JEE Main 2025 Jan 23 Shift 1, Physics Q19: Motion In One Dimension
JEE Main2025Jan 23, Shift 1Physics
Q.
The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is ____ km.
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Solution
$ \text{Distance covered}=\text{Area under the velocity-time graph.} $
$ \text{From }t=0\text{ to }2\text{ s, area of trapezium} $
$ s_1 = \dfrac{1}{2}(200+400)\times2 =600\ \text{m} $
$ \text{From }t=2\text{ s to }30.5\text{ s, velocity is constant at }400\ \text{m/s}. $
$ s_2 = 400\times(30.5-2) = 400\times28.5 = 11400\ \text{m} $
$ \text{Total distance} = s_1+s_2 = 600+11400 = 12000\ \text{m} = 12\ \text{km} $
$ \boxed{12\ \text{km}} $
Concept behind this question
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