Animated Solution for Physics - Kinematics: A boy starts from point A and passes point C of a track ABC shown in the figure. Portion AB of length $l$ is straight and portion BC is a semicircle of radius $r$ ($r<l$). Anywhere on the track, the modulus of the maximum acceleration of the boy is $a$. Find minimum transit time of the boy from A to C.

Imagine you are behind the wheel of a high-performance race car, or perhaps you are programming an autonomous drone. Your goal is simple: get from point A to point C in the absolute minimum time. But physics imposes a strict limit—your tires (or motors) can only provide a maximum total acceleration $a$. This problem is a beautiful exploration of how to balance speed, acceleration, and the geometry of your path to achieve the ultimate lap time.

The track is divided into two distinct segments. First, we have a straightaway $AB$ of length $l$. Here, your path is linear, meaning you don't have to worry about turning. The only acceleration you experience is tangential ($a_t$), which changes your speed.

Second, we have a semicircular curve $BC$ of radius $r$. On this curve, your velocity vector is constantly changing direction, which requires a normal (or centripetal) acceleration $a_n=\frac{v^2}{r}$ pointing towards the center of the curve.

The golden rule of this problem is that the vector sum of these accelerations cannot exceed $a$. Mathematically, this means:

To minimize your overall time, you need to maintain the highest possible speed at all times. Let's look at the curved section $BC$ first. Because you are turning, a portion of your acceleration budget must be dedicated to keeping you on the track. If you go too fast, the required normal acceleration $\frac{v^2}{r}$ will exceed your maximum limit $a$, and you will slide off!

To find the maximum safe speed on the curve, we dedicate all our acceleration to turning, setting $a_n=a$ and $a_t=0$. This gives us:

To traverse the semicircle as quickly as possible, you should enter point B at exactly this speed and maintain it constantly until you reach point C. Since the length of the semicircle is $\pi r$, the time spent on the curve is simply distance divided by speed:

Now, let's tackle the straight path $AB$. You start from rest at A. To cover the distance $l$ quickly, you should slam on the accelerator and speed up at the maximum rate $a$. However, you can't just keep accelerating forever! You must arrive at point B with a speed no greater than the safe entry speed $\sqrt{ar}$.

This means you must accelerate to a peak speed $v_{peak}$, and then hit the brakes, decelerating at the maximum rate $-a$ so that you hit point B at exactly $v_B=\sqrt{ar}$.

Let $s_1$ be the distance you spend accelerating, and $s_2$ be the distance you spend braking. We know that the total distance is $l$:

We can use the classic kinematic equation $v^2-u^2=2as$ for both phases.

During the acceleration phase, starting from rest:

During the deceleration phase, ending at $v_B=\sqrt{ar}$:

Substituting $v_B^2=ar$ and replacing $v_{peak}^2$, we get a fascinating relationship:

We now have a simple system of linear equations. By adding $s_1+s_2=l$ and $s_1-s_2=\frac{r}{2}$, we can solve for the acceleration distance $s_1$:

This allows us to find the absolute maximum speed achieved on the straightaway:

With the peak speed known, calculating the time spent on the straightaway is straightforward. The time to accelerate is $t_1=\frac{v_{peak}}{a}$, and the time to decelerate is $t_2=\frac{v_{peak}-v_B}{a}$. The total time on $AB$ is their sum:

Substituting our expressions for $v_{peak}$ and $v_B$:

Finally, we add the time spent on the straightaway to the time spent on the curve to get the ultimate minimum transit time $T$:

Grouping the common terms yields our elegant final answer:

This result perfectly encapsulates the trade-off between straight-line speed and cornering limits, a principle that governs everything from Formula 1 racing to the trajectory planning of autonomous robots!