Mathematics: Post your doubts here!

[Mr.Physics]

-17.4?

For which one ?

 

[Anum96]

Where is Q 16?

 

[Anum96]

For which one ?

13b

 

[Mr.Physics]

Ans for 13b is 15 ms^-1

 

[Mr.Physics]

Q 16

 

[Mr.Physics]

Where is Q 16?

I uploaded the same pic mistakenly

 

[Anum96]

I'm sorry. My laptop is working really slow. Please wait.

 

[Mr.Physics]

I'm sorry. My laptop is working really slow. Please wait.

Oh okay

 

[Anum96]

Q16. I seem to be doing something wrong with 13, Ill figure that out later.
Q16
Q takes time t-1 and p takes time t
So the diaplacement of Q will be (using SUVAT) [3(t-1) plus 1.8(t-1)^2]
for P it will be [4t plus t^2]

b)Now equate both of them you will get t equals 6

c)use your ans to part b and the substitue 6 for t in s equal t^2 plus 4t
I hope this helps. My laptop is annoying af atm. :'(

 

[fleurisabelle]

'Find the cartesian equation of the locus of the set of points P when P is equidistant from the point (4,1) and the line x=2.'
x=2 is a whole line, how am I to know from which point P is equidistant?

 

[Rizwan Javed]

Q 13b
Anum96

v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.

 

[Anum96]

deceleration is 5 means acceleration is -5
velocity (f) is 12
velocity (I) is 0
x coordinate is 8 meaning distance is 8
2as = v^2 - u^2
2(-5)(8) = 12^2 - u^2
-80 - 144 = -u^2
-224 = -u^2
u = 14.96 ~~ 15

 

[Anum96]

v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.

Bingo!

 

[Rizwan Javed]

Bingo!

Wow :O Hahaha

 

[Mr.Physics]

v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.

deceleration is 5 means acceleration is -5
velocity (f) is 12
velocity (I) is 0
x coordinate is 8 meaning distance is 8
2as = v^2 - u^2
2(-5)(8) = 12^2 - u^2
-80 - 144 = -u^2
-224 = -u^2
u = 14.96 ~~ 15

Thank you guys
I was actually taking 12 ms^-1 as initial velocity

 

[DESTROYER1198]

guys can someone explain this question for me please; Q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w13_qp_31.pdf

 

[Anum96]

 

[Anum96]

I forgot to rotate it

 

[Wâlèé Atèéq]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_1.pdf

Q.9..!! Help..!

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf

Q.9..!! Help..!

dy/dx = 4 (6-2x)^-½

at x = 1, the gradient of the tangent to the curve is:
dy/dx ( at x=1 )= 4(6-2(1))^-½ = 2

gradient of normal = -1 / gradient of tangent
= -1/2

equ of normal at P,

y - 8 = -1/2 (x - 1)
2y + x = 17 -- (1)

When it cuts x-axis at Q, y=0, so x = 17

Q(17,0)

When it cuts y-axis at R, x = 0, y = 17/2
R (0, 17/2 )

mid point =( (17+0)/2 , (8.5+0)/2 ) = (17/2 , 17/4) Ans.

(b) integrate dy/dx

y = -4(6-2x)^½ +c

when x = 1, y = 8, so,

8 = -4(6-2)^½ +c
8= -4*2 + c
c = 16

so y = -4(6-2x)^½ +16