Animated Solution for Physics - Kinematics: A fun drive in an amusement park runs between two spots that are $2.0\text{ km}$ apart. For safety reasons, acceleration of the drive is limited to $\pm 4.0{\text{ m/s}}^2$, and the jerk i.e. rate of change in acceleration, is limited to $\pm 1.0{\text{ m/s}}^3$. If the drive can achieve a maximum speed of $144\text{ km/h}$, find the shortest transit time of the drive between the spots.

Imagine you are the lead engineer designing the ultimate high-speed amusement park ride. Your goal is simple: get the passengers from point A to point B ($2.0\text{ km}$ away) as fast as humanly possible. However, you have strict safety constraints. You can't just launch them like a cannonball. The acceleration is capped at $4.0{\text{ m/s}}^2$, and more importantly, the jerk—the rate at which acceleration changes—is limited to $1.0{\text{ m/s}}^3$.

Jerk is what causes whiplash. It's the sudden jolt you feel when a car slams on the brakes. By limiting jerk, we ensure a thrilling but smooth ride. Let's break down the physics of this journey.

To minimize the total transit time, we must reach our maximum allowed speed of $144\text{ km/h}$ (which is $40\text{ m/s}$) as quickly as possible. But because of the jerk limit, our acceleration-time ($a-t$) graph cannot be a simple rectangle. It must be a trapezoid.

First, we ramp up the acceleration. At a jerk of $j=1.0{\text{ m/s}}^3$, it takes exactly $t_1=\frac{a_{\max }}{j}=4\text{ s}$ to reach the maximum acceleration of $4.0{\text{ m/s}}^2$. During this time, the velocity gained is the area under the $a-t$ triangle:

Before we can cruise at a constant $40\text{ m/s}$, we must smoothly bring the acceleration back down to zero. This ramp-down also takes $4\text{ s}$ and adds another $8\text{ m/s}$ to our speed.

So, just from ramping the acceleration up and down, we gain $8+8=16\text{ m/s}$. But we need to reach $40\text{ m/s}$! This means we must insert a phase of constant maximum acceleration in the middle.

We need an additional $\Delta v_2=40-16=24\text{ m/s}$. At a constant acceleration of $4.0{\text{ m/s}}^2$, the time required is:

Our total acceleration phase takes $t_{\text{acc}}=4+6+4=14\text{ s}$.

To find the distance covered during this phase, we integrate the velocity-time function (or find the area under the $v-t$ curve). The math reveals that the distance covered while accelerating to top speed is exactly $s_{\text{acc}}=280\text{ m}$.

Physics loves symmetry. To safely bring the ride to a halt from $40\text{ m/s}$ using the exact same jerk and acceleration limits, the deceleration profile will be a perfect mirror image of the acceleration profile.

Therefore, the deceleration phase will also take $t_{\text{dec}}=14\text{ s}$ and cover a distance of $s_{\text{dec}}=280\text{ m}$.

We have a total distance of $d=2000\text{ m}$. We spend $280\text{ m}$ speeding up and $280\text{ m}$ slowing down. The remaining distance is covered at our top cruising speed:

At a constant speed of $40\text{ m/s}$, the time spent cruising is:

Finally, we add up the times for all three phases to find the shortest possible transit time:

And there you have it—a perfectly optimized, jerk-limited kinematic journey!