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What is the distance required to stop (i.e., braking distance) of a passenger car when brakes are applied on a 2% downgrade if that vehicle is traveling at 40 mph

User SRaj
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1 Answer

3 votes

Answer:

The value is
d = 161 \ ft

Step-by-step explanation:

From the question we are told that

The gradient is G = 2% = 0.02

The initial speed of the vehicle is u = 40 mph

Generally the breaking speed is mathematically represented as


d = (u^2)/(30 [(a)/(g) - G ])

Here
(a)/(g) is equal to longitudinal friction coefficient which has a value of


(a)/(g) = 0.35

So


d = (40^2)/(30 [0.35 - 0.02 ])

=>
d = 161 \ ft

User Cguedel
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