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

42.0 m/s^2

3.7 m/s^2

3.8 m/s^2

3.5 m/s^2

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation = change in velocity/change in time

a = {velocity final-velocity initial} / (change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity = 2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a = (44-2) / 12 = 42/12

3.5 m/s²

Best Wishes!