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Numerade Educator

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Problem 60 Hard Difficulty

A particle is moving with the given data. Find the position of the particle.

$ v(t) = t^2 - 3\sqrt{t} $, $ \quad s(4) = 8 $

Answer

$$
s(t)=\frac{t^{3}}{3}-2 t \sqrt{t}+\frac{8}{3}
$$

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Video Transcript

So we see that we have a particle moving um with the given data and we want to find the position of the particle. So the velocity is be of tea or we'll call it F. Of X. Is equal to she squared minus three routine. Or three X. Will change this tax as well. This is our graph and we want to find the anti derivative anti derivative S. F. T. So we take the anti driver. This that's going to give us X cube over three and then the three fruit X. That's the same thing as three times X. To the one half. So the way that we would do that would be it would be uh minus three X. To the three halves over times two thirds. So it's gonna be minus two ax to the three halves plus C. Evaluating the position S. Of four equals eight. We get the eight third to see so have plus eight thirds. And then we see that jeep crime of X. He's going to end up giving us the same exact graph of F. Fedex, which means we evaluated this properly. So the anti derivative is G. Of X equals X cubed minus three minus two X 23 have plus 8/3 final answer.