The ball, rolling off the inclined plane, covered a distance of 10 cm in the first second.

The ball, rolling off the inclined plane, covered a distance of 10 cm in the first second. What is the distance traveled by the ball in the fourth second?

t1 = 1 s.

V0 = 0 m / s.

S1 = 10 cm = 0.1 m.

t3 = 3 s.

t4 = 4 s.

S -?

On an inclined plane, the ball moves uniformly, we express its acceleration a by the formula: a = 2 * S1 / t1 ^ 2.

a = 2 * 0.1 m / (1 s) ^ 2 = 0.2 m / s2.

The path of the ball in the fourth second S is expressed by the difference: S = S4 – S3, where S4 is the path of the ball in four seconds t4 = 4 s, S3 is the path of the ball in three seconds t3 = 3 s.

S4 = a * t4 ^ 2/2.

S4 = 0.2 m / s2 * (4 s) ^ 2/2 = 1.6 m.

S3 = a * t3 ^ 2/2.

S3 = 0.2 m / s2 * (3 s) ^ 2/2 = 0.9 m.

S = 1.6 m – 0.9 m = 0.7 m.

Answer: on an inclined plane, the ball will travel a path S = 0.7 m.