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Mathematics: Post your doubts here!

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need help with question 7 ii) october/november 2008 mechanics M1

how come the constant is 5 when we integrated the acceleration ?

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w08_qp_4.pdf
From the glimpse of it (I did not really solve the question but anyway) when you integrate an equation, you always are left with a constant. In this situation, when you integrate the acceleration, you get an equation where you have three unknowns: v, t and c(the constant of integration)! To find c, you substitute the value of v you found in the part (a) as well as the value of corresponding t which leaves you with just an unknown c. Solve the equation to acquire the value of t. Remember! The constant of integration is only ignored or not considered when we're provided with absolute limits.
Hope this clears up! :)
 
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From the glimpse of it (I did not really solve the question but anyway) when you integrate an equation, you always are left with a constant. In this situation, when you integrate the acceleration, you get an equation where you have three unknowns: v, t and c(the constant of integration)! To find c, you substitute the value of v you found in the part (a) as well as the value of corresponding t which leaves you with just an unknown c. Solve the equation to acquire the value of t. Remember! The constant of integration is only ignored or not considered when we're provided with absolute limits.
Hope this clears up! :)
btw what do u mean by absolute limits ?
 
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Gradient of AB = 2
Since gradient of AB = 2m (given),

2m = 2
m = 1

(ii) C is a point of intersection between AC and BC, so the first thing that should occur to you is obtaining and then solving simultaneous equations.

Gradient of AC = -2m = -2(1) = -2
Now with the gradient of AC and co-ordinates of A, you can form an equation in AC.

Likewise, you can form an equation in BC. C will turn out to be (-1, 6).

(iii) We're talking about the perpendicular bisector of AB. Perpendicular gradient of AB is negative reciprocal of 2, which is -1/2. And the co-ordinates of the point will be the midpoint of AB => (9, 10).

Now make the equation of the perpendicular bisector with (9,10) and the gradient of -1/2. You have the equation of BC as well and since they intersect, you can solve them simultaneously to obtain the coordinates of D.
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_4.pdf no7 why its wrong when i do it a=9.25 given the equation and u=5 t=3-0.5=2.5 and the getting dis=ut+0.5at^2 5(2.5)+4.625(6.25)=41.4 +1.25=42.65

in marking scheme its solved by integrations and i cant understand why

Recognize in (ii) that acceleration is no longer a constant( ie not equals to g) and is now a function of time, hence the equations of motions no longer apply; you have to discover the expression for its velocity through one round of integration, and its displacement through a second round of integration.

Hope this helps. Peace.
 
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Gradient of AB = 2
Since gradient of AB = 2m (given),

2m = 2
m = 1

(ii) C is a point of intersection between AC and BC, so the first thing that should occur to you is obtaining and then solving simultaneous equations.

Gradient of AC = -2m = -2(1) = -2
Now with the gradient of AC and co-ordinates of A, you can form an equation in AC.

Likewise, you can form an equation in BC. C will turn out to be (-1, 6).

(iii) We're talking about the perpendicular bisector of AB. Perpendicular gradient of AB is negative reciprocal of 2, which is -1/2. And the co-ordinates of the point will be the midpoint of AB => (9, 10).

Now make the equation of the perpendicular bisector with (9,10) and the gradient of -1/2. You have the equation of BC as well and since they intersect, you can solve them simultaneously to obtain the coordinates of D.

Thank you very much :)
 
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(iii) Notice that f(x) lies between 4-3=1 and 4+3=7 inclusive. Solving for f(x)=k can be interpreted as drawing a horizontal like y=k and finding the number of intersections with the y=f(x) curve. Hence, if no solutions arise, there are no intersections, ie k>7 or k<1.

(iv) A function has an inverse if and only if it is either strictly increasing or decreasing. Try sketching out the graph, you should be able to see that maximum value of A= pi radians, because in between pi/2 and pi radians, the function is strictly increasing. Alternatively, you can find dy/dx= -3 cosx, which is > or equals to zero for the domain defined.

(v) Let y=4-3sinx, then making x the subject gives x= arcsin (4-y)/3
Doing the reflection in the y=x line gives inverse of g to be = arcsin (4-x)/3
Hence, when x=3, g^-1 (3) =arcsin (4-3)/3 =arcsin (1/3).

Hope this helps. Peace.
 
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Please someone help
The question is very simple though bt I m not getting the ryt answer :/
V r suppose to find the gradient
The points given r (p+3,q-5) and (q-5,p+3)
I m getting the gradient as 0 and the answer given in the book is -1 :S
Can someone plz xplain me y?

You probably got zero because you took the y-coordinate of the first point and subtracted it from the x-coordinate of the second point.
For two points (x1, y1) and (x2, y2), the gradient is simply given by m = (y1 - y2)/ (x1 -x2) [ or (y2 - y1)/ (x2 -x1) ]

Hope this helps.
 
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You probably got zero because you took the y-coordinate of the first point and subtracted it from the x-coordinate of the second point.
For two points (x1, y1) and (x2, y2), the gradient is simply given by m = (y1 - y2)/ (x1 -x2) [ or (y2 - y1)/ (x2 -x1) ]

Hope this helps.
silly mistake :p

btw thnx :)
 
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Please Help
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(iii) Notice that f(x) lies between 4-3=1 and 4+3=7 inclusive. Solving for f(x)=k can be interpreted as drawing a horizontal like y=k and finding the number of intersections with the y=f(x) curve. Hence, if no solutions arise, there are no intersections, ie k>7 or k<1.

(iv) A function has an inverse if and only if it is either strictly increasing or decreasing. Try sketching out the graph, you should be able to see that maximum value of A= pi radians, because in between pi/2 and pi radians, the function is strictly increasing. Alternatively, you can find dy/dx= -3 cosx, which is > or equals to zero for the domain defined.

(v) Let y=4-3sinx, then making x the subject gives x= arcsin (4-y)/3
Doing the reflection in the y=x line gives inverse of g to be = arcsin (4-x)/3
Hence, when x=3, g^-1 (3) =arcsin (4-3)/3 =arcsin (1/3).

Hope this helps. Peace.

OMG ! You are great ! Thanks a lot :)
 
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thanks a lot. this constant thingy annoyed me a lot. sometimes they put it and sometimes they don't and i always used to wonder why sometimes they do and sometimes they dont!
lol! Its always there but in limits, they just ignore it! :)
 
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