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No.Great, now I got my confusion cleared. Lets do 2003 now ? And tomorrow, 2011,12,13.. Dont say no
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No.Great, now I got my confusion cleared. Lets do 2003 now ? And tomorrow, 2011,12,13.. Dont say no
As you say.
When the Velocity is zero.guys when we should ignore the integration constant C or assume it is ZERO ?
When the Velocity is zero.
Konsey yaar ka paper hai ji ?Thank you for help, but can you tell me why in the question below in the first equation we ignore the integration constant and in the second equation integration constant is included ?
I hope you can read and understand thathi i really need help im this question
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_42.pdf
Q7, the second part of the part ii
Sorry, i can't understand.Konsey yaar ka paper hai ji ?
Which year paper ?Sorry, i can't understand.
In the first part, we are only finding the distance for the first 60 seconds.Thank you for help, but can you tell me why in the question below in the first equation we ignore the integration constant and in the second equation integration constant is included ?
Which year paper ?
In the first part, we are only finding the distance for the first 60 seconds.
In the third part, we have to consider the distance travelled in the first 60 seconds as well. Thus, we first equate the second equation of displacement to the displacement of the first 60s from the first equation, find c, and then integerate that.
I hope you can read and understand that
we have to use that component of gravity which is influencing the motion of A, since the path is not horizontal, it is g*sinxHi! Yes i can understand what you wrote! thanks!
But what i dont get is why you multiply the 'g and sinx' together? Thats what im confused about. Why do we include that?
What is c in that context? I mean what does it represent?In the first part, we are only finding the distance for the first 60 seconds.
In the third part, we have to consider the distance travelled in the first 60 seconds as well. Thus, we first equate the second equation of displacement to the displacement of the first 60s from the first equation, find c, and then integerate that.
It would give us the distance of the first time interval as wellWhat is c in that context? I mean what does it represent?
Post all your doubts of P1 and M1. Surely I will solve them.
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So of we find the distance moved in the first time interval and add it to the equation of the distance moved in the second time interval, would it give us the same resultant equation of the total distance moved?It would give us the distance of the first time interval as well
Yup. You will have to take the limits of 0 to 60 for the first equation and then 60 to an unknown value of time for the second equation and equal it to the distance. However, i don't think the equation formed would be easily solvable, if solvable at all.So of we find the distance moved in the first time interval and add it to the equation of the distance moved in the second time interval, would it give us the same resultant equation of the total distance moved?
Can't we just integrate the distance without any limits, putting o as the constant and add the distance moved in the first time interval to the equation formed? Can you please try and see if it worksYup. You will have to take the limits of 0 to 60 for the first equation and then 60 to an unknown value of time for the second equation and equal it to the distance. However, i don't think the equation formed would be easily solvable, if solvable at all.
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