Question

select the correct answer. a ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second². what is the final velocity of the ball after 6.0 seconds? a. -9.6 meters/second b. 9.6 meters/second c. -19 meters/second d. 19 meters/second

Explanation:

Answer:

D. 19 meters/second

(Note: Using the kinematic equation \( v = u + at \), where initial velocity \( u = 0 \) (at rest), acceleration \( a = 3.2 \, \text{m/s}^2 \), time \( t = 6.0 \, \text{s} \). So \( v = 0 + 3.2 \times 6.0 = 19.2 \approx 19 \, \text{m/s} \). The direction is down the hill, but if we consider magnitude or the problem’s context (positive direction down), the answer is 19 m/s.)