Question

A rock is rolling down a hill. At position 1, its velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, its velocity is 44.0 m/s. What is the acceleration of the rock?

Answer 1

Answer:

The correct answer is 3.5 m/s²

Explanation:

To determine the average acceleration of the rock,

the change in velocity is divided by the time interval (in seconds)

change in velocity = V₂ - V₁

where V₁ is the initial velocity (2.0 m/s) and V₂ is the final velocity (44.0 m/s)

change in velocity = 44 - 2 = 42 m/s

The time interval is 12 seconds (no need for conversion as it is in seconds already)

average acceleration of the rock = 42 ÷ 12 = 3.5 m/s²

Answer 2

As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

44.0 m/s - 2.0 m/s = 42.0 m/s 

42.0 m/s / 12 s = 3.5 m/s2

the acceleration of the rock would be 3.5 m/s2