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Mechanics pleasee help acceleration

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Reply 20
Original post by NonIndigenous
The funny symbols are as a result of your pdf reader of whatever other software you use to open the document not having the correct fonts installed (don't ask me how to fix it, I don't know lol). The software then replaces certain symbols with others from a different font, that unfortunately may look nothing like the original.

But if you use look closer... you can see that the "T" symbol appears to represent the "=" sign. "H" symbol looks like it means "-". "G" looks like it means "+". The stretched "O" appears to represent the integration symbol. etc.

I think the reason why it sometimes does and sometimes doesn't show the characters depends on whether or not there are spaces between the original characters. That's why sometimes it shows "=" and sometimes it shows "T".

I am getting the same answer both ways. Why is that?
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Original post by h26
So what is the way the mark scheme is showing? lol :tongue: It looks like theyre doing integration of v equation with limits 7 and 2 but why would they do that ? The displacement a t=2 is 0 and then the displacement from t=2 to t=7 is just subbing in t=7 in the integrated v equation. Also, they miss out the constant which is rather weird. Am I missing out something?


I assume you know how to integrate. When you integrate, you end up with a "+C" constant. Then you substitute the upper and lower bound values (7 and 2) for "t", and subtract them from one other. That also means you subtract "C" from itself... effectively cancelling it out.

The displacement at "t=2" being "0" is relative. It depends where you draw the 'baseline', or in other words, where the graph crosses the y-axis. If the graph above instead crossed the y-axis where currently "t=2", the displacement at that point would technically be "0". I probably didn't explain that well earlier, since I've also not worked with things like this for a while now. If that were the case (instead of what it is in the question), the equation they provide would also be different. But it isn't. That might sound confusing, so think of it more simply:

- Just integrate the equation and substitute the upper and lower bounds into the equation. When you subtract the substituted equation of "2" from "7", you effectively obtain the displacement as if it where measured from "2", or in other words, as if "2" were the baseline.
(edited 5 years ago)
Reply 22
Original post by NonIndigenous
I assume you know how to integrate. When you integrate, you end up with a "+C" constant. Then you substitute the upper and lower bound values (7 and 2) for "t", and subtract them from one other. That also means you subtract "C" from itself... effectively cancelling it out.

The displacement at "t=2" being "0" is relative. It depends where you draw the 'baseline', or in other words, where the graph crosses the y-axis. If the graph above instead crossed the y-axis where currently "t=2", the displacement at that point would technically be "0". I probably didn't explain that well earlier, since I've also not worked with things like this for a while now. If that were the case (instead of what it is in the question), the equation they provide would also be different. But it isn't. That might sound confusing, so think of it more simply:

- Just integrate the equation and substitute the upper and lower bounds into the equation. When you subtract the substituted equation of "2" from "7", you effectively obtain the displacement as if it where measured from "2", or in other words, as if "2" were the baseline.

Ahh thank you so much! I understand this now:smile:

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