Asoccer ball starts from rest and accelerates with an acceleration of 0.395 m/s^2 while moving down a 8.50 m long inclined plane. when it reaches the bottom, the ball rolls up another plane, where, after moving 14.75 m, it comes to rest. (a) what is the speed of the ball at the bottom of the first plane (in m/s)? (round your answer to at least two decimal places.) (b) how long does it take to roll down the first plane (in s)?

a) The speed of the ball at the bottom of the first plane is 2.59 m/s

b) It takes the ball 6.56 s to roll down the first plane.

Explanation:

The equations for the position and velocity of the ball are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity at time t

b) First, let´s calculate the time it takes the ball to reach the bottom of the plane using the equation for the position:

x = x0 + v0 · t + 1/2 · a · t²

Placing the center of the frame of reference at the point where the ball starts rolling, x = 0. Since the ball starts from rest, v0 = 0. Then:

x = 1/2 · a ·t²

Let´s find the time when the ball reaches a position of 8.50 m

8.50 m = 1/2 · 0.395 m/s² · t²

t² = 2 · 8.50 m / 0.395 m/s²

t = 6.56 s

a) Now, using the equation of the velocity, we can calculate the velocity of the ball at the bottom of the plane (t = 6.56 s):

v = a · t

v = 0.395 m/s² · 6.56 s = 2.59 m/s

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