Problem

A particle moves along a straight line with equat…

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Problem 43 Medium Difficulty

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) = 80t - 6t^2 $

Answer

The speed when $t=4$ is $|32|=32 \mathrm{m} / \mathrm{s}$

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

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Video Transcript

this problem. Number 43 of the Steward Calculus, eighth edition, Section 2.7. Particle moves along a straight line with equations of motion, Equation of motion as equals F F T, where s is measured in meters and he is is in second find the velocity and speed when t is equal to four. So if f represents s and s is measured in meters, it is a position function, and velocity and speed are related to position by a derivative. So here we use this definition of the derivative and first attempt to find the derivative of F or the velocity function and then use it to calculate the velocity AT T equals to four seconds. So f prime of T is, uh, the limited each purchase zero of the function half evaluated at T plus H which will just be this minus the function evaluated AT T or just the function itself to minus 18 and then minus negative 60 or plus 60 squared and it's all divided by H. Our next step will be to expand the terms in the numerator. We have a T T plus 80 h minus t plus h squared that is T squared plus two th plus h squared. So this will expand to be negative. 60 square minus 12 th and a six h squared minus 80 T plus 60 squared. Okay, this is all over H Now, we, um look for an opportunity to cancel some terms. Positivity and negativity. T think it is 60 squared and positive. 60 squared. And then the remaining three terms each have an H, which can be council with the H and the denominator. So instead of 88 we just have 80 12 th this age councils negative. Six h squared is just negative. Six h, So are reduced. Limit becomes 80 minus 12. Team, uh, minus six h but as a cheaper to 06 h approaches zero. So we're just left over with a T minus 12 t as the velocity function. Now, if we want to know what is the velocity at T equals four seconds. Well, that's just going to be a T minus 12 times four talk. Thanks for his 48. A. T minus 48 is 32. So our answer is 32 m per second. Our velocity is a positive 32 m second and our speed is 32 m per cent and that is the final answer for this problem.

Daniel J.