= Solution
Let $m$ be the car's mass and let $v(t)$ be its forward speed. Initially the driven contact surfaces slip backward relative to the ground because $v<r\Omega$. Thus <kinetic friction> on the car acts forward. On the horizontal surface the total <normal force> is $mg$, so <Newton's second law> gives $m\dot v=\mu mg$. While slipping persists,
$$
v(t)=\mu gt.
$$
The contact slip speed is $r\Omega-v(t)$. Its first zero is the onset of <rolling without slipping>, hence
$$
\boxed{T=\frac{r\Omega}{\mu g}.}
$$
The engine maintains the prescribed wheel speed, so wheel spin-down is not part of this calculation.
On a slope of angle $\alpha>0$, directed uphill, the <normal force> is $mg\cos\alpha$ and gravity contributes $-mg\sin\alpha$ along the slope. Therefore the uphill acceleration while slipping is $g(\mu\cos\alpha-\sin\alpha)$. If it is positive, the onset of <rolling without slipping> occurs at
$$
\boxed{T_\alpha=\frac{r\Omega}{g(\mu\cos\alpha-\sin\alpha)}>T.}
$$
The inequality follows because $\mu\cos\alpha-\sin\alpha<\mu$. If $\mu\cos\alpha\leq\sin\alpha$, the car does not accelerate uphill to $r\Omega$, so there is no finite catch-up time in this sliding-friction model. The finite-time conclusion presumes positive traction sufficient to climb the slope.
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