Coefficient 81094

A car moves along a horizontal road at a speed of 15 m/s. After turning off the engine, the car traveled a distance of 225 m. What was the coefficient of friction for this motion?

Correct answer:

x =  0.05

Step-by-step explanation:

$$\begin{gathered} v=15\text{m/s} \\ s=225\text{m} \\ g=10{\text{m/s}}^2 \\ v=a\cdot t \\ s=\frac{1}{2}\cdot a\cdot t^2 \\ s=\frac{1}{2}\cdot v\cdot t \\ t=2\cdot s/v=2\cdot 225/15=30\text{s} \\ a=2\cdot s/t^2=2\cdot 225/30^2=\frac{1}{2}=0.5{\text{m/s}}^2 \\ F=xmg=m\cdot a \\ x\cdot g=a \\ x=a/g=0.5/10=\frac{1}{20}=0.05 \end{gathered}$$

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