Answer:
[tex]\Delta s = 189.166 m[/tex]
Explanation:
The physical for the system is based on Work-Energy Theorem and Principle of Energy Conservation. The system decelerates because of friction before coming to rest:
[tex]K_{1} = W_{loss,1 \longrightarrow 2}[/tex]
[tex]\frac{1}{2} \cdot m \cdot v^{2} = \mu (t) \cdot m \cdot g \cdot \Delta s[/tex]
The distance before stopping is isolated from expression presented above:
[tex]\Delta s = \frac{ v^{2}}{2 \cdot \mu(t)\cdot g}[/tex]
Where [tex]\mu (t) = 0.1\cdot t[/tex] and [tex]g = 9,807 \frac{m}{s^{2}}[/tex].
By replacing all variables, the needed distance is finally found:
[tex]\Delta s = \frac{(43.5 \frac{m}{s})^{2}}{2 \cdot [0.1\cdot (5.1 sec)]\cdot (9.807 \frac{m}{s^{2}} )}[/tex]
[tex]\Delta s = 189.166 m[/tex]
