Question

A train starting from rest first accelerates uniformly up to a speed of 80 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. 80
• B. 70
• C. 30
• D. 40

Answer 1

Answer: (B)

70

Answer 2

The average speed is calculated using the formula:

$\text{Average speed} = \frac{\text{total distance}}{\text{time taken}}$

During the first phase of acceleration:

$\text{Distance covered} = \frac{1}{2} \times \text{final speed} \times \text{time} = \frac{1}{2} \times 80 \times t = 40t$

During the second phase of constant speed:

$\text{Distance covered} = \text{speed} \times \text{time} = 80 \times 3t = 240t$

Total distance covered:

$40t + 240t = 280t$

Total time taken:

$t + 3t = 4t$

Average speed:

$\text{Average speed} = \frac{280t}{4t} = 70 \, \text{km/h}$