Among the four graphs shown in the figure,the | vedclass

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(B) The average velocity over a time interval $(0, T)$ is defined as the total displacement divided by the total time: $v_{avg} = \frac{x(T) - x(0)}{T

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(B) The average velocity over a time interval $(0, T)$ is defined as the total displacement divided by the total time: $v_{avg} = \frac{x(T) - x(0)}{T}$.For the average velocity to vanish (i.e.,$v_{avg} = 0$),the displacement must be zero,which means $x(T) = x(0)$.This implies that the position of the particle at time $T$ must be the same as its position at time $t = 0$.Looking at the given graphs:In graph $B$,the curve starts at a certain position $x(0)$ at $t = 0$ and returns to the same position $x(T)$ at some later time $T$. This is clearly shown in the solution figure where $OA = BT$,meaning the displacement is zero over the interval $(0, T)$.Therefore,the average velocity can vanish for graph $B$.

Among the four graphs shown in the figure,there is only one graph for which the average velocity over the time interval $(0, T)$ can vanish for a suitably chosen $T$. Which one is it?

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