/*! 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 An object moves along the \(x\) ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 16

An object moves along the \(x\) axis according to the equation \(x(t)=\left(3.00 t^{2}-2.00 t+3.00\right) \mathrm{m} .\) Determine (a) the average speed between \(t=2.00 \mathrm{s}\) and \(t=3.00 \mathrm{s},\) (b) the instantancous speed at \(t=2.00 \mathrm{s}\) and at \(t=3.00 \mathrm{s},\) (c) the average acceleration between \(t=2.00 \mathrm{s}\) and \(t=3.00 \mathrm{s},\) and \((\mathrm{d})\) the instantaneous acceleration at \(t=2.00 \mathrm{s}\) and \(t=3.00 \mathrm{s}\).

Short Answer

Expert verified
The average speed between \( t=2.0s \) and \( t=3.0s \) is obtained from Step 1. \nThe instantaneous speed at \( t=2.0s \) and \( t=3.0s \) is calculated from Step 2. \n The average acceleration between \( t=2.0s \) and \( t=3.0s \) is determined from Step 3.\n The instantaneous acceleration at \( t=2.0s \) and \( t=3.0s \) is computed from Step 4. All of the results are calculated based on the mathematical concepts of kinematic equations and calculus.

Step by step solution

01

Determine the average speed

Average speed is the total distance travelled divided by the time taken. For the given interval, the formula is: \n\(\overline{v}= \frac{ 螖x }{ 螖t }\), where \(\) 螖x \( = x(3.0) - x(2.0) \) and \(\) 螖t \( = 3.0s - 2.0s = 1.0s. \nSubstitute \( x(t) = 3.00t^2 - 2.00t + 3.00 \)m in the place of \( x \) to find the average speed.
02

Compute the instantaneous speed.

Instantaneous speed or velocity is the derivative of position with respect to time, given by \( v(t) = \frac{dx}{dt} \). \nTake the derivative of the function \( x(t) = 3.00t^2 - 2.00t +3.00m \) to get the velocity as a function of time, and substitute \( t = 2.0s \) and \( t = 3.0s \) separately to find the instantaneous speed at these two instances.
03

Determine the average acceleration.

Average acceleration is the change in velocity divided by the time taken. We can calculate this with, \(\) \( \overline{a} = \frac{ 螖v }{ 螖t }\), where \(\) 螖v \( = v(3.0) - v(2.0) \) and \(\) 螖t \( = 3.0s - 2.0s = 1.0s. \nSubstitute the function in the place of \( v \) calculated earlier in Step 2 to find the average acceleration.
04

Compute the instantaneous acceleration.

Instantaneous acceleration is the derivative of velocity with respect to time. It is given by \( a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} \). \n Take the derivative of the velocity function obtained in Step 2, to get the acceleration as a function of time. Substitute \( t = 2.0s \) and \( t = 3.0s \) separately to find the instantaneous acceleration at these two instances.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Speed
To understand average speed, imagine you're on a road trip. The average speed is like looking at the whole trip to see how far you traveled and how long it took. It's the total distance divided by the total time. In our exercise, the formula used is:
  • \( \overline{v} = \frac{ \Delta x }{ \Delta t } \)
  • Where \( \Delta x \) is the change in position from \( x(2.0 \text{s}) \) to \( x(3.0 \text{s}) \).
  • \( \Delta t \) is the time from 2.0 seconds to 3.0 seconds.
Plug in the given position equation \( x(t) = 3.00t^2 - 2.00t + 3.00 \) to find \( x(2.0) \) and \( x(3.0) \), then subtract them for \( \Delta x \). This will help us find how fast the object traveled on average over that time span.
Instantaneous Speed
Instantaneous speed is like what a speedometer reads in a car at a precise moment. It's how fast something is moving exactly at one point in time. To find it, we use the derivative of the position function to get velocity. The formula is:
  • \( v(t) = \frac{dx}{dt} \)
This derivative expression gives us velocity as a function of time. Using this, we substitute specific times, like 2.0 seconds and 3.0 seconds, to find the speed at those exact moments. This tells us the speed of the object precisely when we look at those specific times.
Average Acceleration
Average acceleration is similar to average speed but focuses on how the speed itself changes over time. It's the difference in velocity over a time period. The formula we use is:
  • \( \overline{a} = \frac{ \Delta v }{ \Delta t } \)
  • Here, \( \Delta v \) is the change in velocity from \( v(2.0 \text{s}) \) to \( v(3.0 \text{s}) \).
  • \( \Delta t \) remains as the time between 2.0 seconds and 3.0 seconds.
Take the velocities we found previously and see how much the object's speed changed during that time. Average acceleration provides a sense of how quickly speed is added or taken away over an interval.
Instantaneous Acceleration
Just like instantaneous speed tells us the speed at a specific moment, instantaneous acceleration tells us how the speed is changing right then. It's the rate at which velocity changes at an exact time, calculated using the derivative of velocity with respect to time:
  • \( a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} \)
We find this by taking the derivative of the velocity function. When we substitute specific times into this derivative, we get the acceleration right at those moments, like 2.0 seconds and 3.0 seconds. This way, we know how quickly the object鈥檚 speed is increasing or decreasing right at those times.

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Most popular questions from this chapter

A truck on a straight road starts from rest, accelerating at \(2.00 \mathrm{m} / \mathrm{s}^{2}\) until it reaches a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) Then the truck travels for \(20.0 \mathrm{s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.00 \mathrm{s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

The position of a pinewood derby car was observed at various times; the results are summarized in the following table. Find the average velocity of the car for (a) the first second, (b) the last \(3 \mathrm{s}\), and (c) the entire period of observation.$$\begin{array}{lllllll}\hline t(s) & 0 & 1.0 & 2.0 & 3.0 & 4.0 & 5.0 \\\\\hline x(\mathrm{m}) & 0 & 2.3 & 9.2 & 20.7 & 36.8 & 57.5 \\\\\hline\end{array}$$

A glider on an air track carries a flag of length \(\ell\) through a stationary photogate, which measures the time interval \(\Delta t_{d}\) during which the flag blocks a beam of infrared light passing across the photogate. The ratio \(v_{d}=\ell / \Delta t_{d}\) is the average velocity of the glider over this part of its motion. Suppose the glider moves with constant acceleration. (a) Argue for or against the idea that \(v_{d}\) is equal to the instantaneous velocity of the glider when it is halfway through the photogate in space. (b) Argue for or against the idea that \(v_{d}\) is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time.

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the height it reaches.

Kathy Kool buys a sports car that can accelerate at the rate of \(4.90 \mathrm{m} / \mathrm{s}^{2} .\) She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kathy. If Stan moves with a constant acceleration of \(3.50 \mathrm{m} / \mathrm{s}^{2}\) and Kathy maintains an acceleration of \(4.90 \mathrm{m} / \mathrm{s}^{2},\) find. (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant she overtakes him.
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