91Ó°ÊÓ

When an object speeds up at a constant rate, it is said to experience uniform acceleration. In the context of a train journey, this means the train increases its speed steadily until it reaches its maximum speed. This phase is marked by a consistent increase in velocity over time.
When calculating the distance covered during this phase, the formula used is:

• \[ s_1 = \frac{1}{2} V_{max} \cdot t \]

This equation tells us that the distance (\( s_1 \)) during acceleration is half of the maximum speed (\( V_{max} \)) multiplied by the time (\( t \)) spent accelerating. It's because speed increases linearly from zero to maximum speed.

During the phase of constant velocity motion, the train moves at a steady speed. This means there's no acceleration; the speed remains constant, and thus, the movement is smooth. The time for this phase is significantly longer compared to acceleration and deceleration phases in our problem.
For distance calculation in this phase, we use:

• \[ s_2 = V_{max} \cdot 8t \]

Here, the distance (\( s_2 \)) during constant velocity is simply the maximum speed (\( V_{max} \)) multiplied by the time (\( 8t \)), since the train maintains this speed throughout this duration. This phase covers the majority of the journey's distance.

Uniform deceleration is the process where the train reduces its speed at a steady rate until it comes to a stop. This is the reverse of the acceleration process and, like acceleration, occurs over the same ratio of time (1 unit of time in this problem).
To find the distance during deceleration, just like acceleration, we use:

• \[ s_3 = \frac{1}{2} V_{max} \cdot t \]

This formula indicates that as the speed decreases uniformly from the maximum down to zero, the distance (\( s_3 \)) is half the maximum speed times the duration of deceleration, echoing the symmetry between acceleration and deceleration phases.

The analysis of motion phases—acceleration, constant velocity, and deceleration—helps in calculating average speed over a journey. Dividing the journey into these distinct phases lets us calculate the total distance and total time conveniently.
Total distance (\( S \)) is the sum of distances from all phases:

• \[ S = \frac{1}{2} V_{max} \cdot t + V_{max} \cdot 8t + \frac{1}{2} V_{max} \cdot t = V_{max} \cdot 9t \]

Meanwhile, the total time (\( T \)) is the sum of time from all phases:

• \[ T = t + 8t + t = 10t \]

With these, the average speed (\( V_{avg} \)) is computed as:
- \[ V_{avg} = \frac{S}{T} = \frac{V_{max} \cdot 9t}{10t} = \frac{9}{10} V_{max} \]
This method displays how phase-based calculations simplify the task of determining overall performance characteristics such as average speed, in this case resulting in 54 km/h.