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(C) The rate of change of velocity $\left| \frac{\Delta \vec{v}}{\Delta t} \right|$ represents the magnitude of acceleration,which is equal to the mag

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$4$ to $6 \, s$

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(C) The rate of change of velocity $\left| \frac{\Delta \vec{v}}{\Delta t} ight|$ represents the magnitude of acceleration,which is equal to the magnitude of the slope of the velocity-time graph.Calculate the slope for each interval:$1$. For $0$ to $2 \, s$: $	ext{slope} = \frac{10 - 0}{2 - 0} = 5 \, m/s^2$.$2$. For $2$ to $4 \, s$: $	ext{slope} = 0 \, m/s^2$.$3$. For $4$ to $6 \, s$: $	ext{slope} = \frac{-20 - 10}{6 - 4} = \frac{-30}{2} = -15 \, m/s^2$. Magnitude is $15 \, m/s^2$.$4$. For $6$ to $8 \, s$: $	ext{slope} = \frac{0 - (-20)}{8 - 6} = \frac{20}{2} = 10 \, m/s^2$.Comparing the magnitudes: $5, 0, 15, 10$. The maximum magnitude is $15 \, m/s^2$,which occurs in the interval $4$ to $6 \, s$.

The figure shows a velocity-time graph of a particle moving along a straight line. Identify the region in which the magnitude of the rate of change of velocity $\left| \frac{\Delta \vec{v}}{\Delta t} \right|$ of the particle is maximum.

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