Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The position of a particle with respect to time t along the x-axis is given by x = 9t² - t³ where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x-direction?

32 m

54 m

81 m

24 m

Step-by-Step Solution

Find Velocity ( $v$ ): Velocity is the first derivative of position with respect to time ( $v = dx/dt$ ) . $v = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2$
Condition for Maximum Speed: The speed is maximum when the acceleration ( $a = dv/dt$ ) is zero . $a = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2) = 18 - 6t$ Set $a = 0$ to find the critical time: $18 - 6t = 0 \implies t = 3 \text{ s}$
Calculate Position: Substitute $t = 3$ s into the original position equation. $x = 9(3)^2 - (3)^3$ $x = 9(9) - 27$ $x = 81 - 27 = 54 \text{ m}$

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