(A) Given,$x = t^4 - kt^3$. Velocity $v = \frac{dx}{dt} = 4t^3 - 3kt^2$. For the velocity to be maximum,its derivative with respect to time must be ze

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(A) Given,$x = t^4 - kt^3$. Velocity $v = \frac{dx}{dt} = 4t^3 - 3kt^2$. For the velocity to be maximum,its derivative with respect to time must be zero,i.e.,$\frac{dv}{dt} = 0$. $\frac{dv}{dt} = \frac{d}{dt}(4t^3 - 3kt^2) = 12t^2 - 6kt$. At $t = 2$,$\frac{dv}{dt} = 0$. $12(2)^2 - 6k(2) = 0$. $12(4) - 12k = 0$. $48 - 12k = 0$. $12k = 48$. $k = 4$.

Class 12 Mathematics Applications of Derivatives Maxima and Minima

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The displacement $x$ of a particle at time $t$ is given by $x = t^4 - kt^3$. If the velocity of the particle is maximum at $t = 2$,then $k = $ ..........

A
$4$
B
$-4$
C
$8$
D
$-8$

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Applications of Derivatives

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Maxima and Minima

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The displacement $x$ of a particle at time $t$ is given by $x = t^4 - kt^3$. If the velocity of the particle is maximum at $t = 2$,then $k = $ ..........

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