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A particle moves along a straight line with equation of motion $ s = f(t) $, where $ s $ is measured in meters and $ t $ in seconds. Find the velocity and the speed when $ t = 4 $.

$ f(t) = 10 + \frac{45}{t + 1} $

Velocity is equal to $-1.8 \mathrm{m} / \mathrm{s}$$\\$Speed is equal to 1.8 $\mathrm{m} / \mathrm{s}$

04:54

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Limits

Derivatives

David Base G.

October 27, 2020

Finally, now I'm done with my homework

Campbell University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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Okay, this problem is a position velocity acceleration problem. You're given the position of a particle at twenties, F t is 10 plus 45 over T plus one. The velocity at time T is the derivative of F at time T, Which will be 45 times T plus one to the negative two times a negative. So velocity at four is negative 45 times 5 to the negative, too negative 45/25 which is negative 9/5. The question is written in meters and seconds, so the velocity is in meters per second, acceleration at time T is the derivative of velocity. So that will be positive 90 tons T plus one to the negative three. So acceleration at four is 90 Times 5 to the -3, 90/1, Just 18/25. That will be in leaders per second squared two.

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