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You have been called as an expert witness for a trial in which a driver has been charged with speeding but is claiming innocence. He claims to have slammed on his brakes to avoid rear-ending another car, but tapped the back of the other car just as he came to rest. You have been hired by the prosecution to prove that the driver was indeed speeding. You have received data as follows from the police: skid marks left by the driver are 60.0 m long and the roadway is level. Tires matching those on the car of the driver have been dragged over the same roadway to determine that the coefficient of kinetic friction between the tires and the roadway is 0.620 at all points along the skid mark. The speed limit on the road is 35 mi/h. Construct an argument to be used in court to show that the driver was indeed speeding. (Enter the speed of the driver, in mi/h, just before braking.)

Sagot :

Answer:

v₀ = 60.38 mi / h

Explanation:

For this exercise let's start by using Newton's second law

Y axis

        N-W = 0

        N = W

X axis

         fr = m a

the expression for the friction force is

         fr = μ N

we substitute

        μ mg = m a

        μ g = a

calculate us

         a = 0.620   9.8

         a = 6.076 m / s²

now we can use the kinematics relations

          v² = v₀² - 2 a x

suppose v = 0

          v₀ = [tex]\sqrt{2ax}[/tex]

let's calculate

         v₀ = [tex]\sqrt{2 \ 6.075 \ 60.0}[/tex]

         v₀ = 27.00 m / s

let's slow down to the english system

          v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)

          v₀ = 60.38 mi / h

With this stopping distance, the starting speed should have been 60.38, which is much higher than the maximum speed allowed.

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