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(C) The distance traversed by the particle is equal to the area under the speed-time graph.For the interval $t=2\;s$ to $t=5\;s$:The speed $v$ at time

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$36$

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(C) The distance traversed by the particle is equal to the area under the speed-time graph.For the interval $t=2\;s$ to $t=5\;s$:The speed $v$ at time $t$ is given by $v = at$. Since $v=12\;m/s$ at $t=5\;s$,the acceleration $a_1 = 12/5 = 2.4\;m/s^2$.At $t=2\;s$,$v_2 = 2.4 	imes 2 = 4.8\;m/s$.At $t=5\;s$,$v_5 = 12\;m/s$.The distance $s_1$ is the area of the trapezoid from $t=2$ to $t=5$:$s_1 = \frac{1}{2} 	imes (v_2 + v_5) 	imes (5-2) = \frac{1}{2} 	imes (4.8 + 12) 	imes 3 = \frac{1}{2} 	imes 16.8 	imes 3 = 25.2\;m$.For the interval $t=5\;s$ to $t=6\;s$:The deceleration $a_2$ is the slope of the line from $t=5$ to $t=10$,which is $a_2 = (0-12)/(10-5) = -2.4\;m/s^2$.At $t=6\;s$,the speed $v_6 = v_5 + a_2 	imes (6-5) = 12 + (-2.4) 	imes 1 = 9.6\;m/s$.The distance $s_2$ is the area of the trapezoid from $t=5$ to $t=6$:$s_2 = \frac{1}{2} 	imes (v_5 + v_6) 	imes (6-5) = \frac{1}{2} 	imes (12 + 9.6) 	imes 1 = \frac{1}{2} 	imes 21.6 = 10.8\;m$.Total distance $s = s_1 + s_2 = 25.2 + 10.8 = 36\;m$.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle in $m$ between $t=2\;s$ to $6\;s$.

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