Question

A driver is traveling at 90 mi/h down a 3% grade on good, wet pavement. An accident investigation team 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?

Answer 1

58.7%

Explanation:

First, before calculating the braking efficiency of the car; we need to find the coefficient of the rolling friction of the driving by applying the formula;

`f_r_l`
`= 0.01(1+(((V_1+V_2))/(2) )/(147))`

where;

V₁ = 90 mi/h

V₁ =
`90*(5280)/(3600)` ( since 1 mi = 5280ft and 1 hour is equivalent to 3600 seconds)

V₁ =
`(475200)/(3600)`

V₁ = 132

V₂ = 45 mi/h

V₂ =
`45*(5280)/(3600)`

V₂ =
`(237600)/(3600)`

V₂ = 66

Substituting both data into the above equation, we have;

`f_r_l`
`= 0.01(1+(((132+66))/(2) )/(147))`

`f_r_l`
`= 0.01(1+(((198))/(2) )/(147))`

`f_r_l`
`= 0.01(1+(99 )/(147))`

`f_r_l`
`= 0.01((147+99 )/(147))`

`f_r_l`
`= 0.01((246 )/(147))`

`f_r_l`
`= 0.016735`

Now, to calculate the breaking efficiency of the car
`(n_b)`; we apply the formula:

`S = (Y_b(V_1^2-V_2^2))/(2g(n_b\beta+f_(rl)-sin\alpha g) )`

where;

the braking skid (S) = 410 ft

Value for braking mass factor for automobiles = 1.04

Value for Coefficient (β) of the road adhesion for both good and wet pavement = 0.90

Sin ∝ = 3% = 0.03

coefficient of the rolling friction
`f_r_l` = 0.016735

V₁ = 132 & V₂ = 66

Substituting our parameters in the above formula, we have;

`410 = (1.04 (132^2-66^2))/(2*32.2(n_b(0.9)+(0.016735)-0.03) )`

`410 = (1.04 (17424-4356))/(64.4(n_b(0.9)-(0.013265) )`

`410 = (1.04 (13068))/(57.96n_b-0.854266 )`

`410 = (13590.72)/(57.96n_b-0.854266 )`

`13590.72 = 410 (57.96_(n_b)-0.854266)`

`13590.72 = 23763.6_(n_b)-350.24906`

`13590.72+350.24906= 23763.6_(n_b)`

`13940.96906=23763.6_(n_b)`

`n_b=(13940.96906)/(23763.6)`

`n_b=0.58665`

`n_b` ≅ 0.587

`n_b` = 58.7%

∴ The braking efficiency of the car = 58.7%