How Accurate Is My Gradient Calculation for Maximum Acceleration?

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Peter G.
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For the Scan attachment:
The question asks me to find the maximum acceleration. I used those two points in red in the attachment to calculate the gradient doing difference in y and difference in x. I got 4m/s ^2: does it seems correct, or I should have to tangent it? Because it seemed to be a straight line to me.

The next part of the question asks me to draw the general shape of an acceleration time graph from that velocity time graph. (Answer in reply attachment). I divided the graph in four different parts. In the first section it is accelerating non linearly from 0 to 2.2. I think that line is correct. The second part, from 2.2 to 3 I drew a steeper straight line upwards but I think I should switch that to a straight horizontal line and from 3 to 5.6 I drew another upward straight line and finally, for the last 0.4 seconds a straight line on the x axis. For part one and three of the graph, my lines should be straight or curved?

For the other velocity graph, the scan 0007 they asked us to sketch a displacement time graph for that journey. Does my look right?

Thanks,
Peter G

P.S: Answers in form of attachment in reply to this thread.
 
Attachments
  • Velocity-time graph: v/m s⁻¹ against t/s. Speed rises in a straight line from 0 at t = 0 to about 25 at t = 10 s, stays nearly constant until t = 30 s, then falls linearly to 0 at about t = 35 s.
    scan0007.jpg
    25.8 KB · Views: 419
  • Velocity against time graph: smooth S-shaped curve rising from 0 to 10 m/s over 0 to 6 s, flattening at 10 m/s; two red dots mark points near 2.4 s (4 m/s) and 2.8 s (about 5.7 m/s).
    scan.jpg
    37.7 KB · Views: 446

Discussion

Here are the answers.
 
Attachments
  • Hand-drawn graph on squared paper of acceleration a in m/s^2 against time in seconds, with x-axis marks 1 to 6. The line rises linearly from zero, bends at about t = 2.2, peaks near t = 3, drops, then rises again to about t = 5.5 before falling to zero.
    scan0004.jpg
    33.5 KB · Views: 453
  • Graph of velocity v/m s⁻¹ against time t/s on grey grid paper, with a curve rising from the origin to a peak near t = 30 s, then falling to about t = 35 s.
    scan0006.jpg
    19.9 KB · Views: 458
Let me be clear on this:
scan0004 is answer to scan
scan0006 is answer to scan0007
 
first, let's see scan0007:
the third part of the graph is incorrect, rest is fine.
velocity just decreases but is never negative. How does your displacement reduce?

now let's see scan:
0 to 2.2 is correct,
2.2 to 3 is incorrect,
actually everything after that is incorrect, except that finally acceleration is zero
 
nice to see you are online. let's clear doubts right now
 
oh...I did not check the values. My comments were over the shape

EDIT: by values I did not mean type of graph, I meant the time at which the graph changes, its velocity at that point.
 
Last edited:
Ok, for the scan 0007 I had noticed that, and I corrected it.

For the acceleration graph, I made some changes too and it only asks for the general shape. I will upload my new graphs in 5 minutes.
 
There they are. And by not checking the figures, you didn't check them for the distance time graph or for the acceleration?
 
Attachments
  • Hand-drawn graph on squared paper of distance in metres against t/s, rising in a curve from the origin, becoming a straight line from about t = 10 s to 30 s, then flattening near 35 s.
    scan0009.jpg
    26.2 KB · Views: 450
  • Hand-drawn acceleration-time graph on squared grid paper, a/m s^-2 on the vertical axis and time on the horizontal axis with ticks labelled 1 to 6; the line rises linearly from the origin, levels off, drops, rises again linearly, then falls to zero near t = 6.
    scan0008.jpg
    38.9 KB · Views: 450
scan0009 is right...
scan0008 is again wrong,
2-3 part is correct now,
as you can see after that velocity increases at lower rate which finally becomes constant

and I checked them as per written, why did you doubt that.
 
I noticed something interesting:
scan0009 is basically scan (do not consider units)
integrating scan0007 you get scan,
(you get displacement time graph when you integrate velocity time graph)
so what will you get on differentiating scan.
(you get acceleration time graph when you differentiate velocity time graph)
 
"and I checked them as per written, why did you doubt that." I didn't doubt you, I asked that because I wanted to know if scan 0009 was still right because you said: "oh...I did not check the values. My comments were over the shape" But I guess it was just a misunderstanding.

For scan 0008: So I got the first line right, up to 2.2 seconds and second line, the straight horizontal line, to 3 seconds, right and the last line, last 0.4 seconds is right. How can I correct the third line?

And sorry, I didn't understand your last reply about integration :redface:
 
see about the third line...if you draw tangents at some point you will see that as time increases slope of tangent decreases meaning acceleration decreases. This (and my last post) mean that your acceleration time graph would look like scan0007
 
Ah I see, if I draw tangents to the part of the question graph corresponding to my third line the acceleration is decreasing. I see, thanks! And what about the maximum acceleration in scan? Using both of those red points and doing difference in y / difference x seems fine? I got 4 m/s^2
 
yeah that is fine. you can see from the graph itself. the max acceleration is in that area
 
Thanks a lot for the help. I will fix that acceleration time graph and I am good to go. :approve:
 

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