Question

Braking Distance Example A student drove his 1983 Dodge into a dorm adjacent to the student parking lot. Police found 30' long skid marks leading to the point of impact. The damage assessment found that the speed at the time of impact was 10 mph. The parking lot was level, and the pavement was wet (f = 0.6). The speed limit in the parking lot is 15 mph. How fast was the student traveling at the time that he began to skid? Was he speeding? What would have the impact speed have been had the parking lot been on a 6% uphill grade? How about a 3% downhill grade? What would the impact speed have been given level, icy pavement (f = 0.1) Would there have been a different result, coefficient of friction and vehicle braking capabilities being equal, if he would have been driving a fully loaded newspaper truck?

Answer 1

Following conversion factors are used to solve this problem

1 mph = 0.45 m/s ;   1 ft = 0.0305 m

To get the initial speed, we use the eqn. ,   v² = u² - ( 2 a S )

where v is final speed = 10 mph = 4.5 m/s

u is initial speed to be calculated

a is retardation = $\mu$ $\times$ g , where $\mu$ =0.6 is friction coefficient, g is acceleration due to gravity

S is the distance traveled that equals 30 ' or 30$\times$0.305 = 9.15 m

hence initial speed is calculated as, u= [ 4.5 $\times$ 4.5 + (2 $\times$ 0.6 $\times$ 9.8 $\times$ 9.15 ) ]^1/2$\approx$ 11.31 m/s

initial speed in mph = 11.31/.45 = 25.13 mph

Hence the student is speeding

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g sin u-gcose

when the car is moving 6% upside hill, retardation as shown in figure, gsin(0.06) + $\mu$ gcos(0.06)

retardation = 9.8 $\times$ 0.06 + 0.6 $\times$ 9.8 = 6.47 m/s²

hence initial speed is calculated as, u= [ 4.5 $\times$ 4.5 + (2 $\times$ 6.47 $\times$ 9.15 ) ]^1/2$\approx$ 11.78 m/s = 26.1ph

-ug.cose sine

when the car is moving 6% downside hill, retardation as shown in figure, $\mu$ gcos(0.06)-gsin(0.06)

retardation = 0.6 $\times$ 9.8 - 9.8 $\times$ 0.06 = 5.3 m/s²

hence initial speed is calculated as, u= [ 4.5 $\times$ 4.5 + (2 $\times$ 5.3 $\times$ 9.15 ) ]^1/2$\approx$ 10.83 m/s = 24.06 mph

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impact speed on icy pavement

retardation = $\mu$ $\times$ g = 0.1 $\times$ 9.8 m/s²

v² = u² - 2 a S

v = [ 11.31 $\times$ 11.31 - ( 2 $\times$ 0.1 $\times$ 9.8 $\times$ 9.15 )]^1/2$\approx$ 10.5 m/s = 23.3 mph

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Retardation due to friction is independent of mass, hence above results are independent of mass of vehicle