Since the toboggan is sliding through the horizontal surface of the frozen pond, its acceleration is given by:
[tex]a=\mu g[/tex]The distance that an accelerating object travels while decelerating from a speed v to a stop with an acceleration a is:
[tex]d=\frac{v^2}{2a}[/tex]Replace a=μg as well as v=1.6m/s, g=9.8m/s^2 and μ=0.18:
[tex]d=\frac{(1.6\frac{m}{s})^2}{2(0.18)(9.8\frac{m}{s})}=0.7256...m\approx0.73m[/tex]Therefore, to the nearest hundredth, the distance traveled by the toboggan is 0.73m.
