The displacement of a particle at time t is x | vedclass

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(A) The displacement is given by $x = t^4 - k t^3$. The velocity $v$ is the derivative of displacement with respect to time $t$: $v = \frac{dx}{dt} =

Vedclass · Accepted Answer

$k = 4$

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(A) The displacement is given by $x = t^4 - k t^3$. The velocity $v$ is the derivative of displacement with respect to time $t$: $v = \frac{dx}{dt} = 4t^3 - 3kt^2$. To find the condition for minimum velocity, we find the acceleration $a = \frac{dv}{dt}$: $a = \frac{dv}{dt} = 12t^2 - 6kt$. For the velocity to be minimum at $t = 2$, the derivative of velocity with respect to time must be zero at $t = 2$: $\frac{dv}{dt} \big|_{t=2} = 12(2)^2 - 6k(2) = 0$. $12(4) - 12k = 0$. $48 - 12k = 0$. $12k = 48$. $k = 4$. Thus, the value of $k$ is $4$.

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

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