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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

Miguel
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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.

Cameron Buie
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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.

Ron Gordon
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  • So at $t=2$ the particle is moving and at $t=3$ it doesn't move ... and yet is has been speeding up all the time between those two instants? – hmakholm left over Monica Mar 21 '13 at 01:23
  • Yes, because at the inflection point (at $t=2$), it is not accelerating. It goes from slowing down (velocity decreasing) to speeding up (velocity increasing). During this time, the velocity is negative. So, yes, it makes sense that at $t=3$, it is not moving at that instant. – Ron Gordon Mar 21 '13 at 01:27
  • I don't think that agrees with the usual meaning of "speeding up". Usually "speeding up" means that the absolute value of the velocity increases, not that its signed value in some arbitrarily chosen orientation of a coordinate system does. – hmakholm left over Monica Mar 21 '13 at 01:30
  • You are arguing semantics. In my physics education, this informality in the language was typical. – Ron Gordon Mar 21 '13 at 01:31
  • Yes, of course I am arguing semantics. My point is that you're assigning a wrong, unnatural, unusual semantics to the terms in the word problem. If you have two cars moving along the highway, one towards the north and the other towards the south, and both drivers slam the brakes, would you really say that one of the cars will be "speeding up" while the other "slows down", until each comes to a complete halt? – hmakholm left over Monica Mar 21 '13 at 01:40
  • OK, I missed this from the OP: "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. " That changes things. But, to answer your question, that's a situation where we would be more careful with the language. – Ron Gordon Mar 21 '13 at 01:44
  • Note that since our particle's motion is confined to a line, then @Henning's version--absolute value of velocity increasing--agrees with the OP's version--same (directional) sign on velocity and acceleration. – Cameron Buie Mar 21 '13 at 03:19
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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 $1$.

André Nicolas
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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.

hmakholm left over Monica
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  • I finally understood what it means to speed up in universal language: $s'^2$ is increasing, ie $2s's''=(s'^2)'>0$. It took me some time... – Julien Mar 21 '13 at 01:33
  • so are you saying that using interval notation i should end up with (1,3)(3,4)? – Miguel Mar 21 '13 at 01:39
  • @Miguel On $(2,3)$, $s's''<0$ (concave up=$s''>0$, but begative velocity, as shown by the slope of the tangents). It slows down. – Julien Mar 21 '13 at 01:45
  • @Miguel: Is that for speeding up or slowing down? It is wrong for both, because it includes both times immediately after $1$ (where the particle is speeding up) and times immediately before $3$ (where it is slowing down). – hmakholm left over Monica Mar 21 '13 at 01:45
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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.

Emily
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  • OP was already given the graph. Why would sketching the curve again help? Moreover, the answer as a whole doesn't really clarify anything. – epimorphic Jan 31 '16 at 18:54