A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity and deceleration are
and​

1. Initial velocity = 24 m/s

2. Deceleration = –4 m/s²

Explanation:

From the question given above, the following data were obtained:

Distance travelled (s) = 72 m

Time (t) = 6 s

Final velocity (v) = 0 m/s

1. Determination of the initial velocity.

Distance travelled (s) = 72 m

Time (t) = 6 s

Final velocity (v) = 0 m/s

Initial velocity (u) =?

s = (u + v)t / 2

72 = (u + 0) × 6 / 2

72 = u × 3

Divide both side by 3

u = 72 / 3

u = 24 m/s

2. Determination of the deceleration.

Time (t) = 6 s

Final velocity (v) = 0 m/s

Initial velocity (u) = 24 m/s

Deceleration (a) =?

v = u + at

0 = 24 + (a × 6)

0 = 24 + 6a

Collect like terms

0 – 24 = 6a

–24 = 6a

Divide both side by 6

a = –24 / 6

a = –4 m/s²

option number c is the right answer.