Question

Two cars are traveling on level terrain at 55 mi/h on a road with a coefficient of adhesion of 0.75. The driver of car 1 has a 2.3-s perception/reaction time and the driver of car 2 has a 1.9-s perception/reaction time. Both cars are traveling side by side and the drivers are able to stop their respective cars in the same distance after first seeing a roadway obstacle (perception and reaction plus vehicle stopping distance). If the braking efficiency of car 2 is 0.80, determine the braking efficiency of car 1. (Assume minimum theoretical stopping distance and ignore aerodynamic resistance.)

Answer 1

0.981

Step-by-step 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

Determine the braking efficiency of car 1

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

back to equation 1

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

next we calculate the stopping distances ( d ) using the relation below

ds = d + dr

d = distance travelled during break

dr = distance travelled during reaction time

where : d =
`(v^2intial)/(2ugn)`

for car 1

d1 =
`((55)^2)/(2*0.75*9.81* n1) * ((1609)/(3600) )^2`

∴ d1 =
`(41.10)/(n1)`

for car 2

d2 =
`((55)^2)/(2*0.75*9.8*0.8) * ((1609)/(3600) )^2`

∴ 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 )