A sliding conductor

Let $v$ be the velocity at any point of the movement of the rod. We see that the induced emf is $\epsilon=BLv$ . Then, the current is $I=\dfrac{\epsilon}{R}=\dfrac{BLv}{R}$ and the magnetic force, which goes upwards, has a magnitude of $F_m=ILB=\dfrac{B^2L^2v}{R}$ .

Now, if we apply Newton's 2nd law we get: $mg-F_m=ma$ , which is:

$mg-\dfrac{B^2L^2v}{R}=m\dfrac{dv}{dt}$

This is an easy differential equation, which we can solve as a separable function:

$mgR-B^2L^2v=mR\dfrac{dv}{dt}$

$dt=\dfrac{mRdv}{mgR-B^2L^2v}$

Integrate both sides:

$\displaystyle \int_0^t dt=\int_0^v \dfrac{mR\;\mathrm{d}v}{mgR-B^2L^2v}$

$t=-\dfrac{mR}{B^2L^2}\ln(mgR-B^2L^2v)\Big|_0^v$

$t=\dfrac{mR}{B^2L^2}\ln\left(\dfrac{mgR}{mgR-B^2L^2v}\right)$

Now, solve for $v$ in terms of $t$ :

$\dfrac{B^2L^2t}{mR}=\ln\left(\dfrac{mgR}{mgR-B^2L^2v}\right)$

$e^{\frac{B^2L^2t}{mR}}=\dfrac{mgR}{mgR-B^2L^2v}$

$mgRe^{-\frac{B^2L^2t}{mR}}=mgR-B^2L^2v$

$B^2L^2v=mgR\left(1-e^{-\frac{B^2L^2t}{mR}}\right)$

$v=\dfrac{mgR}{B^2L^2}\left(1-e^{-\frac{B^2L^2t}{mR}}\right)$

Finally, find the distance travelled:

$\displaystyle d=\int_0^t v\;\mathrm{d}t=\dfrac{mgR}{B^2L^2}\left(t+\dfrac{mR}{B^2L^2}e^{-\frac{B^2L^2t}{mR}}\right)\Big|_0^t$

$\boxed{d=\dfrac{mgR}{B^2L^2}\left(t-\dfrac{mR}{B^2L^2}\left(1-e^{-\frac{B^2L^2t}{mR}}\right)\right)}$