## College Physics (4th Edition)

We can find an expression for the speed: $a_r = \frac{v^2}{r}$ $v = \sqrt{a_r~r}$ We can find an expression for the time: $T = \frac{d}{v}$ $T = \frac{\frac{1}{2}(2\pi~r)}{\sqrt{a_r~r}}$ $T = \frac{\pi r}{\sqrt{a_r~r}}$ $T = \pi~\sqrt{\frac{r}{a_r}}$ To minimize the time, we should minimize the radius $r$. The minimum radius is $5.0~m$. We can find the time $T$: $T = \pi~\sqrt{\frac{r}{a_r}}$ $T = \pi~\sqrt{\frac{5.0~m}{3.0~m/s^2}}$ $T = 4.06~s$ The minimum possible time for this U-turn is 4.06 seconds.