Question

A 1500 kg automobile travels at a speed of 85 km/h along a straight concrete highway. Faced with an emergency situation, the driver jams on the brakes, and the car skids to a stop.

(a) What will be the car's stopping distance for dry pavement (µ = 0.85)?

(b) What will be the car's stopping distance for wet pavement (µ = 0.60)?

Answer 1

Initial velocity (u):            = 85km/h                             Final velocity (v)= 0

                                                = (85000/3600) m/s

                                                = 23.61 m/s

Now,

                v²=u²-2fs             [s= stopping distance] [f=retardation]

                0= u²-2fs

                u²=2fs

               

From Newton’s 2^nd law we get,

                                                                F=mf     [F=force; m=mass; f=acceleration]

From law of friction,

                                                F=µmg

                                Now      µmg=mf

                                                f=µg

hence

For dry pavement µ=0.85                             for wet pavement µ=0.60

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s= 33.425m s= 47.35m