You see a body lying across the road and need to stop. If your vehicle was initially traveling at 100 km/h and skids to a stop on a 2.5% upgrade, taking 75 m to do so, what was the coefficient of friction on this surface?
db=(100∗(10003600))2−(0)22∗(9.8)∗(f+0.025)=75m{\displaystyle d_{b}={\frac {\left({100*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*\left({f+0.025}\right)}}=75m\,\!}
(f+0.025)=(27.78)22∗(9.8)∗75{\displaystyle \left({f+0.025}\right)={\frac {\left({27.78}\right)^{2}}{2*\left({9.8}\right)*75}}\,\!}
f=0.50{\displaystyle f=0.50\,\!}