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Mathematics: Post your doubts here!

Esme

Esme

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ahmed abdulla said:
thanks .... can u plz tell me about no.7 WHY we equated it with 0 for accelaration ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_42.pdf
Click to expand...

i found dv/dt first i.e. the acceleration.. and because the question says speed is increasing, so dv/dt > 0
So you'll the get the two values of T for which the speed is increasing, or for which the acceleration is positive however you want to see it. But just understand the question..
 
ZainH

ZainH

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GorgeousEyes said:
Nov012 v.42 no4 :S
Click to expand...

They've given you one of the forces in the y-direction already, 75N . Since they mentioned positive y-direction that means upwards. Anything down will be taking as negative so you'll have to change the sign later.

To find the other two forces use Pythagoras because you have the x-direction and the resultant so you'll get one force as.
68^2-60^2 = √1024 = -32N
Change it to positive = 32N

And the other as
100^2-96^2 = √784 = -28
Change it to positive = 28N

Hope you got it x)
 
Esme

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Silent Hunter said:
cant remember the paper.... did this few weeks ago so cant recall..... just recalled the scenario :p :p
so confirm it should be t-2 ? these type are so confusing :\ anybody recalls a paper similar to this?
Click to expand...

Yeah t-2 it is :) just imagine what the question is saying in your head and you'll get it :D
 
GorgeousEyes

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ZainH said:
They've given you one of the forces in the y-direction already, 75N . Since they mentioned positive y-direction that means upwards. Anything down will be taking as negative so you'll have to change the sign later.

To find the other two forces use Pythagoras because you have the x-direction and the resultant so you'll get one force as.
68^2-60^2 = √1024 = -32N
Change it to positive = 32N

And the other as
100^2-96^2 = √784 = -28
Change it to positive = 28N

Hope you got it x)
Click to expand...
Sorry but I can't get it :(
 
ZainH

ZainH

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Math_zpsa852621e.png
 
AlishaK

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GorgeousEyes said:
Nov012 v.42 no4 :S
Click to expand...
What they have given us:The 'x' components of the following forces:
68 N = -60N
75N= 0N
100N= 96N,
What we gotta find, i) (resolve the forces 100 N and 68N) therefore, 68cos(theta) = -60 , theta= 28.1 deg
100cos (beta) =96, beta=16.2 deg, hence we now know the angles that the forces make with the horizontal axis. We can find 'y' components of the forces.
68 N => -68sin28.1 = -32 N ( negative as it in he the negative y axis direction)
75 N => no components to resolve as it is itself on the y-axis so, 75 N
100 N => -100sin16.2= -27.9 N

ii) List the Fx components of the forces And then the Fy components as we calculated above:
Fx=> -60+0+96 = 36 N
Fy=> -32+75-27.9= 15.1 N
Resultant = (36^2 + 15.1^2)^1/2 (i took the square root of the squares of the sum of the two forces we calculated (Fx and Fy)
Therefore, tan(theta)= y/x => tan (theta) = 15.1/36, theta= 22.8 deg.
Hope u got it. :D
 
GorgeousEyes

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ZainH said:
Math_zpsa852621e.png
Click to expand...
AlishaK said:
What they have given us:The 'x' components of the following forces:
68 N = -60N
75N= 0N
100N= 96N,
What we gotta find, i) (resolve the forces 100 N and 68N) therefore, 68cos(theta) = -60 , theta= 28.1 deg
100cos (beta) =96, beta=16.2 deg, hence we now know the angles that the forces make with the horizontal axis. We can find 'y' components of the forces.
68 N => -68sin28.1 = -32 N ( negative as it in he the negative y axis direction)
75 N => no components to resolve as it is itself on the y-axis so, 75 N
100 N => -100sin16.2= -27.9 N

ii) List the Fx components of the forces And then the Fy components as we calculated above:
Fx=> -60+0+96 = 36 N
Fy=> -32+75-27.9= 15.1 N
Resultant = (36^2 + 15.1^2)^1/2 (i took the square root of the squares of the sum of the two forces we calculated (Fx and Fy)
Therefore, tan(theta)= y/x => tan (theta) = 15.1/36, theta= 22.8 deg.
Hope u got it. :D
Click to expand...



Thaaaanks a millioooooon , May Allaaah bless youu both :)
 
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