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NEET Physics QUESTION #7318
Question 1

The position of a particle is described by $x = 8 + 12t - t^3$ (metres, seconds). The retardation when velocity becomes zero is:

  • $6\text{ m/s}^2$
  • $12\text{ m/s}^2$✔️
  • $24\text{ m/s}^2$
  • Zero
Correct Answer Explanation

$v = \dfrac{dx}{dt} = 12 - 3t^2$. Setting $v = 0$: $t^2 = 4 \Rightarrow t = 2\text{ s}$

$a = \dfrac{dv}{dt} = -6t$. At $t = 2\text{ s}$: $a = -12\text{ m/s}^2$

Retardation $= \mathbf{12\text{ m/s}^2}$