Mathematics: Post your doubts here!

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

Pie-man

or any one? its Q6 (ii)

Thanks

 

[Pie-man]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf

Pie-man

or any one? its Q6 (ii)

Thanks

i did this like a few pages back :

so here is what i did (the correct one):

1- i got the equation for Sp "1/2 x a x (t+2)(squared)"

2- started to get rid of the brackets "1/2 x 1.75 x (t(squared) + 4t +4)
= 0.875(t(squared) + 4t + 4)
= 0.875t(squared) + 3.5t + 3.5

3- got an expression for Sq "1/2 at(squared)"

4- subtracted them from each other to get the distance in-between to be equal to 4.9

0.875t(squared) +3.5t +3.5 - 1/2 x 1.75 x t(squared)
= the "0.875t(squared) is crossed with the "1/2 x 1.75 x t(squared) leaving the equation to be "4.9=3.5t + 3.5"
3.5t=4.9-3.5
t=1.4/3.5 = 0.4

 

[Silent Hunter]

i did this like a few pages back :

so here is what i did (the correct one):

1- i got the equation for Sp "1/2 x a x (t+2)(squared)"

2- started to get rid of the brackets "1/2 x 1.75 x (t(squared) + 4t +4)
= 0.875(t(squared) + 4t + 4)
= 0.875t(squared) + 3.5t + 3.5

3- got an expression for Sq "1/2 at(squared)"

4- subtracted them from each other to get the distance in-between to be equal to 4.9

0.875t(squared) +3.5t +3.5 - 1/2 x 1.75 x t(squared)
= the "0.875t(squared) is crossed with the "1/2 x 1.75 x t(squared) leaving the equation to be "4.9=3.5t + 3.5"
3.5t=4.9-3.5
t=1.4/3.5 = 0.4

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

 

[GorgeousEyes]

Nov01 no7 ii please.

 

[Pie-man]

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

yeah that i have no idea why i did that same mistake the first time i did it .. i was actually waiting to see if you would tell me :/

 

[Silent Hunter]

yeah that i have no idea why i did that same mistake the first time i did it .. i was actually waiting to see if you would tell me :/

me too also cant get it ... maybe another m1 candidate can help us out :\

 

[haha101]

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )

BTW i did it another way

 

[Esme]

thanks .... but why do we minus distance of P from that of Q ? why not the other way round?

I took time taken by P to be 't' and time taken by Q to be (t-2). Then i formed the equations for the distance travelled by both the particles and subtracted distance of Q from distance of P. I got t=2.4. So the time taken by Q is t-2= 2.4-2= 0.4

 

[Esme]

because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )

BTW i did it another way

You're right but they subtracted distance of P from distance of Q.. it has to be Sp-Sq=4.9

 

[haha101]

You're right but they subtracted distance of P from distance of Q.. it has to be Sp-Sq=4.9

oh really ? I didnt seem to understand the method did the Q myself and posted what i thought was logical

 

[Pie-man]

because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )

BTW i did it another way

how did you do it then?

 

[Esme]

all i know is if you have 3 different forces on an object you can use lami
i'm not exactly sure why i got CP2X it just made more sense to me and after getting all the other angles i was sure it was right but it will make more sense if you turn it into a triangle (well for most people i know)

Thanks I'll try to make sense of it

 

[Pie-man]

I took time taken by P to be 't' and time taken by Q to be (t-2). Then i formed the equations for the distance travelled by both the particles and subtracted distance of Q from distance of P. I got t=2.4. So the time taken by Q is t-2= 2.4-2= 0.4

why t-2 and not t+2 as it is 2 seconds late?

 

[Esme]

why t-2 and not t+2 as it is 2 seconds late?

See, P started before Q. So at any instant, P has been travelling for more time Q has. So if time taken by P is t, the time taken by Q to reach that point will be less by 2, hence (t-2)

 

[haha101]

how did you do it then?

First of all made equations for the dist covered by p and q :
Dist of p= o.5*1.75*(2+t)^2 (s=ut+0.5at^2) (u=0)
dist of Q = o.5*1.75*t^2

the difference was 4.9 => dist of P - Dist of Q
o.5*1.75*(2+t)^2 - o.5*1.75*t^2 = 4.9 (took o.5*1.75 common and divided it by 4.9 )

(2+t)^2 - t^2 =5.6

expanded the eq 4 +4t+t^2 -T^2 =5.6 ( T^2 is cancelled )

t =0.4

 

[Esme]

why t-2 and not t+2 as it is 2 seconds late?

You're right too. If the time taken by Q is t, then the time taken by P will be t+2.

 

[Pie-man]

See, P started before Q. So at any instant, P has been travelling for more time Q has. So if time taken by P is t, the time taken by Q to reach that point will be less by 2, hence (t-2)

ok thanks that really helped now i hope they get this on the exam its not as hard as it seems

 

[UnprepareD]

Dear, mathematicians I've hit rock bottom with two of the questions on this paper - Questions 2 and 8.The former I'm absolutely clueless how to do and the later, I'm struggling to substitute du for dx. Any help would be welcome. Thank you .
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[Dug]

Dear, mathematicians I've hit rock bottom with two of the questions on this paper - Questions 2 and 8.The former I'm absolutely clueless how to do and the later, I'm struggling to substitute du for dx. Any help would be welcome. Thank you .
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

Q2)
i) Area of sector = 1/3 (Area of triangle)
(1/2)r²Θ = (1/3)[(1/2)(atanΘ)(a)]
Simplify and rearrange to get tanΘ = 3Θ

ii) Taking initial value = π/4
Θ1 = 1.1694
Θ2 = 1.2391
Θ3 = 1.3185
Θ4 = 1.3232
Θ5 = 1.3240
Θ6 = 1.3242
Θ7 = 1.3242

Θ = 1.32

 

[Talhakhan]

Which Mechanics M1 paper has been the most difficult in the last decade.. I mean which had the lowest grade threshold. I want to attempt that paper plz if anyone have any idea plz share. ;-)