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

The Correct Option is B

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:

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

Answer 2

To calculate the average speed of the train for the entire journey, we need to consider the two phases of motion: acceleration and constant speed.

1. The train accelerates uniformly to a speed of 80 km/h. Let's assume the acceleration occurs over a time \( t \). During this phase, the average speed is half of the final speed due to uniform acceleration from rest. Hence, the average speed during acceleration is: \(\frac{80}{2} = 40 \text{ km/h}\).
2. Next, the train moves at a constant speed of 80 km/h for a time duration of \( 3t \).

The average speed for the entire journey is calculated by dividing the total distance traveled by the total time taken.

Firstly, calculate the distance covered in each phase:

• Distance during acceleration: Since the speed increases uniformly from rest to 80 km/h over time \( t \), the distance covered is: \(d_1 = \text{average speed} \times \text{time} = 40 \times t = 40t \text{ km}\).
• Distance at constant speed: The train travels at 80 km/h for \( 3t \) hours, so the distance covered during this period is: \(d_2 = 80 \times 3t = 240t \text{ km}\).

The total distance \( D \) is: \(D = d_1 + d_2 = 40t + 240t = 280t \text{ km}\).

The total time \( T \) taken for the journey is the sum of the two time intervals: \(T = t + 3t = 4t \text{ hours}\).

Thus, the average speed \( V \) for the entire journey is given by: \(V = \frac{D}{T} = \frac{280t}{4t} = 70 \text{ km/h}\).

Therefore, the average speed of the train over the duration of the journey is 70 km/h.