# When is the velocity zero or changing?

• opus
In summary: Other way round. "Concave" is "bending inward". So if it is open upward, with the... valley-like shape, it would be positive acceleration, and if it is open downward, with the... hump-like shape, it would be negative acceleration.
opus
Gold Member

## Homework Statement

Use the graph (attached below) to determine the time intervals when the velocity is positive, negative, or zero.

## The Attempt at a Solution

The graph is of a position function.
Velocity is the derivative of position.
Velocity will be zero when the derivative of the position function is zero. In other words, where the tangent line to a point on the graph of the position function is horizontal.
On the intervals (2,5) it is clearly horizontal. However, at x=1.5 and x=6, the velocity should also be zero because the derivative at these points would be a horizontal tangent line.

Now in my book, it says that the velocity is zero on (2,5). And that is all.
Is my reasoning off with this? The velocity should be zero on the interval (2,5) as well as at the points x=1.5 and x=6.

#### Attachments

• Screen Shot 2018-10-08 at 12.47.32 PM.png
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I agree w/ you but perhaps they don't consider an "interval" of 0 seconds to be an "interval". If that is in fact their reasoning, I think they are wrong and I can't see any other way ANYONE could consider them right.

opus
phinds said:
I agree w/ you but perhaps they don't consider an "interval" of 0 seconds to be an "interval". If that is in fact their reasoning, I think they are wrong and I can't see any other way ANYONE could consider them right.
Thanks for the reply. And further, for part c, it asks for the positive, negative, and zero acceleration. Again, the solution doesn't account for the actual points where the acceleration is zero i.e. where the graph of the velocity function crosses the x-axis.
I'll play ball with the question, I just wanted to make sure that my reasoning was at least correct.

opus said:
Thanks for the reply. And further, for part c, it asks for the positive, negative, and zero acceleration. Again, the solution doesn't account for the actual points where the acceleration is zero i.e. where the graph of the velocity function crosses the x-axis.
You want to look at concavity of the position function to find out about acceleration. And that particle is surely behaving oddly. It is at rest on ##(2,6)## then suddenly, with positive acceleration, moves in the negative direction. How does that happen?

opus
LCKurtz said:
You want to look at concavity of the position function to find out about acceleration.
Interesting! I never thought about that.

LCKurtz said:
And that particle is surely behaving oddly. It is at rest on (2,6)(2,6) then suddenly, with positive acceleration, moves in the negative direction. How does that happen?
So to answer this question, let me see if I can interpret this graph correctly without doing the math.

By your comment, concavity of a position function relates to acceleration.
So in looking at the graph I will say that on the interval (0,1.5) it in convex which I assume means negative acceleration. This makes sense because at x=1.5 the particle has stopped, meaning it was slowing down to this point.

However I'm not sure if my reasoning holds for intervals (1.5, 2) because that would mean that from a stop, it then gets negative acceleration and my intuition tells me that to move from a stop you would need positive acceleration.

opus said:
Interesting! I never thought about that.So to answer this question, let me see if I can interpret this graph correctly without doing the math.

By your comment, concavity of a position function relates to acceleration.
So in looking at the graph I will say that on the interval (0,1.5) it in convex which I assume means negative acceleration. This makes sense because at x=1.5 the particle has stopped, meaning it was slowing down to this point.

Concave down corresponds to negative acceleration and concave up to positive acceleration.

However I'm not sure if my reasoning holds for intervals (1.5, 2) because that would mean that from a stop, it then gets negative acceleration and my intuition tells me that to move from a stop you would need positive acceleration.
After it stops it starts moving in the negative direction for ##t## in ##(1.5,2)##which is caused by the negative acceleration. All good so far. But that sharp corner in the graph means it suddenly stops and the acceleration jumps to zero. It is the next sharp corner at ##t=5## that is strange. It suddenly goes concave up (positive acceleration) while moving from rest to a negative velocity. So I guess it is a make-believe problem or subject to some strange undefined surroundings.

opus
LCKurtz said:
Concave down corresponds to negative acceleration and concave up to positive acceleration.
By this do you mean "hump-like" is concave up and corresponds to positive acceleration?
And "valley-like" is concave down and corresponds to negative acceleration?
Or do I have it backwards?

opus said:
By this do you mean "hump-like" is concave up and corresponds to positive acceleration?
And "valley-like" is concave down and corresponds to negative acceleration?
Or do I have it backwards?

Other way round. "Concave" is "bending inward". So if it is open upward, with the inward curve facing up like a bowl or in your terminology "valley-like", that is "concave up". And if it is the other way round, with the hump facing up and the inward concave curve on the bottom, that is "concave down".

"Convex" is sticking out, "concave" is bending in.

If it is valley-like or bowl-like, then the slope is negative before the bottom of the bowl and positive after. The velocity is increasing. That is a positive acceleration. So concave up does correspond to positive acceleration. But it's the valley-like picture that is concave up.

opus
RPinPA said:
If it is valley-like or bowl-like, then the slope is negative before the bottom of the bowl and positive after. The velocity is increasing. That is a positive acceleration. So concave up does correspond to positive acceleration. But it's the valley-like picture that is concave up.

Ok I get the first part about the "bowl" having negative slope then positive.
Are we considering the "bowl" to be the position function? Because if we are, I would be inclined (pun intended) to say that the velocity before the bottom of the bowl is negative, then positive after the bottom of the bowl.
And if I think about the concavity of the bowl on the positive velocity side, it looks like " ) ". I'm throwing me self for a bit of a loop here because I can see that the slop would be increasing and thus the velocity in increasing. So then if velocity is increasing we can deduce that acceleration is increasing?

opus said:
Ok I get the first part about the "bowl" having negative slope then positive.
Are we considering the "bowl" to be the position function? Because if we are, I would be inclined (pun intended) to say that the velocity before the bottom of the bowl is negative, then positive after the bottom of the bowl.

I think I just said that. I said "the slope is negative before the bottom of the bowl and positive after." How do you feel that's different from what you said?

opus said:
I'm throwing me self for a bit of a loop here because I can see that the slope would be increasing and thus the velocity in increasing. So then if velocity is increasing we can deduce that acceleration is increasing?

Let's consider something freefalling under the influence of gravity, an acceleration of ##9.8 \text{ m/s}^2## which means an increase of velocity of ##9.8\text{ m/s}## every second. For convenience, let's round that to 10.

Let's say it starts at rest, velocity 0 at time 0. The velocity will increase by 10 m/s each second.
##t = 0\text{ s}, v = 0\text{ m/s}, a = 10\text{ m/s}^2##
##t = 1\text{ s}, v = 10\text{ m/s}, a = 10\text{ m/s}^2##
##t = 2\text{ s}, v = 20\text{ m/s}, a = 10\text{ m/s}^2##
##t = 3\text{ s}, v = 30\text{ m/s}, a = 10\text{ m/s}^2##
##t = 4\text{ s}, v = 40\text{ m/s}, a = 10\text{ m/s}^2##

Is the velocity increasing? Is the acceleration increasing?

Acceleration is the rate at which velocity is increasing. I could easily cook up an example where the acceleration is decreasing while the velocity increases. The fact that velocity is increasing, does not give you information on the rate at which it is increasing, or whether that rate is constant or changing.

opus
opus said:
Ok I get the first part about the "bowl" having negative slope then positive.
Are we considering the "bowl" to be the position function?

No. Geometrically, concave up is like a smile or a bowl which will hold liquid. Frown or turn the bowl upside down for concave down. Note that concavity is not the same thing as slope. You can have very gently or very steeply sloped curves of either concavity.

Because if we are, I would be inclined (pun intended) to say that the velocity before the bottom of the bowl is negative, then positive after the bottom of the bowl.
And if I think about the concavity of the bowl on the positive velocity side, it looks like " ) ". I'm throwing me self for a bit of a loop here because I can see that the slop would be increasing and thus the velocity in increasing. So then if velocity is increasing we can deduce that acceleration is increasing?
You are conflating increasing and positive. If the position is increasing the slope is positive. If the slope is increasing the concavity is positive. With regard to the red sentence, if the velocity is increasing you can deduce the acceleration is positive, not that it is increasing.

opus
opus said:
However I'm not sure if my reasoning holds for intervals (1.5, 2) because that would mean that from a stop, it then gets negative acceleration and my intuition tells me that to move from a stop you would need positive acceleration.
Acceleration in the colloquial sense is different than what the term means in physics and math.

An object speeds up if the velocity and acceleration have the same sign. In particular, if they're both negative, the object is speeding up. Negative acceleration isn't equivalent to "slowing down," nor is positive acceleration equivalent to "speeding up."

opus
LCKurtz said:
You want to look at concavity of the position function to find out about acceleration. And that particle is surely behaving oddly. It is at rest on ##(2,5)## then suddenly, with positive acceleration, moves in the negative direction. How does that happen?
The second derivative at ##t=5## is undefined. Physically, you can think of the object being given a sharp whack at that instant, setting it in motion in the negative direction.

opus
vela said:
Acceleration in the colloquial sense is different than what the term means in physics and math.

An object speeds up if the velocity and acceleration have the same sign. In particular, if they're both negative, the object is speeding up. Negative acceleration isn't equivalent to "slowing down," nor is positive acceleration equivalent to "speeding up."
And, more generally, in more than one dimension, acceleration, being a vector, isn't positive or negative anyway.

opus
RPinPA said:
I think I just said that. I said "the slope is negative before the bottom of the bowl and positive after." How do you feel that's different from what you said?
Apologies if I plagiarized your scientific knowledge. I learn things best by stating them fully from start to finish to see if I have any gaps in my understanding. And since this is over text, I can't point things out on a graph or anything so I post them from start to finish to be as clear as I can. So I guess to answer your question, I don't feel that what I have said is different from what you've said.

RPinPA said:
Let's say it starts at rest, velocity 0 at time 0. The velocity will increase by 10 m/s each second.
t=0 s,v=0 m/s,a=10 m/s2t = 0\text{ s}, v = 0\text{ m/s}, a = 10\text{ m/s}^2
t=1 s,v=10 m/s,a=10 m/s2t = 1\text{ s}, v = 10\text{ m/s}, a = 10\text{ m/s}^2
t=2 s,v=20 m/s,a=10 m/s2t = 2\text{ s}, v = 20\text{ m/s}, a = 10\text{ m/s}^2
t=3 s,v=30 m/s,a=10 m/s2t = 3\text{ s}, v = 30\text{ m/s}, a = 10\text{ m/s}^2
t=4 s,v=40 m/s,a=10 m/s2t = 4\text{ s}, v = 40\text{ m/s}, a = 10\text{ m/s}^2

Is the velocity increasing? Is the acceleration increasing?
Ok this I believe I understand. The acceleration is constant, and the velocity is increasing by 10 meters per second per second.

RPinPA said:
Acceleration is the rate at which velocity is increasing. I could easily cook up an example where the acceleration is decreasing while the velocity increases. The fact that velocity is increasing, does not give you information on the rate at which it is increasing, or whether that rate is constant or changing.
I think I may understand this as well now. If I think about it in terms of one word problem I had where instead of velocity and acceleration, it was bacteria growth, I think I may get the idea.
The gist of the problem was that there were times where the bacteria was still growing (velocity), but the rate at which it was growing was slowing down (acceleration). I think this is kind of the same idea if I understand correctly. I've made the mistake of assuming that to have a positive change in velocity, you need a positive change in acceleration.

LCKurtz said:
No. Geometrically, concave up is like a smile or a bowl which will hold liquid. Frown or turn the bowl upside down for concave down. Note that concavity is not the same thing as slope. You can have very gently or very steeply sloped curves of either concavity.You are conflating increasing and positive. If the position is increasing the slope is positive. If the slope is increasing the concavity is positive. With regard to the red sentence, if the velocity is increasing you can deduce the acceleration is positive, not that it is increasing.
Ahhhh that makes more sense.

vela said:
Acceleration in the colloquial sense is different than what the term means in physics and math.

An object speeds up if the velocity and acceleration have the same sign. In particular, if they're both negative, the object is speeding up. Negative acceleration isn't equivalent to "slowing down," nor is positive acceleration equivalent to "speeding up."
This is confusing if I think about it intuitively. But what you said makes sense. So if both velocity and acceleration are the same sign, it's speeding up.
Now can I say that if they're opposite signs, the object can still speed up, it's just that the rate at which it speeds up could be going down?

I haven't taken and Physics courses yet so I don't have the best handle on this stuff so apologies if I'm being mud brained about it. I'm going to look for some videos to visually see what's happening also.

opus said:
Now can I say that if they're opposite signs, the object can still speed up, it's just that the rate at which it speeds up could be going down?
No. If the acceleration is in opposite direction to the velocity, its speed is slowing down. Again, it is misleading to think of positive and negative acceleration in general, since acceleration is a vector. For example, in uniform circular motion the acceleration is towards the center. It is perpendicular to the velocity and non-zero. But the speed of the object is constant because the acceleration is neither in or opposed to the direction of motion.

opus
Thanks for the reply. So I am going to ask another dumb question. If acceleration and velocity are both vectors, can they be added to each other? I don't know vectors well except from trigonometry where all the vectors were force vectors or velocity vectors and they could be added but they weren't two different measurements like acceleration and velocity.

opus said:
If acceleration and velocity are both vectors, can they be added to each other?

No. They have different units.

opus
olivermsun said:
No. They have different units.
So then, vector wise, how can acceleration act on velocity which is a vector of different units? Maybe this is a physics question which I haven't taken yet so I hope I am not going off on a tangent.

opus said:
So then, vector wise, how can acceleration act on velocity which is a vector of different units?
I don't see how acceleration can "act" on velocity. Acceleration and velocity are attributes of some moving object.

A force can act on an object, causing it to acclerate.

opus
I think this is a good time to call it a day on my questions in this thread. I don't have the prerequisite understandings to ask intelligible questions at this time. But my question was answered in regards to the op so thanks everyone for the help!

opus said:
So then, vector wise, how can acceleration act on velocity which is a vector of different units? Maybe this is a physics question which I haven't taken yet so I hope I am not going off on a tangent.

The velocity of an object changes when the object is accelerated. It's easiest to illustrate how this works out (with correct units and everything!) with a concrete example:

Think of an object which starts out moving to the east in uniform motion at ##1\text{ m/s}##. Now suppose you accelerate the object to the north. Acceleration has units like ##\text{m/s/s} = \text{m/s}^2##; this tells you that you have to keep up the acceleration for a finite time to change the velocity by a finite amount. Let's say the acceleration is ##0.1 \text{ m/s}^2## toward the north. If you keep applying that acceleration for ##10 \text{ s}##, then the object will gain a northward velocity of ##0.1 \text{ m/s}^2 \cdot 10 \text{ s} = 1 \text{ m/s}##, in addition to the eastward velocity that it already had. Using Pythagoras' Theorem, you can see that the initial velocity plus the additional velocity due to the acceleration gives you a final velocity of ##\sqrt{2} \text{ m/s}## to the northeast.

## 1. What is velocity?

Velocity is a measure of an object's speed and direction of motion. It is a vector quantity, meaning it has both magnitude (speed) and direction.

## 2. How is velocity calculated from a position graph?

Velocity can be calculated from a position graph by finding the slope of the line connecting two points on the graph. The slope represents the change in position (y-axis) over the change in time (x-axis), which is equivalent to the object's velocity.

## 3. What does a positive velocity on a position graph indicate?

A positive velocity on a position graph indicates that the object is moving in the positive direction. This means that its position is increasing over time.

## 4. Can an object have a constant velocity on a position graph?

Yes, an object can have a constant velocity on a position graph if its position changes at a constant rate over time. This would appear as a straight line with a constant slope on the graph.

## 5. How is velocity represented on a position graph?

Velocity is represented by the slope of the position graph. A steeper slope indicates a higher velocity, while a shallower slope indicates a lower velocity.

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