The figure shows the velocity (v) of a partic | vedclass

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(D) $1$. The particle changes its direction of motion at $t = T$ because the velocity changes from negative to positive.$2$. The acceleration of the p

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All of the above.

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(D) $1$. The particle changes its direction of motion at $t = T$ because the velocity changes from negative to positive.$2$. The acceleration of the particle is the slope of the $v-t$ graph. Since the graph is a straight line,the slope is constant,meaning the acceleration is constant.$3$. The displacement of the particle is the area under the $v-t$ graph. The area below the $t$-axis (from $0$ to $T$) is equal in magnitude to the area above the $t$-axis (from $T$ to $2T$),but with opposite signs. Therefore,the total displacement is $zero$.$4$. Since all three statements are correct,the correct option is $D$.

The figure shows the velocity $(v)$ of a particle plotted against time $(t)$.

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