Displacement as Function of Time Graph Question

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physicsnobrain
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Homework Statement


The graph below shows the vertical displacement of an object as a function of time.
(a) Construct a graph of the object’s velocity as a function of time.
(b) Construct a graph of the object’s acceleration as a function of time.
(c) Describe a scenario in which a real-life object might move in this way


The Attempt at a Solution



Ok, Here is a picture. Graph A represents the original graph that we construct B and C from. I am not sure if I drew graph B (velocity as a function of time) and C (acceleration as a function of time) correctly. Did I?
Three sketched graphs labelled A, B and C. Graph A: a curve rising from the origin, flattening into a horizontal top, then falling back to the axis. Graph B: a step-like line starting high, dropping to zero, then dropping to a lower level. Graph C: a flat line at zero with two rectangular dips below it.


Also for a potential scenario, I can't come up with anything except for maybe a helicopter that goes up to a certain height, then hovers at that height then comes back down. But I'm not sure about the curvatures on that original graph A so I want to make the scenario detailed.

Thanks!
 
Before I say anything useful I will say that it has been a very long time since I've looked at position (or velocity or acceleration) vs. time graphs so it's possible that I am wrong but: I think your graphs are fine (but you shouldn't trust me entirely).

One thing I will mention is that you should aim to provide titles and label your axes for ease of other readers (ie. us "helpers" or your teacher). Some (many?) teachers will be a stickler for such things and it will be easy to lose marks in a silly way if you don't label your axes carefully.

As for a scenario, I believe the one you have described works well. We could try to be slightly more detailed. On the position vs. time graph, prior to peaking we notice that the slope decreases slightly - this means that the velocity is somewhat decreased.

We can explain this in the following way: A helicopter lifts up off of the ground and moves upwards with some speed. At a certain height the pilot decides to slow his ascent and hover in the air at some height. Then he descends in a similar fashion. (This is basically what you said).

We could make this more elaborate.

Suppose we are on a planet with trees that grow very, very tall. One day, your kitten, Paws, decides to climb one of these trees (she is chasing a delicious looking bird-like creature) and gets stuck very high up. You contact the emergency services and the send a rescue helicopter to save her. The pilot flies his helicopter up and as he approaches the location of Paws he begins to slow down and hover in mid-air when he is near. The passenger, a professional cat rescuer rescues Paws and brings her on board the helicopter and the pilot begins the descent again.
 
physicsnobrain said:
The graph below shows the vertical displacement of an object as a function of time.
(a) Construct a graph of the object’s velocity as a function of time.

Hi physicsnobrain. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Take a ruler and lay it to be tangential at some point on your plot of distance vs time, the steepness of the incline of the ruler here indicates the velocity of the moving object at that moment. The steeper the slope of ruler, the greater the speed.

Slowly move the ruler along the curve, adjusting it so it stays tangential to the curve, and you get a clear picture of how the speed of the object is changing (or not changing, as the case may be). When the ruler's incline is steep, the speed is greater.

Ignore the graphs you have already sketched. Start anew and work out bit by bit how the velocity vs time graph should go.
 
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NascentOxygen said:
Hi physicsnobrain. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Take a ruler and lay it to be tangential at some point on your plot of distance vs time, the steepness of the incline of the ruler here indicates the velocity of the moving object at that moment. The steeper the slope of ruler, the greater the speed.

Slowly move the ruler along the curve, adjusting it so it stays tangential to the curve, and you get a clear picture of how the speed of the object is changing (or not changing, as the case may be). When the ruler's incline is steep, the speed is greater.

Ignore the graphs you have already sketched. Start anew and work out bit by bit how the velocity vs time graph should go.

Are the graphs I drew not correct? Also, since the velocity time is velocity as a function of time, if I drew it steep that would mean there is acceleration, which I believe there isn't judging from the graph A.
 
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NascentOxygen said:
Hi physicsnobrain. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Take a ruler and lay it to be tangential at some point on your plot of distance vs time, the steepness of the incline of the ruler here indicates the velocity of the moving object at that moment. The steeper the slope of ruler, the greater the speed.

Slowly move the ruler along the curve, adjusting it so it stays tangential to the curve, and you get a clear picture of how the speed of the object is changing (or not changing, as the case may be). When the ruler's incline is steep, the speed is greater.

Ignore the graphs you have already sketched. Start anew and work out bit by bit how the velocity vs time graph should go.

Ok, after taking your advice this is now the graph I get of velocity as a function of time constructed from the graph A in my first post.
Hand-drawn graph titled "Velocity (m/s) as a function of time" with vertical axis marked 0 at the midline and horizontal axis labelled t. A segment starts above zero and slopes gently downward, then drops vertically to the zero line; a second segment starts at zero and slopes steeply downward below the axis.


Is it right?
 
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And here is the acceleration as a function of time graph.
Hand-drawn graph titled Acceleration (m/s^2) as Function of time, with a vertical axis marked 0 and horizontal axis labelled t; a line starts below zero, drops, jumps down, then runs flat and slopes further down.


Again, is this right?
 
Does anybody have an idea if my two new graphs work given graph A in my original post?
 
physicsnobrain said:
Ok, after taking your advice this is now the graph I get of velocity as a function of time constructed from the graph A in my first post.
Hand-drawn graph titled "Velocity (m/s) as a function of time", with velocity on the vertical axis and t on the horizontal axis. A straight line slopes downward from a positive value and ends at a vertical drop, then a second segment begins at the zero line and slopes downward below the axis.


Is it right?
That is looking better. Though we don't know that those sloping straight lines should actually be straight lines, it is better than your first post's graph (B).

The right hand side of your plot should not end like you show it, though.

EDITED
 
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NascentOxygen said:
That is looking better. Though we don't know that those sloping straight lines should actually be straight lines, it is better than your first post's graph (B).

The right hand side of your plot should not end like you show it, though.

EDITED

Is this better for velocity as function of time?
Step graph titled Velocity(m/s) as function of time, with time t on the horizontal axis; velocity is positive for an early interval, zero, then negative for a later interval.
 
Also, would the acceleration just be a straight line at zero?
 
physicsnobrain said:
Is this better for velocity as function of time?
Step graph titled Velocity(m/s) as function of time, with t on the horizontal axis and 0 marked on the vertical axis. Velocity is shown above zero, then at zero, then below zero, as horizontal segments with vertical jumps between them.
No, that is not better. A period of constant velocity would show on the displacement-time plot as a straight (though sloping) line. There are no such lines on the distance-time plot given for this question. What you had was good enough for now.

You have noticed how the distance-time graph ends with the object sitting at zero distance for some time?

Only when you have the velocity-time plot right is it worth discussing the acceleration.
 
NascentOxygen said:
No, that is not better. A period of constant velocity would show on the displacement-time plot as a straight (though sloping) line. There are no such lines on the distance-time plot given for this question. What you had was good enough for now.

You have noticed how the distance-time graph ends with the object sitting at zero distance for some time?

Only when you have the velocity-time plot right is it worth discussing the acceleration.

Hand-drawn graph titled Velocity(m/s) as function of time, with v on the vertical axis and t on the horizontal axis. The curve starts positive, declines gently, drops steeply to zero, then falls below zero, steeply and then more gradually.


Ok I drew this velocity as a function of time graph. This is accurate, correct?
 
Pay attention to two regions:

Hand-drawn graph labelled A with a curve rising from the origin, flattening into a horizontal segment, then falling back to the horizontal axis; a red arrow points to the flat section and a red circle marks the right-hand end near the axis.
 
Attachments
  • Hand-drawn curve labelled A starting at the origin, rising to a flat plateau, then falling back to the horizontal axis, with a red arrow pointing to the plateau and a red circle around the right end near the axis.
    15f4y6r.jpg
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NascentOxygen said:
Pay attention to two regions:

Hand-drawn sketch of a curve on axes with the letter A above it, rising from the origin, flattening into a horizontal plateau, then falling back to the horizontal axis. A red arrow points to the start of the plateau and a red circle marks a small flat section at the far right end near the axis.

That second region you circled shouldn't be there. It was a mistake on my part when i drew the graph in paint. Pretend that the graph ends when it hits the x axis.
 
Okay. But how did you decide (on the right hand side of the plot) there is a sudden bend in the line halfway while the object returns to zero level?

The arrow points to a point in time where distance was rapidly changing with time then it abruptly stopped changing. That happens when speed has an abrupt drop to zero.
 
Ok, I see the abrupt change. I went back to the graph I drew before and modified the second part.

Hand-drawn graph titled Velocity(m/s) as function of time, with v on the vertical axis and t on the horizontal axis. The line starts positive, slopes gently down, then drops abruptly to a lower value, crosses zero, and continues downward with a steeper drop before a shallower slope.


Is this more like it?
 
You have the first abrupt change to zero velocity correct, but you are still not showing the second one. When the distance graph finally returns to the t axis, it stays there, so velocity is now zero.
Also, because distance ends up where it started, at zero, the total area under the velocity graph (counting areas below the t axis as negative) should be zero. I.e. the total area above the t axis should be equal to the total area below it.
 
physicsnobrain said:
Ok, I see the abrupt change. I went back to the graph I drew before and modified the second part.

Hand-drawn graph titled Velocity(m/s) as function of time, with a horizontal t axis and a vertical velocity axis marked 0 at the origin; the line starts above zero, slopes gently down, drops sharply to a point below zero, then continues sloping downward.


Is this more like it?
The left hand half of the plot is looking good. Next, that 120 degree bend you show in the right hand half, why did you draw it like that?
 
NascentOxygen said:
The left hand half of the plot is looking good. Next, that 120 degree bend you show in the right hand half, why did you draw it like that?

Well because there is no abrupt change in the second half of the graph, it flows.
 
physicsnobrain said:
Well because there is no abrupt change in the second half of the graph, it flows.
So why did you show an abrupt change in the way velocity changes where there is no abrupt change in the way displacement is changing?

Anyway, you should get back to the graph you drew here. https://www.physicsforums.com/showpost.php?p=4503814&postcount=5

Hand-drawn graph titled 'Velocity (m/s) as a function of time', with t on the horizontal axis and a zero line marked 0. A slightly sloping line segment starts above zero and descends, then a vertical drop occurs, and a steeper downward sloping line continues below zero.


Now that you cleared up the artifact I had circled, that graph from earlier is correct!

Now you can determine the acceleration vs time graph. (The one in your very first post is not correct.)
 
Ok. Now to derive acceleration from this velocity time graph I will find the slope ((vf-vi)/(change in time)) of each segment of the graph.
 
Here is what I get for the acceleration as a function of time graph.
Hand-drawn step graph titled Velocity(m/s) as function of time, with the t axis to the right and 0 marked on the vertical axis. The velocity steps up to a positive level, then drops to a level just below zero, then drops again to a lower level.
 
You used a ruler like I suggested? Aligning it to be tangential to the graph as you slowly slide it from left to right. You won't quite get that graph.

A real life example that fits this trio of graphs would be throwing something like a pillow or a rolled up T-shirt from floor level (awkward, I know) upwards to hit the ceiling.
 
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physicsnobrain said:
Well because there is no abrupt change in the second half of the graph, it flows.
Are you sure? The graph you posted shows the displacement returning to zero and staying there, while the t axis continues briefly.
 
Getting close to finding the correct answer now.
Here is V3.0 of velocity as a function of time:

Hand-drawn graph titled Velocity as a function of time, with velocity on the vertical axis (unlabeled) against t on the horizontal axis, showing a rising line, a sharp drop below zero, then a partial rise back toward zero.
 
Still trying to work this out. First the object increases from 0 velocity to max velocity, then it instantly drops to zero velocity, then 0 velocity decreases to max negative velocity then it instantly hits zero.

Graph titled 'Velocity as a function of time' with v on the vertical axis and t on the horizontal axis; a line rises from zero, drops sharply to zero, then a short section dips below zero and returns to zero.
 
I have now concurred that it is 100% impossible to convert this graph from displacement - time to velocity - time. This is probably my final answer.
 
Ok, this is my final attempt. If this is wrong, I give up.

Hand-drawn graph titled Velocity as a function of time, with v on the vertical axis and t on the horizontal axis. A line rises from zero to a peak, falls back through zero, then dips below and returns to zero.
 
and here is the the graph of acceleration as a function of time I am sure of it.

Graph of acceleration against time, with a horizontal t axis and a vertical axis marked 0. A red step function starts above zero, drops to below zero, then returns to zero.
 
Ok after even further time analysis, I believe this to be the velocity as a function of time graph:
Velocity against time graph with the title 'velocity as function of time': a short rising line segment then a vertical drop to zero, followed by a straight line falling below the t-axis to a vertical drop back to zero.