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We have an accelerating force of ##F = mg\sin \theta## down the slope. The work done by this force over a distance ##x## is ##Fx## and this must equal the total linear and rotational KE. Hence$$mgx \sin \theta = \frac 1 2mv^2 + \frac 1 2 Iw^2 = \frac 1 2 mv^2 + \frac 1 2 kmr^2\frac {v^2}{r^2} = \frac 1 2 mv^2(1 +k)$$This gives us $$v^2 = 2\frac{g\sin \theta}{1 +k}x$$And using $$v^2 = 2ax$$gives$$a = \frac{g\sin \theta}{1 +k}$$Finally, taking ##k = \frac 2 5## and using $$x = \frac 1 2 at^2$$ we get$$x = \frac{5g\sin \theta }{14}\ t^2$$And that only uses what I believe to be high school physics and maths.