KINEMATICS O27

1. Define the “Segment of Stationary Cars”

The convoy consists of 50 identical cars. Initially, all are at rest. Cars start moving one by one from the front (Car 1, then Car 2, etc.).

At any time $t$, the “segment of stationary cars” consists of all cars from index $k$ (the next car to start) to 50 (the last car).
The length of this segment is determined by the positions of the cars currently at rest. Since stationary cars maintain their initial $10 \text{ m}$ spacing, the length depends solely on the number of stationary cars.

2. Analyze the Rate of Change

From the solution to Question 26, we calculated that the time delay between the starting of two adjacent cars is $\Delta t = 5 \text{ s}$.

Process per cycle ($\Delta t = 5 \text{ s}$):

• At the start of the cycle, let $N$ cars be stationary.
• After $5 \text{ seconds}$, the next car in the queue begins to move.
• The “stationary segment” loses one car (the one at the front of the stationary line).

The physical length occupied by one car-interval in the stationary lineup is the initial separation $d = 10 \text{ m}$.

3. Calculation

The length of the stationary segment decreases by $10 \text{ m}$ every $5 \text{ seconds}$.

$$ \text{Average Rate of Change} = \frac{\Delta L}{\Delta t} $$ $$ \text{Average Rate of Change} = \frac{-10 \text{ m}}{5 \text{ s}} = -2 \text{ m/s} $$