Question

Block 1 (2.5kg) is on the surface of a frictionless inclined plane that makes an angle of 20 degrees with the horizontal. Assume g = 10 m/s2

magnitude of the acceleration of block 1 down the incline plane is 3.4 m/s2

How far does block 1 travel down the plane in 1.5 s after it is released?

Answer 1

You already know the acceleration down the plan. They told you what it is.
a = 3.4 m/s^2.

The distance travelled in time t is
S = (1/2) a t^2

Just insert t = 1.5 s and compute your answer.