Based on this graph i have to figuere out when the particle is speeding up and when it is slowing down. My understanding is that when velocity and accelaration have the same sign then we are speeding up. however if velocity and accelaration have different signs then we are slowing down. Using this concept i applied it to this problem and i came up speeding up using interval notation. $(0,1)and(3,4) $ I'm getting the answer wrong can someone please help me understand the missing piece to this puzzle. Thanks Miguel
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$\begingroup$You've got the right idea, but not the right execution. Let's call the direction of positive $x$ "up" and the opposed direction down.
It is moving upward--has a positive velocity--on $(0,1)$ and $(3,4)$. It is moving downward--has a negative velocity--on $(1,3).$ Put another way, the graph's height is increasing on $(0,1)$ as we move to the right--likewise in $(3,4)$--and decreasing on $(1,3)$ as we move to the right.
It is accelerating upward--has a positive acceleration--on $(2,4)$. It is accelerating downward--has a negative acceleration--on $(0,2)$. Put another way, the graph's slope is increasing on $(2,4)$ as we move to the right, and the graph's slope is decreasing on $(0,2)$ as we move to the right.
Now check where the signs match.
$\endgroup$ $\begingroup$A start: Look in the interval $[0,1]$. The position (displacement) is increasing, so the velocity is positive. But the graph is concave down, the acceleration is negative, the thing is slowing down, until it reaches velocity (and speed) $0$ at time $1$.
Continue the analysis beyond $1$.
$\endgroup$ $\begingroup$Naively: At $t=1$ and $t=3$ the particle stands still. Just after these points in time it is moving. Something that goes from standing still to moving must be speeding up, so just to the right of each of $t=1$ and $t=3$ should count as speeding up.
Conversely, just to the left of each of $t=1$ and $t=3$ the particle is moving, but it is going to stand still in a little while. That means that it must be slowing down at those points in time.
$\endgroup$ 4 $\begingroup$Acceleration is a 2nd derivative, so you have to pay attention to where the graph holds water and spills water. Where it holds water, the acceleration is positive, and where it spills, it is negative. So it is slowing down, according to your definition, between $[0,1]$ and $[2,3]$, and speeding up otherwise.
$\endgroup$ 7 $\begingroup$A particle usually speeds up when the velocity and the acceleration have the same signs. It slows down when the acceleration and velocity signs are different. My advice (even though it might take a while) would be to sketch the curve from the given function. Then, you can find where the particle speeds up and slows down.
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