Question

a ball rolling down a hill was displaced 21.9 m while using uniformly accelerating from rest, If the final velocity was 7.14 m/s, what was the rate of acceleration?

Answer 1

a = 1.16 m/s²

Step-by-step explanation:

In order to find the rate of acceleration of the ball, we will use third equation of motion, as follows:

2as = Vf² - Vi²

where,

a = rate of acceleration = ?

s = distance covered by the ball = 21.9 m

Vf = Final Velocity of the ball = 7.14 m/s

Vi = Initial Velocity of the ball = 0 m/s (Since, the ball started from rest)

Therefore,

2(a)(21.9 m) = (7.14 m/s)² - (0 m/s)²

a = (50.97 m²/s²)/(43.8 m)

a = 1.16 m/s²