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JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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The motion of a particle along a straight line is described by the equation: x=8+12t−t3x=8+12 t-t^3, where x{x} is in meter and t{t} in second.

(i) the initial velocity of particle is 12 m/s12 \mathrm{~m} / \mathrm{s}

(ii) the retardation of particle when velocity is zero is 12 m/s212 \mathrm{~m} / \mathrm{s}^2

(iii) when acceleration is zero, displacement is 8 m8 \mathrm{~m}

(iv) the maximum velocity of particle is 12 m/s12 \mathrm{~m} / \mathrm{s}
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Correct option: D
All options are correct
Option D (All options are correct) is correct because all the statements about the particle's motion derived from the given equation are true.

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