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Question

(A) The average velocity $v_{avg}$ is defined as the total change in displacement divided by the total time interval: $v_{avg} = \frac{\Delta x}{\Delt

Vedclass · Accepted Answer

$2.28$

Vedclass · Answer

(A) The average velocity $v_{avg}$ is defined as the total change in displacement divided by the total time interval: $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$.Given $x(t) = \alpha t + \beta t^3$ with $\alpha = 2.0 \ m \ s^{-1}$ and $\beta = 0.01 \ m \ s^{-3}$.At $t_1 = 2.00 \ s$: $x(2) = 2.0(2) + 0.01(2)^3 = 4.0 + 0.01(8) = 4.08 \ m$.At $t_2 = 4.00 \ s$: $x(4) = 2.0(4) + 0.01(4)^3 = 8.0 + 0.01(64) = 8.64 \ m$.Now,calculate the average velocity: $v_{avg} = \frac{8.64 - 4.08}{4.00 - 2.00} = \frac{4.56}{2} = 2.28 \ m \ s^{-1}$.

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