The braking distance d, in meters, of a vehicle traveling at a velocity v, in meters per second, is given by theformula d =2ug*is the coefficient of friction and g is the acceleration due to gravity. On Earth,wheregx9.8 m/s. Complete parts (a) and (b) below.(a) Determine the braking distance of a car traveling at 55 km/hr on dry pavement (u = 0.80). (Be sure to convertthe speed to meters per second.)The braking distance is about meters.(Simplify your answer. Round to the nearest integer as needed.)(b) Determine the braking distance of a car traveling at 63 mi/hr on wet pavement (u = 0.45). Give the finalanswer in feet.The braking distance is aboutfeet.(Round to the nearest ten as needed.)

Name: The braking distance d, in meters, of a vehicle traveling at a veloci
Uploaded: 2024-03-20T06:46:42.000Z
Channel: Bartleby
Description: The braking distance d, in meters, of a vehicle traveling at a velocity v, in meters per second, is given by theformula d =2ug*is the coefficient of friction and g is the acceleration due to gravity. On Earth,wheregx9.8 m/s. Complete parts (a) and (

The given formula for braking distance is d=v22μg, where d is the distance, v is the velocity, μ is the coefficient of friction and g is the acceleration due to gravity. (a) The velocity of the car is 55 km/hr and the coefficient of friction is 0.80. Convert the speed into m/sec as follows. 55km/hr=55×100060×60=55×518=27518m/sec Evaluate the braking distance as follows. d=v22μg=27518220.809.8=7562518215.68=756255080.32=14.885≈15 meters Therefore, the braking distance of the car is about 15 meters.(b) The velocity of the car is 63 mi/hr and the coefficient of friction is 0.45. Convert the speed into ft/sec as follows. 63mi/hr=63×528060×60 1 mi=5280 ft=92.4 ft/sec Convert g into f/sec as follows. g=9.8 m/s2=9.8×3.28084 1 meter=3.28084 ft=32.1522 ft/s2 Evaluate the braking distance as follows. d=v22μg=92.4220.4532.1522 1 meter=3.28084 ft=295.04668≈295 ft Therefore, the braking distance of the car is about 113.9 ft.