A vehicle hits a bridge abutment at a speed estimated by investigations as 20kmph. Skid marks of 30m on pavement
(f=0.35) followed by skid marks of 60m. on the gravel shoulder
approaching the abutment (f=0.50).What was the initial speed of vechile

54.5 kmph

Explanation:

From work-kinetic energy principles, work done by friction on both pavement and gravel shoulder = kinetic energy change of vehicle

ΔK = W = -(f₁d₁ + f₂d₂) where f₁ = frictional force due to pavement = μ₁mg where μ₁ = coefficient of friction of pavement = 0.35, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₁ = distance moved by vehicle across pavement = 30 m and

f₂ = frictional force due to gravel shoulder = μ₂mg where μ₂ = coefficient of friction of pavement = 0.50, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₂ = distance moved by vehicle across gravel shoulder = 60 m

ΔK = 1/2m(v₁² - v₀²) where v₀ = initial velocity of vehicle, v₁ = final velocity of vehicle = 20 kmph = 20 × 1000/3600 = 5.56 m/s and m = mass of vehicle

So,

ΔK = -(f₁d₁ + f₂d₂)

1/2m(v₁² - v₀²) = -(μ₁mgd₁ + μ₂mgd₂)

1/2(v₁² - v₀²) = -(μ₁gd₁ + μ₂gd₂)

v₁² - v₀² = -2g(μ₁d₁ + μ₂d₂)

v₀² = v₁² + 2g(μ₁d₁ + μ₂d₂)

v₀ = √[v₁² + 2g(μ₁d₁ + μ₂d₂)]

substituting the values of the variables into the equation, we have

v₀ = √[(5.56 m/s)² + 2 × 9.8 m/s²(0.35 × 30 m + 0.5 × 60 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(10.5 m + 30 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(40.5 m]

v₀ = √[30.91 (m/s)² + 198.45 (m/s)²]

v₀ = √[229.36 (m/s)²

v₀ = 15.14 m/s

v₀ = 15.14 × 3600/1000

v₀ = 54.5 kmph

So, the initial speed of the vehicle is 54.5 kmph

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2