A car Moving On A straight highway with the speed of 126 km per hour is brought to stop with her a distance of 200 m what is the retardation of the car and how long does it take for the car to stop
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applying third eq of motion
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Speed =126*1000/3600=35m/s=u(Initial velocity)
Final velocity v=0
S=200m
a=?
t=?
We know that,
v2=u2+2aS
So,
0= (35)2+2*a*200
1225= - 400a
a= - 3.0625= - 3m/s2
Also,
v=u+at
0=35+(-3)t
t=35/3=11.666s
t=11.7 s or 12 s
Final velocity v=0
S=200m
a=?
t=?
We know that,
v2=u2+2aS
So,
0= (35)2+2*a*200
1225= - 400a
a= - 3.0625= - 3m/s2
Also,
v=u+at
0=35+(-3)t
t=35/3=11.666s
t=11.7 s or 12 s
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