/*! 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 3 In a time of \(t\) seconds, a pa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a time of \(t\) seconds, a particle moves a distance of \(s\) meters from its starting point, where \(s=3 t^{2}\). (a) Find the average velocity between \(t=1\) and \(t=\) \(1+h\) if: (i) \(h=0.1\), (ii) \(h=0.01\), (iii) \(h=\underline{0.001 \text { . }}\) (b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time \(t=1\).

Short Answer

Expert verified
The instantaneous velocity at \(t=1\) is approximately \(6\) m/s.

Step by step solution

01

Understanding the formula for average velocity

The average velocity of a particle over a time interval from \(t=a\) to \(t=b\) is given by the formula: \( v_{avg} = \frac{s(b) - s(a)}{b - a} \), where \(s(t) = 3t^2\). For part (a), we need to calculate this average velocity for different small intervals around \(t=1\).
02

Calculating distances for different times

Let's calculate \(s(1)\) and \(s(1+h)\) for the given \(h\) values. - \(s(1) = 3(1)^2 = 3\). - For \(h = 0.1\), \(s(1+0.1) = 3(1.1)^2 = 3 \times 1.21 = 3.63\).- For \(h = 0.01\), \(s(1+0.01) = 3(1.01)^2 = 3 \times 1.0201 = 3.0603\).- For \(h = 0.001\), \(s(1+0.001) = 3(1.001)^2 = 3 \times 1.002001 = 3.006003\).
03

Calculating average velocity for h = 0.1

Using the average velocity formula, for \(h=0.1\):\[v_{avg} = \frac{s(1+0.1) - s(1)}{0.1} = \frac{3.63 - 3}{0.1} = \frac{0.63}{0.1} = 6.3\text{ m/s}.\]
04

Calculating average velocity for h = 0.01

Similarly, for \(h=0.01\):\[v_{avg} = \frac{s(1+0.01) - s(1)}{0.01} = \frac{3.0603 - 3}{0.01} = \frac{0.0603}{0.01} = 6.03\text{ m/s}.\]
05

Calculating average velocity for h = 0.001

Finally, for \(h=0.001\):\[v_{avg} = \frac{s(1+0.001) - s(1)}{0.001} = \frac{3.006003 - 3}{0.001} = \frac{0.006003}{0.001} = 6.003\text{ m/s}.\]
06

Estimating instantaneous velocity at t = 1

As \(h\) approaches zero, the average velocity approaches the instantaneous velocity. From the calculations: For \(h = 0.1, 0.01, 0.001\), the average velocities are \(6.3\), \(6.03\), and \(6.003\) respectively. So, by estimating, the instantaneous velocity at \(t=1\) is approximately \(6\text{ m/s}\).

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.

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.

Average Velocity
The concept of average velocity helps us understand the overall rate of change in a particle's position over a specified time interval. It’s calculated by dividing the total change in position (distance covered) by the total time taken. In mathematical terms, we express this as:
  • Average velocity, \( v_{avg} \), is given by:\[v_{avg} = \frac{s(b) - s(a)}{b - a}\]where \(s(t)\) represents the distance function, and \(a\) and \(b\) represent the start and end points of the time interval respectively.
  • For our specific exercise, the particle's distance \(s\) with respect to time \(t\) is \(s = 3t^2\).
This formula implies that to find the average velocity of the particle between two points in time, you find the difference in distance at these two times and divide it by the time difference. For very small intervals, this average can give us insights into the instantaneous velocity, especially as the time interval becomes negligible. Understanding average velocity is crucial to comprehend how motion calculations transform as you begin approximating instantaneous rates.
Particle Motion
Particle motion refers to the movement of a particle (or an object) and is quantified via its position and velocity as it changes over time. In physics, examining how a particle moves can reveal a lot about the forces and directions it experiences.
  • In our exercise, the particle moves according to the distance function \(s = 3t^2\). This quadratic relationship is key to understanding the motion of the particle.
  • Since \(s(t)\) is quadratic, it means that as time increases, the distance covered by the particle doesn’t just increase linearly but at an accelerating rate. The square of the time value increases the impact on the distance.
Therefore, when looking at particle motion, it’s vital to consider the nature of the function depicting its displacement relative to time. This form of analysis allows us to better understand not just the path followed but also how quickly or slowly that path is traversed, which leads us directly into grasping instantaneous velocity as the time interval diminishes.
Distance-Time Relationship
The distance-time relationship describes how the distance covered by a particle varies over time. It's an essential aspect of understanding motion since it provides insights into speed and velocity.
  • In our exercise example, the distance a particle moves over time is given by the function \(s = 3t^2\).
  • This formula tells us that the particle's movement is not constant; instead, its travel distance changes with the square of time, making the motion non-linear.
Understanding this relationship is crucial for determining how fast or slow a particle is moving at any given point. The steeper the curve (rate of change), the faster the particle is moving at that moment. Recognizing how the distance-time curve behaves helps in predicting future positions or retroactively understanding past movement details. When we're estimating instantaneous velocity, this relationship becomes particularly important because as you examine smaller and smaller intervals, this offers approaches to find the precise speed at an exact point in 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

Draw a possible graph of \(y=f(x)\) given the following information about its derivative. \- \(f^{\prime}(x)>0\) for \(x<-1\) \- \(f^{\prime}(x)<0\) for \(x>-1\) \- \(f^{\prime}(x)=0\) at \(x=-1\)

On May 9,2007, CBS Evening News had a \(4.3\) point rating. (Ratings measure the number of viewers.) News executives estimated that a \(0.1\) drop in the ratings for the CBS Evening News corresponds to a \(\$ 5.5\) million drop in revenue. \({ }^{9}\) Express this information as a derivative. Specify the function, the variables, the units, and the point at which the derivative is evaluated.

The cost, \(C=f(w)\), in dollars of buying a chemical is a function of the weight bought, \(w\), in pounds. (a) In the statement \(f(12)=5\), what are the units of the 12 ? What are the units of the 5 ? Explain what this is saying about the cost of buying the chemical. (b) Do you expect the derivative \(f^{t}\) to be positive or negative? Why? (c) In the statement \(f^{\prime}(12)=0.4\), what are the units of the 12 ? What are the units of the \(0.4\) ? Explain what this is saying about the cost of buying the chemical.

For some painkillers, the size of the dose, \(D\), given depends on the weight of the patient, \(W\). Thus, \(D=f(W)\), where \(D\) is in milligrams and \(W\) is in pounds. (a) Interpret the statements \(f(140)=120\) and \(f^{\prime}(140)=3\) in terms of this painkiller. (b) Use the information in the statements in part (a) to estimate \(f(145)\).

Investing \(\$ 1000\) at an annual interest rate of \(r \%\), compounded continuously, for 10 years gives you a balance of \(\$ B\), where \(B=g(r)\). Give a financial interpretation of the statements: (a) \(g(5) \approx 1649 .\) (b) \(g^{\prime}(5) \approx 165\). What are the units of \(g^{\prime}(5)\) ?

See all solutions

Recommended explanations on Math 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.