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Mechanics 1 P42 2014; Discussion!

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Dude. The period between 0-8 and the period between 8-27 have different accelerations, of course they're gonna have different values. :)

Yes, that's perfectly right. So, let me just ask this. If you were to consider the bounds as 8s and 0s, instead of 27 and 8 (just hypothetically saying), what time's distance are you calculating? Can you tell me the answer to that? :)
 
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You don't seem to understand how to apply the limits of integration. Why would you use t=27 as the upper limit? The body only moves according to the variable acceleration formula after t=8s, so when considering variable acceleration the t=0 point was at 8s. The body moved in variable acceleration for 19s only, and so the limits were 19 and 0. Or more simply put, integrate the expression and substitute t=19.

Brother! Finally I get a matching logic :X3:
 
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Haha, let's just calm down for a second now.
The first part is perfectly fine, for first 8 seconds, distance is 8m

BUT, if we consider the equation, it only became active after reaching A, which was reached after T=8. So, what distance are you removing when you consider for s(8)?
If we put in the values, we get 9.6 m, as you showed. But, as you ALSO showed, the first 8 seconds was 8 m, NOT 9.6m. :p
SO what exactly did u remove when u subtracted s(8) ;)
I feel sad for you. Or maybe for me. I dont know.
 
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Yes, that's perfectly right. So, let me just ask this. If you were to consider the bounds as 8s and 0s, instead of 27 and 8 (just hypothetically saying), what time's distance are you calculating? Can you tell me the answer to that? :)
you aare perfectly all correct in tis but bro T is the distance AFTER LEAVING A not in VAIABLE ACCELERATION open your eyes
 
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you aare perfectly all correct in tis but bro T is the distance AFTER LEAVING A not in VAIABLE ACCELERATION open your eyes

T is the total time the particle traveled, considering both the variable and non-variable regions. They said, it reached A at T=8. And then find distance from T=0 to T=27. So T is the time for the entire motion of the particle. :)
 
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Everyone seems to be having trouble with a simple concept. If I remember correctly the expression for s was s= 3/10 T^5/3. Now according to your logic you had to use the limits 8 and 27, as at 8s the body had already covered 8m. Well then substitute the value of t=8s in this expression. You should end up with 8m, if you are correct. But in reality, you get s=9.6
 
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Everyone seems to be having trouble with a simple concept. If I remember correctly the expression for s was s= 3/10 T^5/3. Now according to your logic you had to use the limits 8 and 27, as at 8s the body had already covered 8m. Well then substitute the value of t=8s in this expression. You should end up with 8m, if you are correct. But in reality, you get s=9.6

YES EXACTLY what I asked as well! If I get a satisfactory answer to that, I'll be satisfied.

Anyway, off to revise Physics now....Good luck to y'all :)
 
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I see your point bro. True that.
However, I know for a fact that the Cambridge system is better than the national system we have, so I guess I'm kinda grateful for what I have :)
Meh, true! Still, they have a knack for giving candidates an MI during the exam. FYI, the term MI means myocardial infarction.
 
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