The velocity of a particle is v = v_0 + gt + | vedclass

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(C) We know that velocity $v = \frac{dx}{dt}$,which implies $dx = v \, dt$. Integrating both sides with initial condition $x = 0$ at $t = 0$: $x = \in

Vedclass · Accepted Answer

$v_0 + \frac{g}{2} + \frac{f}{3}$

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(C) We know that velocity $v = \frac{dx}{dt}$,which implies $dx = v \, dt$. Integrating both sides with initial condition $x = 0$ at $t = 0$: $x = \int_{0}^{t} v \, dt = \int_{0}^{t} (v_0 + gt + ft^2) \, dt$. Performing the integration: $x = \left[ v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3} \right]_{0}^{t} = v_0 t + \frac{gt^2}{2} + \frac{ft^3}{3}$. To find the displacement at unit time $(t = 1)$,substitute $t = 1$ into the expression: $x = v_0(1) + \frac{g(1)^2}{2} + \frac{f(1)^3}{3} = v_0 + \frac{g}{2} + \frac{f}{3}$.

The velocity of a particle is $v = v_0 + gt + ft^2$. If its position is $x = 0$ at $t = 0$,then its displacement after unit time $(t = 1)$ is

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