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I have this equation for a rate of change problem below

$s = t^ 4 − 4t^ 3 − 20t^ 2 + 20t,\qquad t \geq 0$

The question asks me

At what time does the particle have a velocity of 20 m/s?

How do i solve this? Basically what steps do I take to find at what time the particle has that velocity?

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    You are just wasting everyone's time by rushing instead of being careful. A desire to learn is commendable; your method of implement this desire is wasteful, and unlikely to actually help you *learn* anything. You aren't taking the time to digest the material, you are obviously not reading your text carefully, and you seem to do no thinking about the problems before posting asking for solutions. Start spending some time on the text and the problems before posting, and when you post, make sure you put the *entire* problem you are having difficulties with, and *say what you have managed to do.*2012-06-30

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If $s$ is the distance as a function of time $t$, then the velocity is given by $v = \dfrac{ds}{dt}$.

Set this derivative $\dfrac{ds}{dt}$ to $20$ and solve the cubic, with the constraint $t>0$, to get the value of $t$.

Note that the cubic you obtain can be factored easily.

Move your cursor over the gray area for the complete answer.

$\dfrac{ds}{dt} = 4t^3 - 12 t^2 - 40 t + 20$ Setting this to $20$ gives us, $4t^3 - 12 t^2 - 40 t + 20 = 20 \implies 4t^3 - 12 t^2 - 40 t = 0 \implies 4t(t^2 - 3t - 10)=0$ This gives us $t(t-5)(t+2) = 0$ Since $t > 0$, we get that $t=5$.

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    @soniccool Setting the original equation to $0$ will only give you the time at which the displacement is $0$. Also, as an aside, given the rate of questions you have been posting, I would suggest you to take some time and read and understand the answers given to your question before posting your next question.2012-06-30
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$s(t)$ is the position at time $t$.

The velocity is the rate of change of the position.

So the velocity at time $t$ is $s'(t)$.

You want to find the value of $t$ at which $s'(t) = 20$.

So... you find $s'(t)$; you set it equal to $20$. And...

Added. And you've added an entirely new question in comments, after requesting here in comments that you be given an answer to the mystery question you kept hidden....

The acceleration is the derivative of the velocity; the velocity is the derivative of the position. So the acceleration is the second derivative of the position (something that I am positive is in whatever book you are trying to learn from; are you actually reading the material and trying to understand it, or are you rushing to the exercises and then just asking for the solutions here?).

To find the times when the acceleration has a particular value $a$, you find the acceleration function by computing $s''(t)$, set $s''(t)$ equal to $a$, and solve for $t$.

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    @soniccool: Why do you want to find the time when the acceleration is $0$? That's not part of the problem you quoted. Where we expected to read your mind **again**? Why waste everyone's time and confuse yourself and others by **not** asking the question you actually want to ask "out loud"?2012-06-30
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The velocity is $\frac{ds}{dt}$, so you find this, set it equal to $20$, and solve for $t$. Don’t worry about the fact that you’ll get a cubic equation to solve: everything works out very nicely.