QJEEM25Ph38d1: JEE Main 2025 Physics - Class 11 JEEM25Ph38d1 Solution: The motion of an airplane is represented by velocity-time graph …

The distance covered by the airplane in the first 30.5 seconds can be determined by calculating the area under the velocity-time graph.

The formula for distance in this scenario is:

$\text{Distance} = \text{Area under the graph}$

Calculating the individual areas:

$300 \, \text{m/s} \times 2 \, \text{s} = 600 \, \text{m}$

$400 \, \text{m/s} \times 28.5 \, \text{s} = 11400 \, \text{m}$

Adding these values gives the total distance:

$\text{Total distance} = 600 \, \text{m} + 11400 \, \text{m} = 12000 \, \text{m}$

Thus, the distance covered by the airplane in the first 30.5 seconds is $12000 \, \text{m}$ , which is equivalent to 12 km.