JEE MainPhysicsModerateKinematics

The acceleration-displacement graph of a particle moving in a
straight line is as shown in figure. Initial velocity of particle is
zero. What is the velocity (in m /s) of the particle when displacement
of the particle is s=16 m.

Show answer and solution

v\ dv = a\ ds

\therefore\int_{0}^{v}vdv = \int_{0}^{12m}ads

\therefore\frac{v^{2}}{2} = area\ under\ a - s\ graph\ froms = 0\ to\ s = 16m

= 2 + 12 + 6 + 12

= 32m/s

v = \sqrt{64}m/s = 8m/s

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx

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