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(A) The displacement of a particle is given by the area under the velocity-time graph.To find the maximum displacement, we calculate the area under th

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$33.3$

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(A) The displacement of a particle is given by the area under the velocity-time graph.To find the maximum displacement, we calculate the area under the graph from $t = 0$ to the time when the velocity becomes zero for the first time.The velocity is zero at $t = 0 \, s$ and again at $t = 4.67 \, s$ (approximately $14/3 \, s$).The area under the graph from $t = 0$ to $t = 2 \, s$ is a triangle: $	ext{Area}_1 = \frac{1}{2} 	imes 2 	imes 10 = 10 \, m$.The area from $t = 2 \, s$ to $t = 4 \, s$ is a rectangle: $	ext{Area}_2 = 2 	imes 10 = 20 \, m$.At $t = 4 \, s$, the velocity starts decreasing. The line passes through zero at $t = 4 + \Delta t$. By similar triangles, $\frac{10}{\Delta t} = \frac{20}{2 - \Delta t}$ is not correct here; rather, the slope is $\frac{-20 - 10}{6 - 4} = -15 \, m/s^2$. The equation of the line for $t > 4$ is $v = 10 - 15(t - 4)$. Setting $v = 0$, we get $10 = 15(t - 4)$, so $t - 4 = 2/3$, or $t = 4.67 \, s$.The area of the small triangle from $t = 4 \, s$ to $t = 4.67 \, s$ is $\frac{1}{2} 	imes (2/3) 	imes 10 = 3.33 \, m$.Total maximum displacement $= 10 + 20 + 3.33 = 33.33 \, m$.

The figure shows a velocity-time graph of a particle moving along a straight line. The maximum displacement of the particle is ........ $m$.

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