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zavuch27 [327]
3 years ago
9

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 velo

city 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
Physics
1 answer:
mr Goodwill [35]3 years ago
5 0

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!

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