Question

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

Answer 1

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`