Kinematics

JEE Main level, single correct, medium

120 of 308

Question 120 of 308

A particle starts moving from time t=0t=0 and its coordinate is given as x(t)=4t3−3tx(t)=4 t^{3}-3 t
A. The particle returns to its original position (origin) 0.866 units later
B. The particle is 1 unit away from origin at its turning point
C. Acceleration of the particle is non-negative
D. The particle is 0.5 units away from origin at its turning point
E. Particle never turns back as acceleration is non-negative
Choose the correct answer from the options given below :

Show answer and solution

Answer: (A) A, B, C Only

Given x(t)=4t3−3tx(t) = 4t^3 - 3t, we find v(t)=12t2−3v(t) = 12t^2 - 3 and a(t)=24ta(t) = 24t.
Statement A: Setting x(t)=0x(t) = 0 gives 4t3−3t=0⇒t(4t2−3)=04t^3 - 3t = 0 \Rightarrow t(4t^2-3) = 0, so t=0t = 0 or t=3/4=0.866t = \sqrt{3/4} = 0.866. The particle returns to origin at t=0.866t = 0.866 units. TRUE.
Statement B: Turning points occur at v(t)=0⇒12t2−3=0⇒t=±0.5v(t) = 0 \Rightarrow 12t^2 - 3 = 0 \Rightarrow t = \pm 0.5. At t=0.5t = 0.5: x=4(0.125)−1.5=−1x = 4(0.125) - 1.5 = -1; distance = 1 unit. At t=−0.5t = -0.5: x=1x = 1; distance = 1 unit. TRUE.
Statement C: For t≥0t \geq 0 (particle starts at t=0t=0), a(t)=24t≥0a(t) = 24t \geq 0. TRUE.
Statement D: At turning points, distance is 1 unit, not 0.5. FALSE.
Statement E: False because acceleration is non-negative only for t≥0t \geq 0, and particle does turn back at t=0.5t = 0.5. FALSE.
Correct statements: A, B, C Only.

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