220 A driver is traveling at 90 mih down a 3 grade on good w

2.20 A driver is traveling at 90 mi/h down a 3% grade on good, wet pavement. An accident investigation tea noted that braking skid marks started 410 ft before a parked car was hit at an estimated 45 mi/h. Ignoring air resistance, and using theoretical stopping distance, what was the braking efficiency of the car?

Solution

Solution:-

Given

V₁ = 90 mph =132 feet/second

V₂ = 45 mph = 66 feet/second

Stopping distance = 410 feet

Grade =3 %

For good wet pavement = f = 0.5

We know that

Stopping sight distance = (V₁ –V₂)²/(2g(f’-0.03))

                        410 = 66²/(2*32*(f’ - 0.03))

                        f’ = 0.196

Braking efficiency = f’ *100/f

Braking efficiency = 0.196*100/0.5

                                = 39.2 %   Answer