the breaks applied to a car produce an acceleration of 16m/s square in the opposite direction to the motion. if the car takes 2 second to stop after the application of break calculate the distance it travel during this time.
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a=-16 m/s sq. v=0m/s t=2seconds.
v=u+at
0=u+-16×2
0=u+(-32)
u=32
by using third equation of motion
v2-u2=2as
-1024=-32s
s=32 m
distance =32m
Hope it is correct and it may help you.
v=u+at
0=u+-16×2
0=u+(-32)
u=32
by using third equation of motion
v2-u2=2as
-1024=-32s
s=32 m
distance =32m
Hope it is correct and it may help you.
vishwajeet15:
what is your name
Mansi
OK thank you and my name is Vishwajeet
and in which standard are you
class 9th
ok
where do you live
asking from you distance in this question
sorry
i think that's the distance
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