Answer:
[tex]a = 1.28 m/s^2[/tex]
Explanation:
As we know that the block is released on the inclined plane from rest
so here the distance moved by the block is given as
[tex]d = \frac{1}{2} at^2[/tex]
so here we will have
d = 3.1 m
t = 2.2 s
now we will have
[tex]3.1 = \frac{1}{2}a(2.2^2)[/tex]
[tex]a = 1.28 m/s^2[/tex]