interval from \(t=1.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s} ?\)

Short Answer

Expert verified
Answer: The average velocity during the time interval from \(t=1.0\text{ s}\) to \(t=4.0\text{ s}\) is \(4.0 \frac{\text{m}}{\text{s}}\).

Step by step solution

01

Calculate Position at the Starting and Ending of the Time Interval

Using the position function, find the position of the particle at \(t=1.0s\) and \(t=4.0s\): \(x(1.0) = 11 + 14(1.0) - 2.0(1.0)^2\) \(x(4.0) = 11 + 14(4.0) - 2.0(4.0)^2\)
02

Compute Positions

Calculate the positions at the given times: \(x(1.0) = 11 + 14 - 2 = 23\text{ m}\) \(x(4.0) = 11 + 56 - 32 = 35\text{ m}\)
03

Find Displacement

Subtract the initial position from the final position to find the displacement: \(\Delta x = x(4.0) - x(1.0) = 35 - 23 = 12\text{ m}\)
04

Calculate Time Interval

Find the time interval between the given times: \(\Delta t = 4 - 1 = 3\text{ s}\)
05

Compute Average Velocity

Divide the displacement by the time interval to find the average velocity: \(v_\text{avg} = \frac{\Delta x}{\Delta t} = \frac{12}{3} = 4.0 \frac{\text{m}}{\text{s}}\) The average velocity during the time interval from \(t=1.0\text{ s}\) to \(t=4.0\text{ s}\) is \(4.0 \frac{\text{m}}{\text{s}}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2},\) and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?

You are trying to improve your shooting skills by shooting at a can on top of a fence post. You miss the can, and the bullet, moving at \(200 . \mathrm{m} / \mathrm{s},\) is embedded \(1.5 \mathrm{~cm}\) into the post when it comes to a stop. If constant acceleration is assumed, how long does it take for the bullet to stop?

You toss a small ball vertically up in the air. How are the velocity and acceleration vectors of the ball oriented with respect to one another during the ball's flight up and down?

The position of a race car on a straight track is given as \(x=a t^{3}+b t^{2}+c,\) where \(a=2.0 \mathrm{~m} / \mathrm{s}^{3}, b=2.0 \mathrm{~m} / \mathrm{s}^{2}\), and \(c=3.0 \mathrm{~m}\). a) What is the car's position between \(t=4.0 \mathrm{~s}\) and \(t=9.0 \mathrm{~s}\) ? b) What is the average speed between \(t=4.0 \mathrm{~s}\) and \(t=9.0 \mathrm{~s} ?\)

An electron, starting from rest and moving with a constant acceleration, travels \(1.0 \mathrm{~cm}\) in \(2.0 \mathrm{~ms}\). What is the magnitude of this acceleration? a) \(25 \mathrm{~km} / \mathrm{s}^{2}\) b) \(20 \mathrm{~km} / \mathrm{s}^{2}\) c) \(15 \mathrm{~km} / \mathrm{s}^{2}\) d) \(10 \mathrm{~km} / \mathrm{s}^{2}\) e) \(5.0 \mathrm{~km} / \mathrm{s}^{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Mechanics and Materials

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Work Energy and Power

Read Explanation

Electricity and Magnetism

Read Explanation
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.

Sign-up for free