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Find the time and distance required to deceleratefrom 100 kph to a standstill at a deceleration of 10 m/s2Also find the effort required to pull thelever and comment whether it is within ergonomic limits for a normal human being. Assume that the rider applied the front brakes only. Motorcycle Weight (with rider): 300 kg Tyre Size: 120/70 R17 Number of Discs: 1 Coefficient of Friction (Brake Pad/Brake Disc): 0.4 Type of Caliper: Fixed Type - 4 piston Caliper Piston diameter: 30 mm Master Cylinder Diameter: 16 mm Effective Diameter of Disc: 300 mm For Brake Lever: Piston to Fulcrum: 30 mm Angle between links at the fulcrum: 90 degrees. Fulcrum to Force Application Point: 90 mm

Find the time and distance required to deceleratefrom 100 kph to a standstill at a deceleration of 10 m/s2Also find the effort required to pull thelever and comment whether it is within ergonomic limits for a normal human being. Assume that 
the rider applied the front brakes only. 
 
Motorcycle Weight (with rider): 300 kg 
Tyre Size: 120/70 R17 
Number of Discs: 1 
Coefficient of Friction (Brake Pad/Brake Disc): 0.4 
Type of Caliper: Fixed Type - 4 piston 
Caliper Piston diameter: 30 mm 
Master Cylinder Diameter: 16 mm 
Effective Diameter of Disc: 300 mm 
For Brake Lever: 
Piston to Fulcrum: 30 mm 
Angle between links at the fulcrum: 90 degrees. 
Fulcrum to Force Application Point: 90 mm

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
Dear student
 
100 kph/ 3.6 ≈ 27.78 m/s 
27.78 m/s ÷ 10 m/s² ≈ 2.78 seconds 

u² - v² = 2ax and x = u²-v²/2a (SUVAT) 
(27.78m/s)²/20 m/s ≈ 38.58 meters 

Ke = ½mv² = 150 kg × (27.78 m/s)² ≈ 115.74 kJ 

120/70 R17 appears to have a circumference of 1.88 m 
and makes about 20.5 rotations in 38.58 meters. 
Using an improbable brake rotor circumference of .92m, 
the brake pad sweeps about 19.3 meters 

115.74 kJ/19.3m ≈ 6 kN 
Hydraulic advantage: 4×30²/16² ≈ 10.5 
Mechanical advantage is at least 3, but not calculable without 
a diagram of the configuration. 
Maximum manual effort 6kN/30 ≈ 200 Newtons 

I have your comment. 
200 Newtons is the weight of 20kg. 
The motorcycle's brake design is probably reasonable 
for a defensive driver braking at ¼ g but it is not suitable 
for braking at 1 g (10 m/s²). 
 
 
Regards
Arun (askIITians forum expert)

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