marble rolling on a smooth show has an initial velocity of 0.4 M per second. if the floor offers the retardation of 0.02 metre per second square ,calculate the time it will take to come to rest
Answers
Answered by
3
use v = u+at
given V= 0 (coming to rest), u = 4m/s and a = -0.02
t = 4/.02 = 200 sec
given V= 0 (coming to rest), u = 4m/s and a = -0.02
t = 4/.02 = 200 sec
jindalnidhi78p9ulkv:
Ur answer is a bit wrong the initial velocity is 0.4 m
thn divide by .4/.02
Pls answer it againnn
you cant divide 0.4/0.2?? ..the ans is 20 sec...figure it out how!
Ok thanks.....
Yes its right
kid...dont just copy paste things...try to get the answer yourself.
Answered by
5
HELLO ,
__________________
we know that
v = u + at
the time taken will be given as
t = (v - u) / a
given
u = 0.4 m/s
v = 0 (comes to rest)
a = -0.02 m/s2 (retardation)
so,
t = (0 - 0.4) / 0.02
thus,
t = 20s
_________________
Hope it will help you :))
Glad to help u dear ☺☺
# Nikky
__________________
we know that
v = u + at
the time taken will be given as
t = (v - u) / a
given
u = 0.4 m/s
v = 0 (comes to rest)
a = -0.02 m/s2 (retardation)
so,
t = (0 - 0.4) / 0.02
thus,
t = 20s
_________________
Hope it will help you :))
Glad to help u dear ☺☺
# Nikky
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