A ball starts from rest and accelerates at 0.520 m/s2 while moving down an inclined plane 9.65 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.40 m, it comes to rest.

(a) What is the speed of the ball at the bottom of the first plane?

(b) How long does it take to roll down the first plane?

(c) What is the acceleration along the second plane?

(d) What is the ball's speed 7.85 m along the second plane?

Nevermind! I got it:

(a) What is the speed of the ball at the bottom of the first plane?

v2 = vi2 + 2 a (x - xi)

(b) How long doe is take to roll down the first plane?

v = vi + a t


(c) What is the acceleration along the second plane?

v2 = vi2 + 2 a (x - xi)

(d) What is the ball's speed 8.0 m along the second plane?

v2 = vi2 + 2 a (x - xi)

To solve this problem, we can use the basic equations of motion. Let's go step by step for each part of the question:

(a) The initial speed of the ball is 0 m/s since it starts from rest. The acceleration of the ball down the inclined plane is given as 0.520 m/s^2. The distance traveled down the inclined plane is 9.65 m. We can use the equation:

v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity
a = acceleration
s = distance

Plugging in the given values, we have:

v^2 = 0^2 + 2 * 0.520 * 9.65

Solving for v, we get:

v^2 = 9.9848

Taking the square root, we find:

v = √9.9848 ≈ 3.16 m/s

Therefore, the speed of the ball at the bottom of the first plane is approximately 3.16 m/s.

(b) To find the time it takes to roll down the first plane, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given that the initial velocity is 0 m/s and the final velocity is approximately 3.16 m/s, and the acceleration is 0.520 m/s^2, we can rearrange the equation to solve for t:

t = (v - u) / a
t = (3.16 - 0) / 0.520

Solving for t, we find:

t ≈ 6.08 seconds

Therefore, it takes approximately 6.08 seconds to roll down the first plane.

(c) On the second plane, the ball comes to rest, which means its final velocity is 0 m/s. The distance traveled on the second plane is given as 14.40 m. We can use the same equation as before to find the acceleration:

v^2 = u^2 + 2as
0^2 = u^2 + 2 * a * 14.40

Simplifying the equation, we find:

u^2 = -2 * a * 14.40

Since the initial velocity, u, is positive, the term on the right side should be negative. Solving for a, we have:

a = -u^2 / (2 * s)
a = -0^2 / (2 * 14.40)
a = 0 m/s^2

Therefore, the acceleration along the second plane is 0 m/s^2.

(d) To find the ball's speed 7.85 m along the second plane, we can use the equation:

v^2 = u^2 + 2as

Given that the acceleration is 0 m/s^2, the initial velocity is 0 m/s (since the ball is at rest), and the distance is 7.85 m, we can solve for v:

v^2 = 0^2 + 2 * 0 * 7.85
v^2 = 0

Therefore, the ball's speed 7.85 m along the second plane is 0 m/s.