Q 11-02-051JEE MainJEE Main 2025 (23 Jan, Shift 1)Medium
The motion of an airplane is represented by the velocity-time graph shown below. The distance covered by the airplane in the first $30.5$ seconds is ______ km.
Answer: (A) $12$
From the graph, the velocity rises uniformly from $200\ \text{m/s}$ at $t = 0$ to $400\ \text{m/s}$ at $t = 2\ \text{s}$ and then stays at $400\ \text{m/s}$.
Distance = area under the graph:
- $0$ to $2\ \text{s}$ (trapezium): $\dfrac{200+400}{2}\times2 = 600\ \text{m}$
- $2$ to $30.5\ \text{s}$: $400\times28.5 = 11400\ \text{m}$
Total $= 12000\ \text{m} = 12\ \text{km}$
Solution by Sreeraj P, M.Sc Physics