The position x of a particle at any time t is | vedclass

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(C) Given the position function: $x(t) = 4t^3 - 3t^2 + 2$. Velocity $v(t)$ is the first derivative of position with respect to time: $v(t) = \frac{dx}

Vedclass · Accepted Answer

$42 \, ms^{-2}$ and $36 \, ms^{-1}$

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(C) Given the position function: $x(t) = 4t^3 - 3t^2 + 2$. Velocity $v(t)$ is the first derivative of position with respect to time: $v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^3 - 3t^2 + 2) = 12t^2 - 6t$. Acceleration $a(t)$ is the derivative of velocity with respect to time: $a(t) = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 6t) = 24t - 6$. At $t = 2 \, s$: Velocity $v(2) = 12(2)^2 - 6(2) = 12(4) - 12 = 48 - 12 = 36 \, ms^{-1}$. Acceleration $a(2) = 24(2) - 6 = 48 - 6 = 42 \, ms^{-2}$. Therefore,the acceleration is $42 \, ms^{-2}$ and the velocity is $36 \, ms^{-1}$.

The position $x$ of a particle at any time $t$ is given by $x(t) = 4t^3 - 3t^2 + 2$. The acceleration and velocity of the particle at $t = 2 \, s$ are respectively:

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