Problem

The displacement (in feet) of a particle moving i…

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Problem 15 Medium Difficulty

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion
$ s = 1/t^2 $, where $ t $ is measured in seconds. Find the velocity of the particle at times $ t = a $, $ t = 1 $, $ t = 2 $, and $ t = 3 $.

Answer

So $v(1)=\frac{-2}{1^{3}}=-2 \mathrm{m} / \mathrm{s}, v(2)=\frac{-2}{2^{3}}=-\frac{1}{4} \mathrm{m} / \mathrm{s},$ and $v(3)=\frac{-2}{3^{3}}=-\frac{2}{27} \mathrm{m} / \mathrm{s}$

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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 fifteen of this tour Calculus eighth edition, section two point seven. The displacement and meters of a particle moving in a straight line is given by the equation of motion as equals one over T squared Ortiz measured in seconds. Find the velocity of the particle Times T equals a tickles. One tickles too and t equals three. So for this for this problem, the position functioned. The displacement is given here we know dysfunction. The velocity will be the slope of this function of slope function. I mean, we'LL use definition one Where are slow. We're gonna call our velocity and its element as t approaches A This is the first part we're going to do Thi is equal today Parties of progeny out of the function F fifty minus the function evaluated a today divided by the difference of Tina Our function f ft is given as one over t squared and dysfunction. Evaluated A is one of a squared all over eighteen minutes saying we can get rid of the functions in the numerator by multiplying by the that was common denominator, which is? He scored a squared. So we're gonna want to apply that to the top of the bottom T squared a squared multiplied by the first term. Just he was in a squared. He scored a squared multiplied by the second term just leaves a T's great! And in the denominator we'LL have this term t minus saying, provided by this T squared a squirter. Our next step is too factor thie greater. It's a difference of squares, meaning that hey, screw my C squared is the same as a minus. Team A plus team on the right, by the quantity T minus a ah t squared a squared. Now we can cancel here we can replace a minus T with negative team, anything that's equivalent, and then we can cancel team and with T minus A. Those two terms are equivalent as T approaches a and we're left with negative. A pristine divided by T squared a squared. We're still looking, too. Evaluate this limit as T approaches e. So we're going to show t approach in a. If he approaches a For this term, a plus T approaches to aim as T approaches a for the autumn term T score T squared perches eight of the fourth, so in the end. If we simplify, we it native to over a cube. And this is our velocity for the first part where t is equal today. So we have that and we're in uses equation to figure out the velocity at Ty's equal to one. It is equal to two and he is equal to three just by playing into this function party's equal to one. We have negative silver Ah, one cube which is clearly negative too. Meters per second forty is equal to two. We have ah, all of T zero two is equal to two because this limited week finger down is for one team party, so is too. We have native to over two cute, which is negative. Yeah, two over eight. We're negative one over for So our velocities thinking one or four meters per second and finally for tea is equal to three is equal to three. We have negative too. Over three cute which is twenty seven till the while seizing two over twenty seven meters per second. Ah, and this completes our problem

Daniel J.

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Problem

The displacement (in feet) of a particle moving i…

Problem 15 Medium Difficulty

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion
$ s = 1/t^2 $, where $ t $ is measured in seconds. Find the velocity of the particle at times $ t = a $, $ t = 1 $, $ t = 2 $, and $ t = 3 $.

Answer

So $v(1)=\frac{-2}{1^{3}}=-2 \mathrm{m} / \mathrm{s}, v(2)=\frac{-2}{2^{3}}=-\frac{1}{4} \mathrm{m} / \mathrm{s},$ and $v(3)=\frac{-2}{3^{3}}=-\frac{2}{27} \mathrm{m} / \mathrm{s}$

Related Courses

Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Grace H.

Caleb E.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview