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Problem 66 Hard Difficulty

Repeat Exercise 65 for the position function
$x(t)=(t-1)(t-3)^{2}, 0 \leq t \leq 5$

Answer

$3\left(\frac{7}{3}\right)^{2}-14\left(\frac{7}{3}\right)+15$

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Video Transcript

So let's say we have a particle. This position It's described by T minus one times T minus three. Squared four two between zero First, let's expand this out. What we have. Cute Linus, 70 Squared 15 to Brothersnew, My Velocity two. Canton T Brother, These three. T squared. Well, as for 12 Positive tune Acceleration Term Tio Who's 16 Line us for two now when question Part B is to find the open T novels on which the particles moving to the right. In other words, when it's no greater than or equal to zero, well, you have to use the quadratic formula. No, nice is three question miners. Squad room The Squared Choose 14 squared minus four times three comes to 15 all over today. Inches three. So that's six. Oh, just three clothes for minors. It's cool. 14 squared watches When that thinks minus 12. 12 10 to 15. Ciro three 15 That's it. It's no thanks. That's 93. Also my squared of 16 which is for six. So what? Suman one. The last question is find the bluff of the particle acceleration cereal, So acceleration two 16 minus for tune. That's people to zero. That means that for Tim into the 16 I turn clients Tuesday. Two for two or fix or certainly we're three. 50 three times seven over three squared my 14 times three close to that's how fast the particles coming from the acceleration. I hope that answers any questions.