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Author Topic: MATH AS QUESTIONS>>>  (Read 4367 times)

Offline astarmathsandphysics

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Re: MATH AS QUESTIONS>>>
« Reply #15 on: October 15, 2009, 08:53:12 am »
The points A and  B have position vectors a and b respectively, realative to an origin O. The vectors a and b are given by:
a= 2i-2j+k
b= 6i-2j+3k

i] obtain the value of a.b and hence find angle AOB, correct to the nearest degree.


Vector c has magnitude 12units and is such that c=pa, where p is a constant.
Vector d has magnitude 14 units and is such that d=qb, where q is a constant.

ii] Find the values of p and q
iii] Find the magnitude of d-c
i)a.b=2*6+-2*-2+1*3=11
cosx=\frac {a.b}{|a||b|}=\frac {11}{sqrt(2^2+(-2)^2+1^2) sqrt(6^2+(-2)^2+3^2)}0.5238
x=cos^-1 (0.5238)=58
p=|c|/|a|=12/3=4 q=|d|/|b|=14/7=2
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Offline Eamyzz

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Re: MATH AS QUESTIONS>>>
« Reply #16 on: October 15, 2009, 09:14:13 am »
Quote from: astarmathsandphysics on October 15, 2009, 08:53:12 am
The points A and  B have position vectors a and b respectively, realative to an origin O. The vectors a and b are given by:
a= 2i-2j+k
b= 6i-2j+3k

i] obtain the value of a.b and hence find angle AOB, correct to the nearest degree.


Vector c has magnitude 12units and is such that c=pa, where p is a constant.
Vector d has magnitude 14 units and is such that d=qb, where q is a constant.

ii] Find the values of p and q
iii] Find the magnitude of d-c
i)a.b=2*6+-2*-2+1*3=11
cosx=\frac {a.b}{|a||b|}=\frac {11}{sqrt(2^2+(-2)^2+1^2) sqrt(6^2+(-2)^2+3^2)}0.5238
x=cos^-1 (0.5238)=58
p=|c|/|a|=12/3=4 q=|d|/|b|=14/7=2



Thank you  vry much
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Offline Eamyzz

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Re: MATH AS QUESTIONS>>>
« Reply #17 on: October 15, 2009, 09:23:21 am »
do u have any tricky question any tip u wanna give b4 ma exam mr. paul =)
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Offline astarmathsandphysics

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Re: MATH AS QUESTIONS>>>
« Reply #18 on: October 15, 2009, 09:53:40 am »
Get there early. Be at peace. Maybe read an exam paper but probably not answer any questions
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Offline slvri

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Re: MATH AS QUESTIONS>>>
« Reply #19 on: October 15, 2009, 10:35:35 am »
Quote from: Eamyzz on October 15, 2009, 07:57:51 am
yes wat abt u??
nope not me.......im giving AS and A2 math in june 2010
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Offline Eamyzz

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Re: MATH AS QUESTIONS>>>
« Reply #20 on: October 15, 2009, 02:39:41 pm »
Quote from: astarmathsandphysics on October 15, 2009, 09:53:40 am
Get there early. Be at peace. Maybe read an exam paper but probably not answer any questions
thnk u mr. paul =)
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Offline Eamyzz

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Re: MATH AS QUESTIONS>>>
« Reply #21 on: October 15, 2009, 02:40:11 pm »
Quote from: slvri on October 15, 2009, 10:35:35 am
nope not me.......im giving AS and A2 math in june 2010

gd luck slvri  ;) ;)
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Offline Eamyzz

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Re: MATH AS QUESTIONS>>>
« Reply #22 on: October 15, 2009, 02:51:48 pm »
hey all hw was da paper?? =)
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Offline sanity_master

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Re: MATH AS QUESTIONS>>>
« Reply #23 on: October 18, 2009, 03:31:39 pm »
i got a question in M1

N 03.....#7 (IV)


i solved almost 3/4 of it....but the final part doesnt wanna get right....so would some1 plz solve IV
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Offline Ghost Of Highbury

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Re: MATH AS QUESTIONS>>>
« Reply #24 on: October 18, 2009, 04:01:30 pm »
Quote from: sanity_master on October 18, 2009, 03:31:39 pm
i got a question in M1

N 03.....#7 (IV)


i solved almost 3/4 of it....but the final part doesnt wanna get right....so would some1 plz solve IV
i can use my knowledge of physics and additional mathas to solve this.

first find the (t) where velocity of B is maximum
for this differentiate velocity and equate to 0

d/dx (0.06t - 0.00012t2) = 0.06 - 0.00024t

equate this to 0

0.06 - 0.00024t = 0
0.06 = 0.00024t
t = 0.06/0.00024 = 250s

now

at 250 seconds find the distance covered by B

for finding the distance integrate the velocity formula

<= integration symbol

<(0.06t - 0.00012t2) = 0.03t2 - 0.00004t3 + c

when t = 0 ; s =0 therefore c = 0

so apply t=250 in the formula

s = 0.03t2 - 0.00004t3

s = 1875 - 625 = 1250

-
now find the distance traveled by A

u gotta use the diagram for this.

first find the equation of the line AB for this .

u wud get the equation

y = 0.012x + 3.6

substitute x=250 here

y = 6.6

name this point C

find the area of the trapezium ACXY

u wud get the area(distance) as 855

now find the area of the triangle AXO = 240

thus, the total area (distance) = 240 + 855 = 1095

now 1250 - 1095 = 155m

hope that helped

find the diagram attached
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Offline astarmathsandphysics

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Re: MATH AS QUESTIONS>>>
« Reply #25 on: October 18, 2009, 04:04:36 pm »
v_B=0.06t-000.12t^2 so v is max when a=0  0=a=\frac{dv}{dt} =0.06-0.00024 t so t=\frac {0.06}{0.00024} =250
s_B =\int_0^250 0.06t-0.00012t^2 dt =[0.03t^2-0.00004t^3]_0^250 =[0,03*250^2-0.00004*250^3]-01250m
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Offline astarmathsandphysics

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Re: MATH AS QUESTIONS>>>
« Reply #26 on: October 18, 2009, 04:05:44 pm »
Thanks. I was answering this but always happy when someone beats me tpo it.
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Offline sanity_master

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Re: MATH AS QUESTIONS>>>
« Reply #27 on: October 18, 2009, 04:27:16 pm »
really thank u guys......i did all wt u have done...the problem is about the slope of the equation adi.....my calculator is not getting the acceleration as 0.012

so would u plz show me hw to get that one....maybe i am missing something


nd Thanks again for the really hard work....:)
« Last Edit: October 18, 2009, 04:32:17 pm by sanity_master »
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Offline sanity_master

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Re: MATH AS QUESTIONS>>>
« Reply #28 on: October 18, 2009, 04:29:13 pm »
OMG>......fool me

i got wt im missing......small mistake

i forgot to subtract the 4.8 from the 7.2 :D LOL hahahahahha

silly mistake for an AL student :D hahaha

Thanks guys!!

really appreciate it.....ur the best!!
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Offline Ghost Of Highbury

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Re: MATH AS QUESTIONS>>>
« Reply #29 on: October 18, 2009, 06:05:42 pm »
Quote from: sanity_master on October 18, 2009, 04:29:13 pm
OMG>......fool me

i got wt im missing......small mistake

i forgot to subtract the 4.8 from the 7.2 :D LOL hahahahahha

silly mistake for an AL student :D hahaha

Thanks guys!!

really appreciate it.....ur the best!!

ahh..thats k..happens...and ur welcome.. :)
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