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

the acceleration of the rock is 3.5 m/s2

Answer 2

`3.5 (m)/(s^(2) )`

Step-by-step explanation:

Hello.

The acceleration is the change of speed in a time elapsed from point A to B

`a=(V_(f) -V_(i) )/(t_(2)-t_(1))`

step 1

position 1

`velocity: 2.0(m)/(s)\\ time : t_(1)`

position 2

`velocity: 44.0(m)/(s)\\ time : t_(1) +12 =t_(2)\\`

step 2

replace

`a=(V_(f) -V_(i) )/(t_(2)-t_(1))\\\\a=(44-2)/(t_(2)-t_(1) )=(44-2)/(t_(1)+12-t_(1) )\\a=(42(m)/(s) )/(12 s) }\\a=3.5(m)/(s^(2) )`

the acceleration of the rock is

`3.5 (m)/(s^(2) )`

I hope it helps.