Correct Answer - Option 2 : 72.83%
Concept:
Braking distance:
(i) It is the distance travelled by vehicle after the application of brakes.
(ii) It is obtained by equating the work done in stopping the vehicle and kinetic energy stored in the vehicle.
Let us consider, F is the maximum frictional force developed and L is the braking distance, then the work done against friction in stopping the vehicle will be F × L.
Where, frictional force, F = f × W
Where W is the total weight of vehicle and f is the coefficient of friction
So, work done in stopping the vehicle = f × WL
While the kinetic energy stored in vehicle = \(\frac{1}{2}m{v^2} = \frac{1}{2}\frac{{W{v^2}}}{g}\) (∵ W = mg)
Now by using the law of conservation of energy,
Work done in stopping the vehicle = Kinetic energy stored in the vehicle
\(fWL = \frac{1}{2}\frac{{W{v^2}}}{g}\)
\(L = \frac{{{v^2}}}{{2gf}}\)
Braking distance, \(L = \frac{{{v^2}}}{{2gf}}\)
If the braking efficiency is η,
Braking distance, \(L = \frac{{{v^2}}}{{2gf\eta }}\)
Calculation:
Given,
v = 50 kmph = 50/3.6 = 13.89 m/s (∵ 1 kmph = 1/3.6 m/s)
L = 18 m
f = 0.75
∵ we know that, braking distance, \(L = \frac{{{v^2}}}{{2gf\eta }}\)
\(18 = \frac{{{{13.89}^2}}}{{2\; \times \; 9.81\; \times \;0.75\; \times \;\eta }}\)
η = 0.7284 = 72.84 %
