need help: During an emergency stop, the skid marks are....

[Guest]

During an emergency stop, a motorist makes skid marks 14.8 m long, all four wheels being locked, on a surface where the coefficient of friction is 0.7. He claims that prior to applying the brakes he was not exceeding the speed limit of 50 km/h. Does the evidence support the claim?

 

[stapel]

We can help you with the math, but you'll need to provide the other information. Specifically, what formulas and/or algorithms have you been given for making the necessary computations? How far have you gotten in applying them?

Please reply with specifics. Thank you.

Eliz.

 

[skeeter]

\(\displaystyle v_f^2 = v_0^2 - 2a(\Delta x)\)

since the magnitude of acceleration is \(\displaystyle a = \mu_k g\) and \(\displaystyle v_f = 0\),

\(\displaystyle v_0 = \sqrt{2(\mu_k g)(\Delta x)}\)

using \(\displaystyle g = 9.8 m/s^2\)

\(\displaystyle v_0 = \sqrt{2(6.86)(14.8)} = 14.25 m/s\)

I'll leave the conversion of m/s to km/hr to you.