Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The motion of a particle along a straight line is described by the equation x = 8 + 12t - t³, where x is in metre and t is in second. The retardation of the particle when its velocity becomes zero is:

24 ms⁻²

zero

6 ms⁻²

12 ms⁻²

Step-by-Step Solution

Find Velocity ( $v$ ): Velocity is defined as the rate of change of position with respect to time ( $v = \frac{dx}{dt}$ ) . Given $x = 8 + 12t - t^3$ . $v = \frac{d}{dt}(8 + 12t - t^3) = 0 + 12 - 3t^2$
Find Time ( $t$ ) when Velocity is Zero: Set $v = 0$ : $12 - 3t^2 = 0$ $3t^2 = 12 \implies t^2 = 4 \implies t = 2 \text{ s}$
Find Acceleration ( $a$ ): Acceleration is defined as the rate of change of velocity with respect to time ( $a = \frac{dv}{dt}$ ) . $a = \frac{d}{dt}(12 - 3t^2) = -6t$
Calculate Retardation: At $t = 2$ s, acceleration $a = -6(2) = -12 \text{ ms}^{-2}$ . Retardation is the negative of acceleration (or the magnitude of deceleration). Therefore, retardation is $12 \text{ ms}^{-2}$ .

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