A-level Futher Mathematics Help....Mechanics

[amina hasan]

I am writing to you with regard to my Edexcel A-level Further Mathematics mechanics modules.The modules that I have chosen are M2,M3,M4,S2,FP1,FP2.  I really need help with the mechanics modules M3 and M4 .I currently reside in Dubai, UAE and am doing Further Mathematics privately (self-studying) since no school in Dubai offers Futher Mathematics. I have managed so far on my own with the help of notebooks of previous students who have already done these modules.Unfortunately nobody has actually done mechanics 3 and 4 so I have no tutor and no notebooks to rely on. I am finding the Mechanics modules very tedious and  have very little time to cover them. I have already done S2,FP1, FP2. Last year I completed the A-levels Mathematics and Pure Mathematics and so this was the only easiest combination of modules for Further Mathematics available to me. Is there any help that you can offer me online , like answering some of my questions or would you know of anybody in Dubai who could be of any help?..

[astarmathsandphysics]

I am your man. Post your questions.

[amina hasan]

Hey..
Could you help me with this question.(Mechanics 3)-
1.A particle P moves in a straight line with acceleration,at time t seconds, proportional to (t+to)^-3, where t0 is positive and constant. The intial speed of P at t=0 is um/s.Show that the speed of P approches a limiting value as t tends to infinity. Given that this limiting value is 2um/s, show that at time t seconds the particle will have travelled a distance
ut(2t+to)/(t+to) metres.

[astarmathsandphysics]

Hey..
Could you help me with this question.(Mechanics 3)-
1.A particle P moves in a straight line with acceleration,at time t seconds, proportional to (t+to)^-3, where t0 is positive and constant. The intial speed of P at t=0 is um/s.Show that the speed of P approches a limiting value as t tends to infinity. Given that this limiting value is 2um/s, show that at time t seconds the particle will have travelled a distance
ut(2t+to)/(t+to) metres.

dv/dt=k(t+to)^-3

We have to integrate. Add 1 to power and divide by new power. v=(k(t+to)^-2)/-2+c

Thent=0 v=u so u=(k(0+to)^-2)/-2+c=-k/(2(to)^2)+c so c=u+ k/(2(to)^2)

v=k((t+to)^-2)/-2+u+ k/(2(to)^2)=0.5k(((to)^-2)-(t+to)^-2))+u
As t tends to infinity second term in brackets tends to zero so v tends to 0.5k((to)^-2)+u we solve 2u=0.5k((to)^-2)+u so k=2to^2

dx/dt=1/2*2uto^2(-1/(t+to)^2+1/to^2)+u
       
        =u-to^2/(t+to)62+u= 2u-uto^2/(t+to)^2

Integrate x=2ut+uto^2/(t+to)+ c

x=0 t= 0 so c=-uto
and x=2ut+uto^2/(t+to)-uto
=u(2t+to^2/(t+to)-t0?
=u/(t+to)(2t^2+2tto-tto-to^2)
=u/(t+to)(2t^2+tto)=ut(2t+t0)/(t+to)

[amina hasan]

thanks....alot

[astarmathsandphysics]

No probs. Tricky question. Post more questions to keep me working.

[amina hasan]

Hey...I have some more questions....14,18,19,20,22,23....hmm...that's about six .....I am sending you pictures of the questions because some of them had diagrams ....let me know if you're not able to read them properly ...thanks  again...

[amina hasan]

hey...I just remembered ...I didn't give you the answers....
14.  19.8N,0.2N
18.  2.04m/s^2
19.  8.49m/s^2
20.  44.6N,0.172
23.  10m/s^2 upwards
......................

[astarmathsandphysics]

Hey...I have some more questions....14,18,19,20,22,23....hmm...that's about six .....I am sending you pictures of the questions because some of them had diagrams ....let me know if you're not able to read them properly ...thanks  again...

Had internet probs all day. Will do them in about 3.5 hours time

[wasimlhr]

do u guys donate to this man for his help and time ?


u should setup donate button!

[saifalan]

Would astar take charity ??  HArd to tell

[astarmathsandphysics]

I would  not. Everything on the internet should be  free and the best website is the one that make everything easiest to get.

[astarmathsandphysics]

Finished your questions. I dont have a pdf scaneer so have to go to the copy shop. Will be up in half hour. Cant read question3

[amina hasan]

ohhh...I can't thank you enough.... ....thankkk you sooooo much..... this really is the best website....

[astarmathsandphysics]

here