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?

  • $\begingroup$ Also how do i find at what time the acceleration is 0 then? $\endgroup$ – soniccool Jun 30 '12 at 19:32
  • 4
    $\begingroup$ Don't add parts to question in comments; and don't add them without being very clear that they are being added after you have already received answers (least of all after you've accepted answers). That makes it look like the answers are incomplete (or incorrect). Are you in such a hurry to (not-)learn that you must post your problems at a rush without bothering to spend the time to make them complete and clear? Then you are wasting your time and everyone else's. Hardly "cool". $\endgroup$ – Arturo Magidin Jun 30 '12 at 19:36
  • $\begingroup$ Sorry im just trying to learn $\endgroup$ – soniccool Jun 30 '12 at 19:36
  • 5
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Jun 30 '12 at 19:39

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 $t=5$.

  • $\begingroup$ Awesome this was great. How do i find at what time the acceleration is 0 then? $\endgroup$ – soniccool Jun 30 '12 at 19:31
  • $\begingroup$ To find the time at which the acceleration is $0$, set the second derivative of $s$ to $0$ and solve the resulting quadratic. $\endgroup$ – user17762 Jun 30 '12 at 19:32
  • $\begingroup$ Alrighty so i find the second derivative and set it to 0 and solve. Let me try it. $\endgroup$ – soniccool Jun 30 '12 at 19:34
  • $\begingroup$ Wait are you sure i have to get the second derivative? Cant i just set the original to 0? $\endgroup$ – soniccool Jun 30 '12 at 19:36
  • $\begingroup$ @soniccool: Of course not. $s(t)$ is the position. Setting the position equal to $0$ and solving for $t$ will tell you when the position is $0$, not when the acceleration is $0$. Will reading "War and Peace" tell you the plot of "The Brothers Karamazov"? Unlikely. If you want to know when the aceleration is $0$, you need to set the acceleration (second derivative) to $0$ and solve. $\endgroup$ – Arturo Magidin Jun 30 '12 at 19:46

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.


$s(t)$ 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 $t$.

  • $\begingroup$ So find the derivative of it and then set it equal to zero correct? $\endgroup$ – soniccool Jun 30 '12 at 19:21
  • 1
    $\begingroup$ @soniccool: Are you trying to find the value of $t$ at which the velocity is equal to $0$? If not, then why would you set the derivative equal to $0$? $\endgroup$ – Arturo Magidin Jun 30 '12 at 19:23
  • $\begingroup$ Oh so you set it to 20 and find what you get? $\endgroup$ – soniccool Jun 30 '12 at 19:26
  • $\begingroup$ @soniccool: You don't "find what you get". So figue out what value of $t$ will make that equation true and make sense for your problem. That is, you solve the resulting equation. There's nothing to "get" from the equation. $\endgroup$ – Arturo Magidin Jun 30 '12 at 19:26
  • 1
    $\begingroup$ @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"? $\endgroup$ – Arturo Magidin Jun 30 '12 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.