Question

a marble rolling on a smooth floor has an initial velocity of 0.4 m/s. if the floor offers a retardations of 0.02m/s^2 ,calculate the time it will take to come to rest

Answer 1

Given conditions ⇒

Retardation = 0.02 m/s².
∴ Acceleration(a) = -0.02 m/s²

Initial velocity (u) = 0.4 m/s.
Final Velocity (v) = 0 m/s.

Using the Formula,

 Acceleration = Change in the Velocity/Time
 ∴ a = (v - u)/t
∴ t = (v - u)/a
⇒ t = (0 - 4)/-0.02
⇒ t = -4/-0.02
∴ t = 200 s.


Hence, the time taken by it to come to the rest is 200 seconds.


Hope it helps.