Recent content by LukeEvans

How does that look?

Right. m = g - vvn

m = g - 3vn I'll be honest I don't know where the 3 has come from.

how about: m = g * vn or m = g*(v+vn)

m = g - (v*vn) possibly?

m = g - vn would be an uneducated guess :-p

A body of mass m (kg) is falling vertically under gravity g m/s2 in a medium whose resistance to the speed of the body, v m/s, is proportional to vn (n positive). If the body was released from rest and has terminal velocity, vt m/s, use Newton's 2nd law of motion to show that its...

Wow thank you. Glad I was close, and thanks for the heads up too!

Ah I believe I understand now. Thank you for helping me understand this, or at least figure out what to do in an exam! I have one more issue - should I be asked to integrate 4e2t + 2 how would I deal with the e? My best guess would be that the integral of that would be: 2t + e2t + C

Actually as a side note, I'm mulling a question over. Would the derivative of the position (distance) give the velocity? As giving the anti derivative of velocity gives us position, would the derivative of position give us the velocity?

Is the next step to include the 8m (which is the information loaded in the last paragraph?)? So it should be: d(t) = 2t2 + t -8− sin(t)

I can see it's almost looking like it...

x(t)+x0 = 2t2+ t - sin(t) is this what you mean? I've got to show that f( t ) = 2 t2 + t - 8 - sin t

x(0) = -sin(0) + 0 + 0 + c or just x(0) = 0 + c Correct?

x(0) would be distance zero? As in the datum point 0?