A train of length `l=350m` starts moving rectilinearly with constant acceleration `w=3.0*10^-2 m//s^2`, `t=30s` after the start the locomotive headlight is switched on (event 1), and `tau=60s` after that event the tail signal light is switched on (event 2). Find the distance between these events in the reference frames fixed to be train and to the Earth. How and at what constant velocity V relative to the Earth must a certain reference frame K move for the two events to occur in it at the same point?

Explanation

To solve the problem step by step, we will analyze the motion of the train and calculate the distances for the two events in both the Earth and train reference frames.

### Step 1: Calculate the distance covered by the train when the headlight is switched on (Event 1)

The train starts from rest and accelerates with a constant acceleration \( w = 3.0 \times 10^{-2} \, \text{m/s}^2 \). The time until the headlight is switched on is \( t = 30 \, \text{s} \).

Using the equation of motion:
\[ s_1 = \frac{1}{2} w t^2 \]
Substituting the values:
\[ s_1 = \frac{1}{2} \times (3.0 \times 10^{-2}) \times (30^2) \]
\[ s_1 = \frac{1}{2} \times (3.0 \times 10^{-2}) \times 900 \]
\[ s_1 = 13.5 \, \text{m} \]

### Step 2: Calculate the distance covered by the train when the tail signal light is switched on (Event 2)

The time until the tail signal light is switched on is \( \tau = 60 \, \text{s} \) after Event 1, which means the total time from the start is \( t + \tau = 30 + 60 = 90 \, \text{s} \).

Using the same equation of motion:
\[ s_2 = \frac{1}{2} w (t + \tau)^2 \]
Substituting the values:
\[ s_2 = \frac{1}{2} \times (3.0 \times 10^{-2}) \times (90^2) \]
\[ s_2 = \frac{1}{2} \times (3.0 \times 10^{-2}) \times 8100 \]
\[ s_2 = 121.5 \, \text{m} \]

### Step 3: Calculate the distance between the two events in the Earth reference frame

The distance between the two events is:
\[ \Delta s = s_2 - s_1 \]
Substituting the values:
\[ \Delta s = 121.5 \, \text{m} - 13.5 \, \text{m} = 108 \, \text{m} \]

### Step 4: Calculate the distance between the two events in the train reference frame

In the train reference frame, the distance between the two events is equal to the length of the train, which is:
\[ \text{Distance in train frame} = l = 350 \, \text{m} \]

### Step 5: Calculate the relative velocity \( V \) for the reference frame K

To find the constant velocity \( V \) such that both events occur at the same point in reference frame K, we can use the time interval \( \tau = 60 \, \text{s} \) and the distance \( \Delta s = 108 \, \text{m} \):
\[ V = \frac{\Delta s}{\tau} = \frac{108 \, \text{m}}{60 \, \text{s}} = 1.8 \, \text{m/s} \]

### Summary of Results
- Distance between events in Earth frame: \( 108 \, \text{m} \)
- Distance between events in train frame: \( 350 \, \text{m} \)
- Required velocity \( V \) for reference frame K: \( 1.8 \, \text{m/s} \)