Answer:
The deceleration, a = -1.6 m/s²
Explanation:
Given,
The initial velocity, u = 14 m/s
The time taken by the stone is, t = 2.5 s
The distance between the point, s = 30 m
Using the 2nd equations of motion
s = ut + 1/2 at²
∴ a = 2(s - ut)/t²
Substituting the values
a = 2(30 - (14)(2.5))/2.5²
= -1.6 m/s²
Hence, the deceleration of the stone, a = -1.6 m/s²
