JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Kinematics

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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 .

(A)9
(B)6
(C)3
(D)12✓ Correct

Answer: (D)

Solution

The distance covered equals the area under the velocity–time graph. From the graph, the velocity starts at

200 m/s

t = 0

, rises linearly to

400 m/s

t = 2 s

, and then stays constant at

400 m/s

up to

t = 30.5 s

.

Area of the initial trapezoidal segment (0 to 2 s):

\frac{( 200 + 400 )}{2} \times 2 = 600 m

Area of the constant-velocity segment (2 s to 30.5 s):

400 \times (30.5 - 2) = 400 \times 28.5 = 11400 m

Total distance:

600 + 11400 = 12000 m = 12 km

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option

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