/*! 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 32 Find the position and velocity o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 32

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. $$a(t)=\frac{20}{(t+2)^{2}}, v(0)=20, s(0)=10$$

Short Answer

Expert verified
Question: Determine the position and velocity functions for an object moving along a straight line with an acceleration function of $$a(t) = \frac{20}{(t+2)^{2}}$$, initial velocity of 20 m/s, and initial position of 10 m. Answer: The position function of the object is given by $$s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2}$$, and the velocity function is given by $$v(t) = -\frac{20}{t+2} + 30$$.

Step by step solution

01

Integrate the acceleration function to find the velocity function

To find the velocity function, we need to integrate the acceleration function with respect to time. The acceleration function is given as $$a(t) = \frac{20}{(t+2)^{2}}$$. By integrating with respect to time, we obtain the result: $$\int a(t) dt = v(t) = \int \frac{20}{(t+2)^2}dt$$ Now we need to find the antiderivative of the acceleration function. We use substitution method. Let $$u = t + 2$$ and $$du = dt$$. $$v(t) = \int \frac{20}{u^2} du = -\frac{20}{u} + C_1 = -\frac{20}{t+2} + C_1$$ Next, we use initial velocity condition to find $$C_1$$. The initial velocity is given as: $$v(0) = 20 = -\frac{20}{0+2} + C_1$$ Solving for $$C_1$$, we get $$C_1 = 30$$. Therefore, the velocity function is: $$v(t) = -\frac{20}{t+2} + 30$$
02

Integrate the velocity function to find the position function

Next, we need to find the position function, which is done by integrating the velocity function with respect to time: $$\int v(t) dt = s(t) = \int \left(-\frac{20}{t+2} + 30\right) dt$$ Now, we separate the integral into two different integrals: $$s(t) = \int\left(-\frac{20}{t+2}\right) dt + \int 30 dt$$ Integrating both functions individually and combining their results: $$s(t) = -20 \ln|t+2| + 30t + C_2$$ Now, we use the initial position condition to find $$C_2$$. The initial position is given as: $$s(0) = 10 = -20\ln{(0+2)} + 30(0) + C_2$$ Solving for $$C_2$$, we get $$C_2 = 10 + 20\ln{2}$$. Therefore, the position function is: $$s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2}$$ The position function s(t) and the velocity function v(t) of the object moving along the straight line are given by: $$s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2}$$ and $$v(t) = -\frac{20}{t+2} + 30$$

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 91影视!

Key Concepts

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

Integrating the Acceleration Function
When an object moves along a straight line, its acceleration at any given moment can describe how quickly it is speeding up or slowing down. To find out how this acceleration affects the velocity of the object over time, we integrate the acceleration function with respect to time.

Integrating an acceleration function like \( a(t) = \frac{20}{(t+2)^2} \) essentially involves finding the area under the acceleration-time curve from a starting point to time \( t \). This process gives us the velocity function minus the constant of integration. Using mathematical methods such as substitution can simplify the integration process. In our exercise, the result after integration and using the initial conditions is the velocity function \( v(t) = -\frac{20}{t+2} + 30 \).

This step uncovers how the object's speed changes over time and is critical for understanding its movement.
Finding the Velocity Function
The velocity function is directly obtained by integrating the acceleration function. It reveals at what speed and in which direction the object is moving at any point in time. For analytical purposes, recognizing that the velocity is the first derivative of the position function and the integral of the acceleration function is essential. By integrating the specific acceleration function given in the problem, we derive the object鈥檚 velocity.

It is through this calculated velocity function that we can explain the object's motion and predict its future speed. Moreover, applying the initial velocity, we find the particular solution to the problem. So, the velocity function isn't just a formula, it presents a narrative of the object's journey over time.
Initial Velocity and Position
Understanding initial conditions is crucial in problems that involve motion. The initial velocity, represented as \( v(0) \), and the initial position, symbolized as \( s(0) \), serve as starting points in our calculations.

When we integrate the acceleration function to find velocity, or velocity to find position, we always get a generic function plus a constant denoted as \( C \). This constant \( C \) can only be determined by applying the initial conditions provided in the problem. Intuitively, you can think of it as calibrating your results to match the specific scenario of the object鈥檚 motion at the start of observation. In our example, the initial conditions led to the constants \( C_1 = 30 \) and \( C_2 = 10 + 20\ln{2} \), fitting the velocity and position functions to the object's particular situation.
Position Function Calculus
The position function is the final step in understanding the entire picture of an object's movement. By integrating the velocity function with respect to time, we get the position function, which tells us the location of the object at any time \( t \).

Calculus comes into play once more as we perform this integration, looking out for any constants that emerge, and then apply our initial conditions to determine the specific position function. The formula \( s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2} \) gives the precise position of our object at any selected time. With this function in hand, you can visualize the path and position of the object as it moves along the straight line, serving as the roadmap of its journey.

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

Assume \(f\) and \(g\) are even, integrable functions on \([-a, a],\) where \(a>1 .\) Suppose \(f(x)>g(x)>0\) on \([-a, a]\) and the area bounded by the graphs of \(f\) and \(g\) on \([-a, a]\) is \(10 .\) What is the value of \(\int_{0}^{\sqrt{a}} x\left(f\left(x^{2}\right)-g\left(x^{2}\right)\right) d x ?\)

Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in \(\mathrm{mi} / \mathrm{hr}\) ). Assume \(t\) is measured in hours. Theo: \(v_{T}(t)=10,\) for \(t \geq 0\) Sasha: \(v_{S}(t)=15 t,\) for \(0 \leq t \leq 1\) and \(v_{S}(t)=15,\) for \(t>1\) a. Graph the velocity functions for both riders. b. If the riders ride for 1 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). c. If the riders ride for 2 hr, who rides farther? Interpret your answer geometrically using the graphs of part (a). d. Which rider arrives first at the \(10-, 15-\), and 20 -mile markers of the race? Interpret your answer geometrically using the graphs of part (a). e. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{mi}\) and the riders ride for 20 mi. Who wins the race? f. Suppose Sasha gives Theo a head start of \(0.2 \mathrm{hr}\) and the riders ride for 20 mi. Who wins the race?

A tsunami is an ocean wave often caused by earthquakes on the ocean floor; these waves typically have long wavelengths, ranging between \(150\) to \(1000 \mathrm{km} .\) Imagine a tsunami traveling across the Pacific Ocean, which is the deepest ocean in the world, with an average depth of about \(4000 \mathrm{m}.\) Explain why the shallow-water velocity equation (Exercise 71 ) applies to tsunamis even though the actual depth of the water is large. What does the shallow-water equation say about the speed of a tsunami in the Pacific Ocean (use \(d=4000 \mathrm{m}\) )?

Determine whether the following statements are true and give an explanation or counterexample. a. A pyramid is a solid of revolution. b. The volume of a hemisphere can be computed using the disk method. c. Let \(R_{1}\) be the region bounded by \(y=\cos x\) and the \(x\) -axis on \([-\pi / 2, \pi / 2] .\) Let \(R_{2}\) be the region bounded by \(y=\sin x\) and the \(x\) -axis on \([0, \pi] .\) The volumes of the solids generated when \(R_{1}\) and \(R_{2}\) are revolved about the \(x\) -axis are equal.

Inverse hyperbolic tangent Recall that the inverse hyperbolic tangent is defined as \(y=\tanh ^{-1} x \Leftrightarrow x=\tanh y,\) for \(-1
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

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.