A ball rolls down an incline at a constant acceleration of 2 m/s^2. What is the velocity of the ball after 8.5 s?

How far does the ball roll in 10s? How far does the ball roll in the seventh second?

a = (Vf - Vi)/t

d = Vi*t + (1/2)a*t^2

where Vf = final velocity, and Vi = initial velocity.

9m/s

To find the answers to these questions, we will use the equations of motion for an object with constant acceleration.

1. Velocity after a certain time:
The equation that relates initial velocity (u), time (t), acceleration (a), and final velocity (v) is:
v = u + at

Given:
Initial velocity (u) = 0 (as the ball starts from rest)
Acceleration (a) = 2 m/s^2
Time (t) = 8.5 s

Substituting the given values into the equation, we can solve for the final velocity (v):
v = 0 + 2 * 8.5

v = 17 m/s

Therefore, the velocity of the ball after 8.5 seconds is 17 m/s.

2. Distance traveled in a certain time:
The equation that relates initial velocity (u), time (t), acceleration (a), and distance traveled (s) is:
s = ut + (1/2) * a * t^2

a. Distance traveled in 10 seconds:
Given:
Initial velocity (u) = 0 (as the ball starts from rest)
Acceleration (a) = 2 m/s^2
Time (t) = 10 s

Substituting the given values into the equation, we can solve for the distance traveled (s):
s = 0 * 10 + (1/2) * 2 * (10^2)

s = 100 m

Therefore, the ball rolls 100 meters in 10 seconds.

b. Distance traveled in the seventh second:
To find the distance traveled in the seventh second, we need to calculate the distance traveled in 7 seconds and subtract the distance traveled in 6 seconds.

Distance traveled in 7 seconds:
s1 = 0 * 7 + (1/2) * 2 * (7^2)
s1 = 49 m

Distance traveled in 6 seconds:
s2 = 0 * 6 + (1/2) * 2 * (6^2)
s2 = 36 m

Therefore, the ball rolls 13 meters (49 m - 36 m) in the seventh second.

In summary:
- The velocity of the ball after 8.5 seconds is 17 m/s.
- The ball rolls 100 meters in 10 seconds.
- The ball rolls 13 meters in the seventh second.