Question

A ball rolling down a hill has an acceleration of 4.12m/s/s. What would the change in velocity be if it rolled down the hill at this constant acceleration for 22.2 seconds?

Answer 1

The change in velocity of a ball accelerating down a hill at 4.12 m/s² for 22.2 seconds is 91.464 m/s.

Step-by-step explanation:

The question regards calculating the change in velocity of a ball that has a constant acceleration down a hill. To find the change in velocity (Δv), we can use the formula Δv = a * t, where 'a' is the acceleration and 't' is the time the ball accelerates for. Given that the acceleration 'a' is 4.12 m/s² and time 't' is 22.2 seconds, the change in velocity can be calculated as follows:

Δv = 4.12 m/s² * 22.2 s = 91.464 m/s.

So, the ball's velocity will change by 91.464 meters per second after rolling down the hill for 22.2 seconds at the stated constant acceleration.