An object moves along a straight line. The object's position at time t is givenby s(t). Find the position, velocity, acceleration, and speed at the specifiedtime.2t,t = 3t + 3°а. s(1)b. s(t),t = 1t + 2'= t + –

a. Given position function is s(t)=2tt+3 When t=3, the position is s(t=3)=2(3)3+3=66=1 For velocity,…

Answered: An object moves along a straight line.…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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5. Use the graph provided below to answer this question. Assume this function is named f(x). Label all local extrema, absolute extrema, and inflection points. Describe where this function is increasing and decreasing using interval notation. Describe where this function is concave up and concave down using interval notation.
A graphing calculator is recommended. A particle moves according to a law of motion s = f(t), t 2 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.) f(t) = 3 - 9t2 + 24t
In order to keep the songbirds in the back yard happy, one person puts out 14g of seeds at the end of each week. During the week, the birds find and eat 70% of the seeds. The DTDS for the amount of seeds in the back yard at time t where t is measured in weeks and seeds are counted just after a new supply is given by Xt+1 = 0.3х, + 14 Which of the following is an equation for the general solution of this DTDS, based on an initial seed quantity of 50g? None of the above/below Xt = (0.7)'(50) + 50 Xị = (0.3)'(30) + 64 Xt = (0.3)'(30) + 20 X; = (14)'(50) + 20 Xị = (0.7)'(30) + 20 = lx
T varies jointly with x and cube of z. If x=2 and z=12 then T=6. Find the value of T when x=18 and z=2.

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An object moves along a straight line. The object's position at time t is given by s(1). Find the position, velocity, acceleration, and speed at the specified time. 2t a. s(t) 5 b. s(t) = t + = 3 %3D t + 3 1+ 2!= 1