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Mathematics: Post your doubts here!

Saad Mughal

Saad Mughal

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usama321 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_41.pdf
Can someone help me with the last question last part. After getting the value of constant c for the equation of velocity by integerating the acceleration equation, why do we once again find the value of c when we once again integrate velocity to get displacement? Don't we usually just ignore the constant when taking limits? Is it because there were two different equations, and had there been only one, we would not have needed to find out the constant for the displacement. Kindly someone help me. Thanks

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_ms_41.pdf
Click to expand...
We ignore the constant c when taking limits, here you are integrating to form an equation for displacement from P to A.
At P, the time t was 10 and displacement s was 100, hence the constant c could not be 0.
When you integrate the equation again to get a formula for s, you are using the time you calculated using the formula you got for velocity.
Basically, since there was displacement and time taken at the start of P (equation for which you are using), the constant c cannot be 0.
You can't integrate with limits because there are two different equations, one from O to P and one from P to A (as you said), whenever there is one equation, you can integrate with limits.
 
Last edited:
Saad Mughal

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Thought blocker said:
View attachment 41651 HELP
Click to expand...
(i) The body is in equilibrium, there are three force acting downwards and two forces acting upwards. According to the conditions for equilibrium.
Sum of upward forces = Sum of downward forces.
Q + R = 120g.
Taking moments about Q,
(60+20)(0.8)(g) = (0.4)(40)(g) + (1.6)(R)
1.6R =48g
R = 30g
Therefore, Q = 120g - 30g = 90g.

Reaction at Q = 90(10) = 900 N.
Reaction ap R = 30(10) = 300 N.

(ii) Reaction at Q = 2 * Reaction at R.
Hence, Q + R = 120g
3R = 120g
R = 40g

Taking moments about Q,
Let x be the distance from Q to x.
The body is in equilibrium, therefore Arthur has to sit in between Q and R. Hence,
60g(x) + 20g(0.8) = 40g(0.4) + 40g(1.6)
60g(x) = 64g
x = 16/15 = 1.07 m.
 
Thought blocker

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Saad Mughal said:
(i) The body is in equilibrium, there are three force acting downwards and two forces acting upwards. According to the conditions for equilibrium.
Sum of upward forces = Sum of downward forces.
Q + R = 120g.
Taking moments about Q,
(60+20)(0.8)(g) = (0.4)(40)(g) + (1.6)(R)
1.6R =48g
R = 30g
Therefore, Q = 120g - 30g = 90g.

Reaction at Q = 90(10) = 900 N.
Reaction ap R = 30(10) = 300 N.

(ii) Reaction at Q = 2 * Reaction at R.
Hence, Q + R = 120g
3R = 120g
R = 40g

Taking moments about Q,
Let x be the distance from Q to x.
The body is in equilibrium, therefore Arthur has to sit in between Q and R. Hence,
60g(x) + 20g(0.8) = 40g(0.4) + 40g(1.6)
60g(x) = 64g
x = 16/15 = 1.07 m.
Click to expand...
ty
 
Namehere

Namehere

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Saad Mughal said:
(i) The body is in equilibrium, there are three force acting downwards and two forces acting upwards. According to the conditions for equilibrium.
Sum of upward forces = Sum of downward forces.
Q + R = 120g.
Taking moments about Q,
(60+20)(0.8)(g) = (0.4)(40)(g) + (1.6)(R)
1.6R =48g
R = 30g
Therefore, Q = 120g - 30g = 90g.

Reaction at Q = 90(10) = 900 N.
Reaction ap R = 30(10) = 300 N.

(ii) Reaction at Q = 2 * Reaction at R.
Hence, Q + R = 120g
3R = 120g
R = 40g

Taking moments about Q,
Let x be the distance from Q to x.
The body is in equilibrium, therefore Arthur has to sit in between Q and R. Hence,
60g(x) + 20g(0.8) = 40g(0.4) + 40g(1.6)
60g(x) = 64g
x = 16/15 = 1.07 m.
Click to expand...

Is this M1 or M2?
 
AnujaK

AnujaK

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Hi everyone:)
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Please do check it out!
https://sites.google.com/site/fromatoscom/home
 
Saad Mughal

Saad Mughal

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Suchal Riaz said:
So are you doing Edexcel or CIE? I clearly mentioned in my post that it is not part of 'CIE syllabus'. Thought blocker, who asked the question, is doing CIE board.
Click to expand...
I never said it was in the CIE syllabus either. I clearly stated I have no idea which year the question is from because a fellow member asked it, I googled after that and found it to be M1 Edexcel and so I said so.
 
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