A train starting from rest first accelerates uniformly up to a speed of \(80 \mathrm{~km} / \mathrm{h}\) 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 \(\mathrm{km} / \mathrm{h}\) ) _______.

Given:

Final speed = \(80 \, \text{km/h}\)

Time of acceleration = \(t\)

Time at constant speed = \(3t\)

Solution:

During uniform acceleration from rest,

\(\text{Average speed} = \frac{0 + 80}{2} = 40 \, \text{km/h}\)

Distance covered during acceleration:

\(s_1 = 40t\)

Distance covered at constant speed:

\(s_2 = 80 \times 3t = 240t\)

Total distance:

\(s = 40t + 240t = 280t\)

Total time:

\(T = t + 3t = 4t\)

Average speed:

\(\displaystyle v_{\text{avg}} = \frac{280t}{4t} = 70 \, \text{km/h}\)

Answer: \(70 \, \text{km/h}\)