a car moving along a straight highway with speed of 90kmph to brought to rest with a distance of 200m (a) what is the retardation of the car and (b) how long does it take for the car stop
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Answered by
4
Answer:
1st I convert 90km/h into m/s
=5/18×90
=25m/s
v=0m/s
u=25m/s
According to the equation of motion
v^2-u^2=2as
0-25^2=2×a×200
-625=400a
a=400/625
a= -0.64m/s^2
Since,negative acceleration=retardation
retardation=-0.64m/s^2
Now,a=(v-u)/t
-0.64=(0-25)/t
-0.64= -25/t
t=0.64/25
t=0.0256 sec
Answered by
0
Answer:
a= - 1.5 m/s² and t= 16.6s
u=90km/s
=90 × 5/18 m/s
=25m/s
v=0
s=200m
v²-u²=2as
0²-25²=2a(200)
-25×25=400a
(-25×25)/400=a
a=-25/16
a= - 1.5m/s²
a=(v-u)/t
-1.5=(0-25)/t
t= -25/-1.5
t=25/1.5
t=16.6s
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