The velocity of a particle moving along the x | vedclass

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(B) The velocity of the particle is given by $v(t) = 1 - 3t^2 + 2t^3$. We know that velocity is the rate of change of position,$v = \frac{dx}{dt}$. Th

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(B) The velocity of the particle is given by $v(t) = 1 - 3t^2 + 2t^3$. We know that velocity is the rate of change of position,$v = \frac{dx}{dt}$. Therefore,the displacement $x$ can be found by integrating the velocity with respect to time: $x(t) = \int v(t) \ dt = \int (1 - 3t^2 + 2t^3) \ dt$. Performing the integration: $x(t) = t - t^3 + \frac{2t^4}{4} + C = t - t^3 + \frac{t^4}{2} + C$. Given that at $t = 0$,$x = 0$,we substitute these values to find the constant $C$: $0 = 0 - 0^3 + \frac{0^4}{2} + C \Rightarrow C = 0$. So,the position function is $x(t) = t - t^3 + \frac{t^4}{2}$. At $t = 2 \ s$,the position is: $x(2) = 2 - (2)^3 + \frac{(2)^4}{2} = 2 - 8 + \frac{16}{2} = 2 - 8 + 8 = 2 \ m$.

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$,then at $t = 2 \ s$,its position is: (in $m$)

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