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Exam mechanics question (1 Viewer)

b3kh1t

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With the mechanics question of Jac and Gil parachuting, how can Gil reach 3vT when vT is the terminal velocity??
 

hscishard

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Vt was the terminal velocity with the parachute. Gil initially went without the parachute opened

Couldn't finish the ii) one. Thought it ate too much of my time
 

apollo1

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With the mechanics question of Jac and Gil parachuting, how can Gil reach 3vT when vT is the terminal velocity??
haha. i didnt even notice that lol.

I let initial v for jac = 1/3 VT
and final = 2/3 vt
initial for gil = 3vt
final = 3/2 vt

and then i sub these in into the equation and the expressions i got were equal so i said the time to do blahh blah blah is equal.

lol i dont think thats right
 

b3kh1t

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yeh I completed the question but I was wondering how that could happen when it is the terminal velocity. btw I just subbed the velocities they gave then halved it and subbed it, then minus the time to find how long it took to reach it and the difference in times for Gil and Jac were equal
 

Implying

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that question took so much time. I got to part (ii) and realised you needed to use partial fractions and was like that, went and did questions 7 & 8 which left me a couple of minutes to finish (ii) and pick up 2 marks in (iii) because that also took quite a long time. Didn't help that the question was relatively confusing

but yeah, only lost 1 mark in it
 
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khorne

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holy fuck that question took so much time. I got to part (ii) and realised you needed to use partial fractions and was like fuck that, went and did questions 7 & 8 which left me a couple of minutes to finish (ii) and pick up 2 marks in (iii) because that also took quite a long time. Didn't help that the question was relatively confusing

but yeah, only lost 1 mark in it
LOL there's a standard integral you could have quoted 1/(a^2-x^2) =1/2a ln(a+x/a-x)

saved so much time
 

AAEldar

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LOL there's a standard integral you could have quoted 1/(a^2-x^2) =1/2a ln(a+x/a-x)

saved so much time
Ahhh I knew there was one somewhere, but I couldn't remember it >.< Bah.
 

hup

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i got to part iii wrote half a page of working and just sacked the rest might get 1 or 2 lol
 

Nympha

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LOL there's a standard integral you could have quoted 1/(a^2-x^2) =1/2a ln(a+x/a-x)

saved so much time
Since when is that a standard integral? The only one that looks vaguely like that is the 1/(a^2 + x^2) one.. there isn't one with a negative in the middle on the sheet...?
 
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khorne

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Since when is that a standard integral? The only one that looks vaguely like that is the 1/(a^2 + x^2) one.. therefore isn't one with a negative in the middle on the sheet...?
Its not on the sheet, but you can cite it as its a standard integral. I wrote it down in temrs of a and x and then applied it to the v one.

Corono and Cambridge both use this I am fairly sure
 

Wight

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I did ii) with partials...doesn't take as long as you would expect.
 

Implying

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LOL there's a standard integral you could have quoted 1/(a^2-x^2) =1/2a ln(a+x/a-x)

saved so much time
oh, you're kidding me :D

Even last night, i was like, for intergrals you need to remember the shortcuts for semicircles and the last two on the intergrals sheet

:(
 
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Nympha

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Its not on the sheet, but you can cite it as its a standard integral. I wrote it down in temrs of a and x and then applied it to the v one.

Corono and Cambridge both use this I am fairly sure
Oh okay. Damn.. wish I had known this :( I thought we were only allowed to quote the integrals on the sheet. ah well. pissed off, cause I'd done a question that looked almost identical to that in terms of integration and didn't have time to bash it with partial fractions. Fml, kissing my band 6 goodbye :(
 

cssftw

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here's what I did:

I found the time it took Jack to go from (1/3)VT to (2/3)VT

The tricky thing was, u had to use the result from (ii) but as u r doing a definite integral u forget about the +C.

So to calculate time for Jack

on the dv side of the integral --> initial = 1/3 VT and final = 2/3VT
On the dt side of the itnegral --> intial = 0, final =T (where T= time taken for his speed to double)

then u integrate it and the time it took Jack to double his speed was ((VT/2g)ln(5/2)) seconds

The u do the integration for Gill.

But on the dv side, you have initial = 3VT and final = H (where H is her velocity atT)

and on the dt side, you have initial = 0, and final = ((VT/2g)ln(5/2))

when you calculate for H --> you get (3/2)VT hence calculating the answer
 

Implying

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I ran out of time. I got the 3/2 vt bit and was about to finish the question when they called pens down. So i finished off with a quick 'by inspection, this is equal to whatever'
 

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