C

Step 1: Calculate the distance traveled by the train when the headlight is switched on.
The distance traveled by the train after 30 seconds can be calculated using the equation of motion:
$$ s = ut + \frac{1}{2} a t^2 $$
Here, the initial velocity (u) is 0, acceleration (a) is $3.0 \times 10^{-2}$ m/s², and time (t) is 30 seconds.
So,
$$ s_1 = 0 \cdot 30 + \frac{1}{2} \cdot (3.0 \times 10^{-2}) \cdot (30)^2 $$
$$ s_1 = \frac{1}{2} \cdot (3.0 \times 10^{-2}) \cdot 900 = 13.5 \text{ m} $$

Step 2: Calculate the distance traveled by the train after the headlight is switched on until the signal light is switched on.
The time interval between the two events is 60 seconds, starting from when the headlight is turned on. The train will accelerate during this period, and we need to determine the distance it travels during this interval. The initial velocity at t = 30 s (when the headlight is switched on) can be calculated as follows:
$$ v = u + at = 0 + (3.0 \times 10^{-2}) \cdot 30 = 0.9 \text{ m/s} $$
The distance traveled during the next 60 seconds can be calculated as follows:
$$ s_2 = v \cdot t + \frac{1}{2} a t^2 $$
$$ s_2 = (0.9 \text{ m/s}) \cdot 60 + \frac{1}{2} \cdot (3.0 \times 10^{-2}) \cdot (60)^2 $$
The first term is: $$ 0.9 \cdot 60 = 54 ext{ m} $$, and the second term is:
$$ \frac{1}{2} \cdot (3.0 \times 10^{-2}) \cdot 3600 = 54 ext{ m} $$
So,
$$ s_2 = 54 + 54 = 108 ext{ m} $$

Step 3: Calculate the total distance from the position of event 1 to event 2.
The total distance traveled by the train from the start until event 2 is:
$$ s_{total} = s_1 + s_2 = 13.5 + 108 = 121.5 ext{ m} $$

Step 4: Finding the relative motion for the frames.
To analyze how reference frame K moves such that events 1 and 2 occur at the same location, we equate the distances calculated in both frames:
$$ V = \frac{s_{total}}{(t_1 + t_2)} = \frac{121.5 ext{ m}}{90 ext{ s}} = 1.35 ext{ m/s} $$

Final Answer:
The distance between the two events in the Earth’s frame is 121.5 m, and the velocity V at which frame K must move relative to the Earth is 1.35 m/s.