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

Answer:

$3.5 \frac{m}{s^{2} }$

Explanation:

Hello.

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

$a=\frac{V_{f} -V_{i} }{t_{2}-t_{1}}$

step 1

position 1

$velocity: 2.0\frac{m}{s}\\ time : t_{1}$

position 2

$velocity: 44.0\frac{m}{s}\\ time : t_{1} +12 =t_{2}$

step 2

replace

$a=\frac{V_{f} -V_{i} }{t_{2}-t_{1}}\\\\a=\frac{44-2}{t_{2}-t_{1} }=\frac{44-2}{t_{1}+12-t_{1} }\\a=\frac{42\frac{m}{s} }{12 s} }\\a=3.5\frac{m}{s^{2} }$

the acceleration of the rock is  

$3.5 \frac{m}{s^{2} }$

I hope it helps.