Recent content by GFreeman64

If we were to say that: F(v)=\frac{m(1+av)}{ka} Then the integration seems to work and is equal to klog(1+au). For now I'll assume this is correct. For the second part I have taken the equation t+T(v)=T(u) to give: t+klog(1+av)=klog(1+au) Rearranging this for v gives...

Hmm, so in your example here, if f(y)=y then it would also be true that f(x)=x. Is then as simple as saying that if: F(u)=-\frac{1+au}{ka} then F(v)=-\frac{1+av}{ka}? If we substitute this expression for F(v) into the original integral it seems to almost work, apart from the fact that we...

Ah hah! So if we were to differentiate both sides with respect to u, the RHS would become ka/(1+au). Then using the theorem given in the hint the LHS would become -1/F(u). So that gives us: F(u)=-\frac{1+au}{ka}. However the question asked for the force in terms of v, so how might I go...

Hello everyone, This is my first time in Physics forums so forgive me if I stray from the convention for asking questions. Me and a friend are completely stumped on our first assignment for the year and we are not sure where else to turn! Homework Statement A particle of mass m can...