(B) Given the displacement function: $x = t^3 - 6t^2 + t$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{d

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(B) Given the displacement function: $x = t^3 - 6t^2 + t$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + t) = 3t^2 - 12t + 1$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 1) = 6t - 12$. For acceleration to be zero: $a = 0$. $6t - 12 = 0$. $6t = 12$. $t = 2$ unit time.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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$A$ particle is moving in a straight line. At time $t$,the distance of the particle from its starting point is given by $x = t^3 - 6t^2 + t$. Its acceleration will be zero at

WBJEE 2009 All WBJEE

A
$t = 1$ unit time
B
$t = 2$ unit time
C
$t = 3$ unit time
D
$t = 4$ unit time

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Applications of Derivatives

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Rate of Change of Quantities

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$A$ particle is moving in a straight line. At time $t$,the distance of the particle from its starting point is given by $x = t^3 - 6t^2 + t$. Its acceleration will be zero at

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$A$ point is moving on $y=4-2x^2$. The $x$-coordinate of the point is decreasing at the rate of $5 \text{ units/s}$. Then,the rate at which the $y$-coordinate of the point is changing when the point is at $(1, 2)$ is

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