A 0.525 ball starts from rest and rolls down a hill with uniform acceleration, traveling 154 during the second 10.0 of its motion.

How far did it roll during the first 4.00 of motion?

To find the distance the ball rolled during the first 4.00 seconds of motion, we can use the equations of motion for uniformly accelerated motion.

The first equation we will use is:
s = ut + (1/2)at^2

Where:
s = distance traveled
u = initial velocity (0, as the ball starts from rest)
a = acceleration
t = time

First, let's find the acceleration of the ball.

We know that the ball traveled a distance of 154 meters during the second 10.0 seconds of motion.
Using the equation above, we can rewrite it as:
154 = 0 + (1/2)a(10.0^2)

Simplifying this equation gives us:
154 = 50a

Solving for the acceleration, we get:
a = 3.08 m/s^2

Now that we have the acceleration, we can find the distance traveled during the first 4.00 seconds.

Plug in the values into the equation of motion:
s = ut + (1/2)at^2
s = 0 + (1/2)(3.08)(4.00^2)
s = 24.64 meters

Therefore, the ball rolled a distance of 24.64 meters during the first 4.00 seconds of motion.