yes NOVEMBER 2014 P41 mechanicsAre you sure the ms says 0.1 m/s? Which paper is this?
http://maxpapers.com/wp-content/uploads/2012/11/9709_w14_qp_41.pdf
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yes NOVEMBER 2014 P41 mechanicsAre you sure the ms says 0.1 m/s? Which paper is this?
Firstly, equations of motion are to be used only when acceleration is constant. Particle P is moving at a constant speed, but Q is not. Your using v=u+at to find v = 3.3 for P is correct. For Q however, none of the equations of motion can be used, because Q's acceleration is not constant. You will have to integrate the acceleration to find its velocity function. When you integrate, the velocity function will include a +c, the arbitrary constant. By knowing that when t=20, v=3.3 for Q, you're able to substitute this into velocity function, to find the arbitrary constant, and thus able to find v when t=0.View attachment 52128
in part 1 i used v=u+at but the ans turned out to be wrong why??
i used vP= 1.3 +(0.1*20)=3.3m/s
ince vP=vQ so 3.3= vQ
thus, at t=20s 3.3= uQ+(0.016*20) uQ= 3.3-0.32=2.98m/s
Whereas, in markscheme the formula s=ut+1/2at^2 is used which gives the ans 0.1m/s
Can someone explain why cant we use v=u+at here???
how can i figure out from the question that the acceleration is not constant?Firstly, equations of motion are to be used only when acceleration is constant. Particle P is moving at a constant speed, but Q is not. Your using v=u+at to find v = 3.3 for P is correct. For Q however, none of the equations of motion can be used, because Q's acceleration is not constant. You will have to integrate the acceleration to find its velocity function. When you integrate, the velocity function will include a +c, the arbitrary constant. By knowing that when t=20, v=3.3 for Q, you're able to substitute this into velocity function, to find the arbitrary constant, and thus able to find v when t=0.
x=3 will be used and the equation will be equal to 0 as dy/dx is 0.View attachment 52134
to find f '(x) integration= y= -18/x^2 +c ...why do we have to use x=0 to find c why not the x of stationary point that is 3?
For P the acceleration is constant as the acceleration is just a number but for Q the expression involves time too. This means it varies according to time and hence is not constant.how can i figure out from the question that the acceleration is not constant?
thnksx=3 will be used and the equation will be equal to 0 as dy/dx is 0.
For P the acceleration is constant as the acceleration is just a number but for Q the expression involves time too. This means it varies according to time and hence is not constant.
5(i) I hope its clearCan someone help me to solve may June 2014 paper 31 question 5(i), 8, 10(ii)
http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_31.pdf
Can someone help me to solve may June 2014 paper 31 question 5(i), 8, 10(ii)
http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_31.pdf
thx a lot.. can u help me to solve 5(ii) too? i stuck at thr again... ><5(i) I hope its clear
Ill do 8 and 10(ii) for you later
haha.. is okay.. but can i know how 8(iii) get the initial x=0.5pi ?8) I had to squeeze it in a bit xD I also think my sketch is bad.
If you're not given the initial x when doing iteration, you start with the middle x, here x is between 0 and pie, so 0.5piehaha.. is okay.. but can i know how 8(iii) get the initial x=0.5pi ?
oh... Thx a lot.. Help me a lotIf you're not given the initial x when doing iteration, you start with the middle x, here x is between 0 and pie, so 0.5pie
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_42/
can someone help me solve question 5?
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