Mechanics 1 P42 2014; Discussion!

[A star]

EXACTLY

i make silly mistake and the wierdest of wierd ideas coe to my head and i have to ask myself to rethink . but that didnt even come near my brain xD

 

[Oishee Asif]

THEY GAVE US THE LIMITS IN THE QUESTION: 0 TO 27.
Finished.

 

[Snowysangel]

yar its basicaly cosX=B/H so to find X you need to do X=cos^-1(B/H) but majically my brain malfunctioned and i worte X=cos(B/H) i used a wrong formulae

I didn't even write that. I made pseudo angles and then directly write Ta=B/H...will they cut marks for that?

 

[thementor]

Took the words out of my mouth

 

[A star]

I didn't even write that. I made pseudo angles and then directly write Ta=B/H...will they cut marks for that?

nana all you need to show are the angles in the equation. i only wrote it to explain it to you

 

[wizteddy13]

Ok, first they said the particle has linear acceleration of 0.25 ms-2 until it reaches point A, at T=8. After that point, the particle has a non-linear acceleration, shown by the equation of v. SO that equation of v was applicable ONLY after the particle had reached point A at t=8.
Now, let us question what the question asked. The distance traveled by the particle from T=0 to T=27. Good. Moving on.
If we are to integrate the v equation, we obtain an s equation. Still all good. But now comes the snag.

If we consider limits of 27 and 8, and USE the integrated v equation (a.k.a the s equation), this applies only AFTER the particle had reached A. Thus. when we take the s(27) limit, we are considering 27 seconds AFTER P reached, which would be 35 seconds. And then, if we subtract s(8), that distance too is the distance the particle moved 8 seconds AFTER it had reached A.

Now, if we look at it another way, after reaching point A, the equation is viable, yes. Then, the particle traveled 19 MORE seconds after reaching point A, and 8 seconds BEFORE reaching point A, for a total of 27 seconds. Thus, if we use the s equation we obtained and use t=19, we get the distance traveled 19s after the particle reached A. And then we add the distance moved by particle before reaching A, i.e. 8m.

Well, that's my logic anyways.

 

[Oishee Asif]

THIS IS THE SOLUTION TO NUMBER 4ii
SHUT THE HELL UP

 

[A star]

Ok, first they said the particle has linear acceleration of 0.25 ms-2 until it reaches point A, at T=8. After that point, the particle has a non-linear acceleration, shown by the equation of v. SO that equation of v was applicable ONLY after the particle had reached point A at t=8.
Now, let us question what the question asked. The distance traveled by the particle from T=0 to T=27. Good. Moving on.
If we are to integrate the v equation, we obtain an s equation. Still all good. But now comes the snag.

If we consider limits of 27 and 8, and USE the integrated v equation (a.k.a the s equation), this applies only AFTER the particle had reached A. Thus. when we take the s(27) limit, we are considering 27 seconds AFTER P reached, which would be 35 seconds. And then, if we subtract s(8), that distance too is the distance the particle moved 8 seconds AFTER it had reached A.

Now, if we look at it another way, after reaching point A, the equation is viable, yes. Then, the particle traveled 19 MORE seconds after reaching point A, and 8 seconds BEFORE reaching point A, for a total of 27 seconds. Thus, if we use the s equation we obtained and use t=19, we get the distance traveled 19s after the particle reached A. And then we add the distance moved by particle before reaching A, i.e. 8m.

Well, that's my logic anyways.

bro you know right we are arguing for one mark .

 

[Ahmedlool]

Solution to #3.

So do i loose all the marks for q3 if i used lamis rule and didnt count the mass??

 

[Oishee Asif]

So do i loose all the marks for q3 if i used lamis rule and didnt count the mass??

You're supposed to use Lami's rule when there isn't any mass >_<

 

[wizteddy13]

bro you know right we are arguing for one mark .

yeah, I guess so. Oh well

 

[Ahmedlool]

so i lost 6 marks?

 

[thementor]

Ok, first they said the particle has linear acceleration of 0.25 ms-2 until it reaches point A, at T=8. After that point, the particle has a non-linear acceleration, shown by the equation of v. SO that equation of v was applicable ONLY after the particle had reached point A at t=8.
Now, let us question what the question asked. The distance traveled by the particle from T=0 to T=27. Good. Moving on.
If we are to integrate the v equation, we obtain an s equation. Still all good. But now comes the snag.

If we consider limits of 27 and 8, and USE the integrated v equation (a.k.a the s equation), this applies only AFTER the particle had reached A. Thus. when we take the s(27) limit, we are considering 27 seconds AFTER P reached, which would be 35 seconds. And then, if we subtract s(8), that distance too is the distance the particle moved 8 seconds AFTER it had reached A.

Now, if we look at it another way, after reaching point A, the equation is viable, yes. Then, the particle traveled 19 MORE seconds after reaching point A, and 8 seconds BEFORE reaching point A, for a total of 27 seconds. Thus, if we use the s equation we obtained and use t=19, we get the distance traveled 19s after the particle reached A. And then we add the distance moved by particle before reaching A, i.e. 8m.

Well, that's my logic anyways.

THIS IS THE SOLUTION TO NUMBER 4ii
SHUT THE HELL UP

Well. In an alternate reality, I am drinking coffee.

 

[Oishee Asif]

yeah, I guess so. Oh well

Dude.
Look above.
My solution.

 

[A star]

so i lost 6 marks?

yup cause no m1 nor b1 nor A1 marks

 

[Ahmedlool]

oishee asif do you know if they can atlest give me like 3 marks or are the whole 6 marks taken. btw thnks for all the help!

 

[Oishee Asif]

so i lost 6 marks?

:S
I... I don't wanna sound like a complete *****, but... but... *looks around*
*nods head slowly*

 

[wizteddy13]

THIS IS THE SOLUTION TO NUMBER 4ii
SHUT THE HELL UP

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.
SO what exactly did u remove when u subtracted s(8)

 

[Oishee Asif]

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.
SO what exactly did u remove when u subtracted s(8)

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

 

[17194]

it was 71.3 confirmed
the first distance was 8 the rest was found by integrating from 27 to 8 it came as 63.3 , cheers !

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.