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? 42.0 m/s

2 3.7 m/s2
3.8 m/s2
3.5 m/s2

Answer 1

acceleration = change in velocity / time

acceleration = (44 -2) /12

acceleration = 3.5

Answer 2

Answer: Option (d) is the correct answer.

Step-by-step explanation:

Acceleration is the change in velocity with respect to time.

Mathematically, a =
`(dv)/(dt)`

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 =
`(dv)/(dt)`

=
`(44.0 m/s - 2.0 m/s)/(12 seconds)`

=
`(44.0 m/s - 2.0 m/s)/(12 seconds)`

=
`(42.0 m/s)/(12 seconds)`

=
`3.5 m/s^(2)`

Thus, we can conclude that acceleration of rock is
`3.5 m/s^(2)`.