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? 42.0 m/s
2 3.7 m/s2
3.8 m/s2
3.5 m/s2

Respuesta :

acceleration = change in velocity / time

acceleration = (44 -2) /12

acceleration = 3.5

Answer: Option (d) is the correct answer.

Explanation:

Acceleration is the change in velocity with respect to time.

Mathematically,         a = [tex]\frac{dv}{dt}[/tex]

where               a = acceleration

                        dv = change in velocity

                        dt = change in time

It is given that initial velocity is 2.0 m/s and final velocity is 44.0 m/s. Chenge in time is 12 seconds. Therefore, calculate the acceleration as follows.

                     a = [tex]\frac{dv}{dt}[/tex]

                        = [tex]\frac{44.0 m/s - 2.0 m/s}{12 seconds}[/tex]

                        = [tex]\frac{44.0 m/s - 2.0 m/s}{12 seconds}[/tex]

                        = [tex]\frac{42.0 m/s}{12 seconds}[/tex]

                        = [tex]3.5 m/s^{2}[/tex]

Thus, we can conclude that acceleration of rock is [tex]3.5 m/s^{2}[/tex].