/*! 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 89 Given the following acceleration... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 89

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=-32 ; v(0)=20, s(0)=0$$

Short Answer

Expert verified
Question: Determine the position function of an object moving along a line with an acceleration function of \(a(t) = -32\), an initial velocity of \(v(0) = 20\), and an initial position of \(s(0) = 0\). Answer: The position function of the object is \(s(t) = -16t^2 + 20t\).

Step by step solution

01

Integrate the acceleration function to find the velocity function

Integrate the acceleration function, \(a(t) = -32\), with respect to time: $$v(t) = \int a(t) dt = \int (-32) dt = -32t + C_1$$ We are given the initial velocity: \(v(0) = 20\). Use this condition to find the constant \(C_1\): $$v(0) = -32(0) + C_1 \Rightarrow 20 = C_1$$ So the velocity function is: $$v(t) = -32t + 20$$
02

Integrate the velocity function to find the position function

Integrate the velocity function, \(v(t) = -32t + 20\), with respect to time: $$s(t) = \int v(t) dt = \int (-32t + 20) dt = -16t^2 + 20t + C_2$$ We are given the initial position: \(s(0) = 0\). Use this condition to find the constant \(C_2\): $$s(0) = -16(0)^2 + 20(0) + C_2 \Rightarrow 0 = C_2$$ So the position function is: $$s(t) = -16t^2 + 20t$$ The position function of the object moving along a line is: $$s(t) = -16t^2 + 20t$$

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!

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

Given the following velocity functions of an object moving along a line, find the position function with the given initial position. Then graph both the velocity and position functions. $$v(t)=2 t+4 ; s(0)=0$$

Find the solution of the following initial value problems. $$v^{\prime}(x)=4 x^{1 / 3}+2 x^{-1 / 3} ; v(8)=40$$

Locate the critical points of the following functions and use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=x^{3}+2 x^{2}+4 x-1$$

Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation \(a(t)=v^{\prime}(t)=g,\) where \(g=-9.8 \mathrm{m} / \mathrm{s}^{2}\) a. Find the velocity of the object for all relevant times. b. Find the position of the object for all relevant times. c. Find the time when the object reaches its highest point. What is the height? d. Find the time when the object strikes the ground. A stone is thrown vertically upward with a velocity of \(30 \mathrm{m} / \mathrm{s}\) from the edge of a cliff \(200 \mathrm{m}\) above a river.

Find the solution of the following initial value problems. $$y^{\prime}(\theta)=\frac{\sqrt{2} \cos ^{3} \theta+1}{\cos ^{2} \theta} ; y\left(\frac{\pi}{4}\right)=3$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Probability and 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.