Recent content by Ashley1nOnly

Right so all he did was integrate the velocity again to get position and then solved for t. the initial x is zero and final x is 900 miles

So I integrate the acceleration above once. a=(cvt^2+cv^2)/m Why can't I solve for the position, set x0= 0 and x=900 and then solve for t

to find out how long it takes to get to the destination a= (cvt^2+cv^2)/m integrate it twice to get position and then solve for t

To find the F, I made another post with is a continuation of this if you could help

Right because they were all constants, I kept looking at that but the answer is usually always infinity or zero so I was hesitant to pick that.

Homework Statement Looking for F for 200 km/hr Homework Equations F=ma The Attempt at a Solution F-cv^2 = ma F= ma +cv^2 v=200 km/hr F= ma +c(200 km/hr)^2 so now I need to find a, from previous work I found that t= (m/c)*(1/2 vter)*(ln[ (vter +v)/(vter-v)]) next I would solve for v...

How do I calculate the F?

v= vt*e^((t*c*2*vt)/m) -vt) / (e^((t*c*2*vt)/m) +1) where x = (t*c*2*vt)/mv must go to 0

t= m/(c*2*vt) *ln [ (vt+v )/ (vt-v)]

Right I totally skipped over that. I can't divide like that. first integral ln[ vt/ (vt-v)] second integral ln[ 1 + v/vt] ln[ vt/ (vt-v)]+ln[ 1 + v/vt] =ln [ (vt+v )/ (vt-v)]

But now I have two v's I just need one

t= m/(c*2*vt) * ln(1- vt^2 / v^2) v= sqrt ( e^((t*c*2*vt )/m) *(1-vt^2)

so final answer for the integration part is ln(1- vt^2 / v^2)

applying limits first integral -ln|vt -v| - [-ln |vt-0|] -ln|vt -v| +ln|vt| ln| vt/ (vt-v) | ln| (1-vt/v)| second integral ln|vt+v| -ln|vt+0| ln|vt+v|-ln|vt| ln| (vt + v )/(vt) | ln| (1+ v/vt) first integral + second integral ln| (1-vt/v)| + ln| (1+ v/vt)| ln | (1-vt/v) / (1+ v/vt)|

so when you integrate the first one you get -ln|vt-v| and the second one ln|vt+v| without applying the limits