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ehild said:Now what about Orodruin's hint? What differential equation do you get for y=x/(L+vt)?

ehild

[itex]\frac{dy}{dt}=\frac{\dot x}{L+vt}-\frac{xv}{(L+vt)^2}[/itex]

(With initial value y=0 when t=0)

Which could be written as:

[itex]\frac{dy}{dt}=\frac{\dot x-yv}{L+vt}[/itex]

(I got stuck here for a few minutes but then thought of this )

Maybe I should plug in [itex]\dot x=u+vy[/itex] and then I get:

[itex]\frac{dy}{dt}=\frac{u}{L+vt}[/itex]

[itex]\int dy=\int \frac{u}{L+vt}dt=\frac{u}{v}ln(L+vt)+C=y[/itex]

From our initial condition we see that C=0

Now we just set [itex]y=1[/itex] and solve for [itex]t[/itex]!

[itex]ln(1+t)=100000[/itex]

[itex]t=e^{100000}-1[/itex]