Recent content by Keval

okay here's trying this new method i get lost pretty early :\ F=ma mg-F_d=ma using F_d=-\frac{1}{2}\rho A C_d v^2 mg-kv^2=ma using k=-\frac{1}{2}\rho A C_d mg-k(\frac{ds}{dt})^2=m\frac{dv}{dt} mg-k\frac{d^2s}{dt^2}=m\frac{dv}{dt} now what? o.o, I'm not sure if i shoud be using v=dx/dt, or...

Hey just got back from uni and saw thiss could you use some latex to show me? I'm sort of confused, am still reading up on it thought

someone check this out for me ?? -\frac{1}{k} \int \frac{dx}{x^2-\frac{g}{k}} to simplify i let\frac{g}{k}=m^2 to get -\frac{1}{k}\int \frac{dx}{x^2-m^2}=-\frac{1}{k}\int \frac{dx}{(x-m)(x+m)} Using partial fractions i got \int \frac{dx}{x^2-m^2}=\int...

Question is in orange, answer is in black. I got no idea how they got this answer :\ The way I am trying is using a substition of kx^2 = gsin\theta Just the one question i can't get my head aroung in this exercise step by step method would be appreciated

\Delta t \left(gt-\frac{k}{m}v^2 \right)= \Delta v If your trying to work out the change in velocity how can you use v^2 on the left when that's wht your trying to work out for this part. and your saying once i get that sorted i use \frac{\Delta v }{\Delta t }=s

We are taught for free until we start university which i only started on wednesday which we pay £3225 a year Can you show me the differential way maybe ill undestand that

i understand that Delta[t/v] means change in time or velocity

Well here in england pre-uni we were taught parts of calculus in our syllabus which was Basic Differentiation + Integration, Differentiation + Integration of Exponentions Logs and Trig, Volumes of revolution using int, Integration of partial fractions and further trig, Solving first order...

I understood upto here what assumption did you make to get this mg-kv^2 = m\frac{dv}{dt} to this \Delta t \left(gt-\frac{k}{m}v^2 \right)= \Delta v also the above is same as this right? \frac{dt}{dt} \left(gt-\frac{k}{m}v^2 \right)= \frac{dv}{dt} the greek delta confuses me...

ahh i understand the constant bit now and why the v^2 is seperate your last latex is messed up :s

Huh i don't follow mg- \frac{1}{2}\rho{A}{C_d} = m\frac{dV}{dT}?

so you want mg-k=m\frac{dv}{dt}

1stly wasnt the equation a double minus as k= -0.5pv^2ACd and mg - Fd = ma => mg - -k = ma so mg + k = ma? and I am confused by your last line, how do i get an explicit equn in displacemnt sorry I am so terrible atm

Im good with calculus but I've never applied calculus to mechanics all I've heard is that vdot = dV/dt or something lol

yeah i figured but i need one that takes into consideration the drag :\