Solution 8
Optimization Strategy (Bang-Bang Control):
To minimize transit time, we must maximize average velocity.
To minimize transit time, we must maximize average velocity.
- Curved Section BC: The velocity is strictly limited by the total acceleration constraint. Since centripetal acceleration $a_n = v^2/r$, and total $a \ge a_n$, the maximum speed on the curve is $v_{max,curve} = \sqrt{ar}$. To pass fastest, we travel this whole section at this constant max speed.
- Straight Section AB: We start from rest. We want to spend as much time as possible at high speeds. The best strategy is to accelerate at maximum rate $a$ to a peak velocity $v_p$, and then immediately decelerate at maximum rate $a$ to meet the curve’s entry speed requirement ($v_B = \sqrt{ar}$).
1. Calculating the Motion on Straight AB
Let the peak velocity reached be $v_p$.
- Phase 1 (Accel): $0 \to v_p$. Distance $d_1 = \frac{v_p^2}{2a}$.
- Phase 2 (Decel): $v_p \to v_B$. Distance $d_2 = \frac{v_p^2 – v_B^2}{2a}$.
Total length $l = d_1 + d_2$:
$$ l = \frac{v_p^2}{2a} + \frac{v_p^2 – ar}{2a} \quad (\text{Using } v_B = \sqrt{ar}) $$ $$ 2al = 2v_p^2 – ar \implies v_p = \sqrt{a\left(l + \frac{r}{2}\right)} $$Now, calculate time for AB ($t_{AB}$):
$$ t_{AB} = t_{acc} + t_{dec} = \frac{v_p}{a} + \frac{v_p – v_B}{a} = \frac{2v_p – v_B}{a} $$ $$ t_{AB} = \frac{2\sqrt{a(l+r/2)} – \sqrt{ar}}{a} = 2\sqrt{\frac{l+r/2}{a}} – \sqrt{\frac{r}{a}} $$Rearranging the term under the radical to match the answer key format ($\frac{2l+r}{2a}$):
$$ t_{AB} = 2\sqrt{\frac{2l+r}{2a}} – \sqrt{\frac{r}{a}} $$
2. Calculating the Motion on Curve BC
The path length is a semicircle: $s = \pi r$. Speed is constant $v_B = \sqrt{ar}$.
$$ t_{BC} = \frac{\pi r}{\sqrt{ar}} = \pi \sqrt{\frac{r}{a}} $$
3. Total Minimum Time
$$ T = t_{AB} + t_{BC} $$
$$ T = \left( 2\sqrt{\frac{2l+r}{2a}} – \sqrt{\frac{r}{a}} \right) + \pi \sqrt{\frac{r}{a}} $$
Grouping the $\sqrt{r/a}$ terms:
$$ T = (\pi – 1)\sqrt{\frac{r}{a}} + 2\sqrt{\frac{2l+r}{2a}} $$