A train starting from rest first accelerates uniformly up to a speed of $$80 \text{ km/h}$$ for time $$t$$, then it moves with a constant speed for time $$3t$$. The average speed of the train for this duration of journey will be (in km/h) :

A train starts from rest, accelerates uniformly to 80 km/h in time $$t$$, then moves at constant speed for time $$3t$$. We need the average speed.

Calculate the distance during acceleration.

During uniform acceleration from 0 to 80 km/h over time $$t$$, the average speed is $$\frac{0 + 80}{2} = 40$$ km/h.

$$ d_1 = \text{average speed} \times \text{time} = 40t \text{ (in km, with } t \text{ in hours)} $$

Calculate the distance during constant speed.

$$ d_2 = 80 \times 3t = 240t \text{ km} $$

Calculate the average speed.

$$ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d_1 + d_2}{t + 3t} = \frac{40t + 240t}{4t} = \frac{280t}{4t} = 70 \text{ km/h} $$

The correct answer is Option (4): 70.