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Laws Of Motion

Question
CBSEENPH11016543

Driver of a car is driving at the speed 9-8 m/s. He saw a child standing in the middle of road and applied the brakes when the car was at a distance 4.9m away from the child. Show that the minimum retarding force of the brakes on the car to just save the child must be half the weight of the car.

Solution
The car must stop by travelling the maximum distance of 4.9m, in order to save the child.
Let M be mass of the car.
Here,      
Initial velocity of the car, u = 9.8 m/s
Final velocity, v = 0
MAximum distance travelled,  straight S subscript max space equals space 4.9 space straight m
Therefore minimum retardation of the car is,
             straight r subscript min space equals space minus fraction numerator straight v squared minus straight u squared over denominator 2 straight S subscript max end fraction
                   equals negative fraction numerator 0 squared minus left parenthesis 9.8 right parenthesis squared over denominator 2 cross times 9.8 end fraction space equals space 4.9 space straight m divided by straight s squared
The minimum retarding force on the car is,
                      straight M cross times straight r subscript min space equals space 4.9 straight M
Now,
      fraction numerator Retarding space force over denominator Weight end fraction space equals space fraction numerator 4.9 straight M over denominator 9.8 straight M end fraction equals 1 half 
Thus,
       Retarding force = 1 half Weight