JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Kinematics

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The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between

t = 0

t = 4 s

(A)30 m✓ Correct
(B)10 m
(C)13 m
(D)11 m

Answer: (A)

Solution

The distance travelled equals the area under the velocity-time graph.

The graph rises linearly from

v = 0

t = 0

v = 10 m/s

t = 2 s

, stays constant at

10 m/s

from

t = 2 s

t = 4 s

, and then decreases afterward.

From

t = 0

t = 2 s

(triangular region):

A_{1} = \frac{1}{2} \times 2 \times 10 = 10 m

From

t = 2 s

t = 4 s

(rectangular region):

A_{2} = 2 \times 10 = 20 m

Total distance from

t = 0

t = 4 s

A_{1} + A_{2} = 10 + 20 = 30 m

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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