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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given, a= -10m/sec square ,t=5sec &v=0
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
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
vishu543:
coz when u had told me that yur calculation is wrong and then i had calculate two times and got the same answer
no worry dear
its my pleasure to help u
Yess after telling ur calculation wrong I asked my friend this question he answered same as urgent answer
Ur answer
ok
not to worry
I have one more question
yaa ask
If two bodies are in the circular path of radii1:2 take some time to complete their circles what is the ratio of their angular speeds
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