Some cars have an FCW, which stands for
1 answer:
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Answer:
0.981
Explanation:
velocity of cars ( v1 , v2 ) = 55 mi/h
coefficient of adhesion ( u ) = 0.75
Reaction time of driver of car 1 = 2.3 -s
Reaction time of driver of car 2 = 1.9 -s
breaking efficiency of car 2 ( n2 ) = 0.80
<u>Determine the braking efficiency of car 1</u>
First determine the distance travelled during reaction time ( dr )
dr = v * tr ------- ( 1 )
tr ( reaction time )
v = velocity
note : 1 mile = 1609 m , I hour = 60 * 60 secs
<em>back to equation 1</em>
for car 1
dr1 =( 55 * 2.3 * 1609 ) / ( 60 * 60 )
= 56.53 m
for car 2
dr2 = ( 55 * 1.9 * 1609 ) / ( 60 * 60 )
= 46.70 m
<em>next we calculate the stopping distances ( d ) using the relation below</em>
ds = d + dr
d = distance travelled during break
dr = distance travelled during reaction time
where : d =
<em>for car 1 </em>
d1 =
∴ d1 =
<em>for car 2 </em>
d2 =
∴ d2 = 51.38
since the stopping distance for both cars are the same
d1 + dr1 = d2 + dr2
( 41.10 /n1 ) + 56.53 = 51.38 + 46.70
solve for n1
hence n1 = 0.981 ( braking efficiency of car 1 )