The position (x) of a particle moving along t | vedclass

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(A) The average acceleration is defined as the change in velocity divided by the time interval: $a_{	ext{avg}} = \frac{\Delta v}{\Delta t}$.From the

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$-5$

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(A) The average acceleration is defined as the change in velocity divided by the time interval: $a_{	ext{avg}} = \frac{\Delta v}{\Delta t}$.From the position-time graph:For the interval $t=0$ to $t=2 \, s$,the velocity $v_1 = \frac{40 - 0}{2 - 0} = 20 \, m/s$.For the interval $t=6$ to $t=8 \, s$,the velocity $v_2 = \frac{0 - 40}{8 - 6} = -20 \, m/s$.The change in velocity $\Delta v = v_2 - v_1 = -20 - 20 = -40 \, m/s$.The total time interval $\Delta t = 8 - 0 = 8 \, s$.Therefore,$a_{	ext{avg}} = \frac{-40}{8} = -5 \, m/s^2$.

The position $(x)$ of a particle moving along the $x$-axis varies with time $(t)$ as shown in the figure. The average acceleration of the particle in the time interval $t=0$ to $t=8 \, s$ is ........... $m/s^2$.

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