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(C) The displacement $x$ is the area under the velocity-time graph. $1$. From $t = 0$ to $t = 2 \ s$,velocity increases linearly,so displacement $x$ i

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(C) The displacement $x$ is the area under the velocity-time graph. $1$. From $t = 0$ to $t = 2 \ s$,velocity increases linearly,so displacement $x$ increases parabolically (concave up). $2$. From $t = 2 \ s$ to $t = 4 \ s$,velocity is constant $(10 \ m/s)$,so displacement $x$ increases linearly. $3$. From $t = 4 \ s$ to $t = 5 \ s$,velocity decreases from $10 \ m/s$ to $0$,so displacement $x$ increases with decreasing slope (concave down) until it reaches a maximum at $t = 5 \ s$. $4$. From $t = 5 \ s$ to $t = 6 \ s$,velocity becomes negative,so displacement $x$ starts decreasing. $5$. From $t = 6 \ s$ to $t = 8 \ s$,velocity increases from $-20 \ m/s$ to $0$,so displacement $x$ continues to decrease but with an increasing slope (concave up) until it returns to a certain value at $t = 8 \ s$. Comparing these features with the given options,option $C$ correctly represents the displacement-time graph.

The figure shows a velocity-time graph of a particle moving along a straight line. The correct displacement-time graph of the particle is shown as:

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