Question

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

Answer 1

Answer:

a = 1.16 m/s²

Explanation:

In order to find the acceleration of the ball we will use 3rd equation of motion.

2as = Vf² - Vi²

where,

a = acceleration = ?

s = displacement = 21.9 m

Vf = Final Velocity = 7.14 m/s

Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)

Therefore, using the values, we get:

2a(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