A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration $$a_{0}$$. The angle between the inclined plane and ground is θ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ___ .

The plane is accelerating left with acceleration a₀. So in the plane’s frame, a pseudo force acts on the block towards the right with magnitude m a₀.

Now resolve forces along the incline.

1. Component of gravity along incline (down the plane):
   g sinθ
2. Component of pseudo acceleration along incline:
   The pseudo acceleration is horizontal to the right. Its component along the incline (down the plane) is:
   $$a₀\cosθ$$

So total effective acceleration of the block along the incline becomes:
$$a=g\sinθ+a₀\cosθ$$

Now find the distance the block has to travel along the incline.

Given base = L
Incline length = $$L/\cosθ$$

So displacement along the plane:
$$s=L/\cosθ$$

Initial velocity = 0

Use equation of motion:
$$s=(1/2)at^2$$

Substitute:
$$L/\cosθ=(1/2)(g\sinθ+a₀\cosθ)t^2$$

Solve for t:
$$t=\sqrt{\ \frac{2L}{(\cosθ(g\sinθ+a₀\cosθ))}}$$

Now simplify the denominator:

$$\cosθ·g\sinθ=(g/2)\sin2θ$$
$$\cos^2θ=(1+\cos2θ)/2$$

So denominator becomes:
$$(g/2)\sin2θ+(a₀/2)(1+\cos2θ)$$

Factor 1/2 out:

$$=(1/2)[g\sin2θ+a₀(1+\cos2θ)]$$

So finally:

$$t=\sqrt{\ \frac{4L}{(g\sin2θ+a₀(1+\cos2θ))}}$$