the break applied to a car it produces an acceleration of - 10 m/s square . if the car takes 5 second to stop after applying break calculate the distance covered by the car before coming to rest.
the break applied to a car it produces an acceleration of - 10 m/s square . if the car takes 5 second to stop after applying break calculate the distance covered by the car before coming to rest.
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using first equation v=u+at
0=u+(-10)×5 then,u=50m/sec
now using third euation
v2=u2+ 2as
(0)2=(50)2×2(-10)×s
then,s= 125m
hope it's ryt answer