How Accurate Is My Gradient Calculation for Maximum Acceleration?

[Peter G.]

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.

 

[Peter G.]

Here are the answers.

 

[ashishsinghal]

Let me be clear on this:
scan0004 is answer to scan
scan0006 is answer to scan0007

 

[Peter G.]

Yes, correct.

 

[ashishsinghal]

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

 

[ashishsinghal]

nice to see you are online. let's clear doubts right now

 

[ashishsinghal]

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.

 

[Peter G.]

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.

 

[Peter G.]

There they are. And by not checking the figures, you didn't check them for the distance time graph or for the acceleration?

 

[ashishsinghal]

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.

 

[ashishsinghal]

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)

 

[Peter G.]

"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

 

[ashishsinghal]

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

 

[Peter G.]

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

 

[ashishsinghal]

yeah that is fine. you can see from the graph itself. the max acceleration is in that area

 

[Peter G.]

Thanks a lot for the help. I will fix that acceleration time graph and I am good to go.