The position of an object moving along the X- | vedclass

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(B) The position is given by $x(t) = \alpha + \beta t^2$.Given $\alpha = 8 \ m$, the equation becomes $x(t) = 8 + \beta t^2$.The average velocity $v_{

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$2$

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(B) The position is given by $x(t) = \alpha + \beta t^2$.Given $\alpha = 8 \ m$, the equation becomes $x(t) = 8 + \beta t^2$.The average velocity $v_{avg}$ between $t_1 = 2 \ s$ and $t_2 = 4 \ s$ is defined as $v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$.Calculate the position at $t_1 = 2 \ s$: $x(2) = 8 + \beta(2)^2 = 8 + 4\beta$.Calculate the position at $t_2 = 4 \ s$: $x(4) = 8 + \beta(4)^2 = 8 + 16\beta$.The change in position is $\Delta x = x(4) - x(2) = (8 + 16\beta) - (8 + 4\beta) = 12\beta$.The time interval is $\Delta t = 4 - 2 = 2 \ s$.Given $v_{avg} = 12 \ m/s$, we have $12 = \frac{12\beta}{2}$.$12 = 6\beta$, which implies $\beta = 2 \ m/s^2$.

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