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Problem 19 Easy Difficulty

Velocity and acceleration from position Consider the following position functions.
a. Find the velocity and speed of the object.
b. Find the acceleration of the object.
$$\mathbf{r}(t)=\left\langle 1, t^{2}, e^{-t}\right\rangle, \text { for } t \geq 0$$


a. $\mathbf{v}(t)=\left\langle 0,2 t,-e^{-t}\right\rangle$
$|\mathbf{v}(t)|=\sqrt{4 t^{2}+e^{-2 t}}$
b. $a(t)=\left\langle 0,2, e^{-t}\right\rangle$


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Video Transcript

so start by getting the velocity from taking the derivative of the position function. And so when we do that, the X component one through he swear comes to t you have the minus teaming up this minus each minus t And I guess that's the velocity. If we want to get speed, we just take the magnitude of the velocity. And so that becomes I'm not gonna write zero squared, because that's irrelevant. Um, but we have to t squared plus minus e to the minus two and so well, together. Uh, ignore that. This word. So all together that's gonna be equal to the square root of four Chiefs Weird. Plus each the minus two. There's no great. We'd like that further. So I'm gonna leave it. Us is on. The acceleration is equal to the derivative of the velocity function. And so when you do that component wise, you're going to get still Zero for the first component that to t becomes just too. And then another minus comes down from the minus t to the minus in front of the eat cancels out, leaving us with just e to the minus t

University of California, Davis
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