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.