Animated Solution for Physics - Kinematics: Comprehension Passage

Imagine a very long highway, and on it, 50 identical cars are parked in a perfectly straight line. There is exactly a $10\text{ m}$ gap between each car. This isn't just a traffic jam; it's a beautifully orchestrated physics experiment waiting to happen.

As soon as the order to move is given, the first car starts immediately. But the second car doesn't move right away. It waits patiently until the gap between it and the first car stretches to exactly $35\text{ m}$. This simple rule creates a fascinating ripple effect down the entire convoy. To truly understand this motion, we need to look at the kinematics of the very first car.

The first car starts at $t=0$ with a constant acceleration of $a=2{\text{ m/s}}^2$. Using the second equation of motion, its position as a function of time is given by:

Now, when does the second car start? It waits until the gap is $35\text{ m}$. Since it was initially parked $10\text{ m}$ behind the first car, the first car must travel an additional $25\text{ m}$ to create that $35\text{ m}$ gap. We can set up the equation:

This $5\text{ s}$ delay is the master key to the entire problem! Every single car will follow this exact same behavior with the car in front of it. The third car starts $5\text{ s}$ after the second, the fourth $5\text{ s}$ after the third, and so on. The entire motion profile of the convoy is simply shifted by $5\text{ s}$ for each subsequent car.

As the cars start moving one by one, the line of stationary cars is getting shorter. Think about the boundary between the moving cars and the stopped cars. This boundary shifts backward by one car every $5\text{ s}$.

Since the cars are spaced $10\text{ m}$ apart initially, shifting back by one car means moving back by $10\text{ m}$. Therefore, the speed at which this boundary propagates backward is:

This means the length of the segment consisting of stationary cars is decreasing at a steady rate of $2\text{ m/s}$.

The cars won't accelerate forever. They hit a speed limit of $72\text{ km/h}$. Let's convert this to standard SI units:

When both the first and second cars reach this maximum speed, what will be the gap between them? Because the first car started $5\text{ s}$ earlier, it has been traveling at the maximum speed for $5\text{ s}$ longer relative to the second car. In those $5$ extra seconds, it covers an extra distance:

But remember, there was already a $10\text{ m}$ gap from the very beginning. So, the total separation between adjacent cars at maximum speed becomes:

Finally, the order is given to stop the convoy. The first car applies its brakes immediately. The second car applies its brakes after some time delay, let's call it $\tau$. During this delay, the second car keeps moving forward at $20\text{ m/s}$, rapidly closing the $110\text{ m}$ gap.

Once both cars are braking at the same constant deceleration $a_d$, their braking distances will perfectly cancel each other out. We want the final gap to return to the original $10\text{ m}$. We can write this as:

Solving this gives $\tau =5\text{ s}$. Notice something incredible? The deceleration $a_d$ completely vanished from the equation! The final separation depends only on the reaction delay.

However, we must ensure the cars don't crash during that $5\text{ s}$ delay. The gap must remain positive. The position of the first car while braking is $110-\frac{1}{2}{a}_d{t}^2$. At $t=5\text{ s}$, this gap is:

The problem states the maximum braking capacity is $4{\text{ m/s}}^2$. Since $4$ is well below $8.8$, any deceleration up to $4{\text{ m/s}}^2$ is perfectly safe and will result in exactly a $10\text{ m}$ final separation. Thus, the deceleration can be $2{\text{ m/s}}^2$, $4{\text{ m/s}}^2$, or any value $\le 4{\text{ m/s}}^2$.