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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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Video Transcript

suppose the position of a particle at time T is defined by the function S of T, which is equal to one over t squared. And here we want to find the velocity at time T, which is equal to a 1, 2 and three to do this, we find the derivative of S at a point T, which is equal to a. Now by definition of the derivative at the point we have s prime of a. This is equal to limit as T approaches A. Of S F T minus S. Of a fish all over t minus A. So from here we have limits. S T approaches A. Of S F T, which is one over T squared. This minus S of a. Which is one over a squared all over t minus A. Now combining the numerator, we have limit S. T approaches A. Of we have a common denominator of a square T squared and then we have a squared minus d squared. This times the reciprocal of t minus A, which is one over t minus A. And then from here we get limit. SD approaches a. We can factor out a squared minus d squared into a minus t, times a plus t. This all over a square times t squared and then times the reciprocal of t minus A, which is one over t minus E. And then know that the a minus T. We can rewrite this into negative of t minus A. And so we can cancel this along with the T -8 in the Denominator. And so we have limit as T approaches a of the negative of a plus T over a square times T squared. And so evaluating at T. Which is equal to a. We have negative of a plus A All over a square times a square. Distrust negative to a All over 8 to the fourth power, or this is just negative 2/8 to the third power. And so this is the velocity of the particle at time T. Which is a. Now we will use this to find the velocity At Times 1, 2 and three. So if the velocity which is just S prime of E. is -2 over a cube, this is the velocity at time equals A. Then when T. Is that's a one, we have s prime of one, this is just negative two over 1 to the third power or negative two. And when he is too we have S. Prime of two, which is just negative too, Over 2 to the 3rd power, That's negative to over eight or -1/4. And when these three we have s prime of three, That's equal to negative two over 3 to the 3rd power. Or that's just negative two over 27. So these are the velocities at specific times T, Which is 1, 2 and three