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A particle $P$ is moving along a straight line. The fixed point $O$ lies on the line. At time $t\geq0$ seconds, the displacement of $P$ from $O$ is $s$ meters where $s = t^3 -9t^2 + 33t - 6$. Find the minimum speed of $P$.

Edit: So I've tried to differentiate and solve for t when the displacement between them is equal to zero and I got this: $3t^2 - 18t + 33 = 0$. I divided the equation by three to get $t^2 - 6t + 11 = 0.$ I then tried solving for t but I keep got stuck at $(t - 3) ^ 2 = -11 + 9$.

Edit #2: I got it now. The answer is 6

teft24
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    Welcome to MSE! What attempts have you done so far? Where are you stuck? Please use mathjax to show your work and review the meta read for tips on providing context in your question to better assist the community. – Jessie Mar 30 '21 at 17:38
  • This is not a homework and answer site... Please tell us what you've tried so we know where to help you. – Adam Rubinson Mar 30 '21 at 17:44
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    Keep in mind that $ds/dt$ gives you the velocity at time $t$, but you don't want the velocity to be $0$; you want the change in velocity (that is, the acceleration) to be $0$. Find out when that happens, then evaluate the velocity at that time. (ETA: And confirm that the velocity is indeed at a minimum at that point.) – Brian Tung Mar 30 '21 at 18:00
  • Incidentally, this problem is a perfect illustration of why context matters, I think. – Brian Tung Mar 30 '21 at 18:01
  • Thank you for helping me! I understand now. – teft24 Mar 30 '21 at 20:56

2 Answers2

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Hints: If $s$ is the displacement, then $v = ds/dt$ is the velocity. You want to minimize velocity, not displacement, so you want to find where $dv/dt = 0$, not where $ds/dt=0$. Note you will also have to check any boundary points on the domain ($t=0$).

MPW
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The speed is the (absolute value of) the derivative of the position with respect to time, which you have computed. Now to minimize the speed you need to take another derivative to get the acceleration and set that to zero.

Ross Millikan
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