Question

A marble Rolling on a smooth floor has an initial velocity of 0.4m/s If the floor offers a retardation. of 0.02m/s^2, calculate the time in which it will come to rest..?

Answer 1

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Given,

Initial velocity =u = 0.4m/s

Final velocity v= 0(As the marble comes to rest)

Retardation= -a = -0,02m/s

Acceleration = a =-0.02m/s^2

Time=t?

V= u +at

0=0.4-0.02×t

0.02-r=0.4

t= 0.4/0.02

=20s

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Answer 2

Hye !!

Refer to the attachment !!