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(D) The position of the particle at any time $t$ is given by the area under the velocity-time graph from $t = 0$ to $t = 5\,s$.$1$. From $t = 0$ to $t

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$9$

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(D) The position of the particle at any time $t$ is given by the area under the velocity-time graph from $t = 0$ to $t = 5\,s$.$1$. From $t = 0$ to $t = 2\,s$,the graph is a triangle with base $2\,s$ and height $2\,m/s$. Area = $\frac{1}{2} 	imes 2 	imes 2 = 2\,m$.$2$. From $t = 2\,s$ to $t = 4\,s$,the graph is a rectangle with width $2\,s$ and height $2\,m/s$. Area = $2 	imes 2 = 4\,m$.$3$. From $t = 4\,s$ to $t = 5\,s$,the graph is a rectangle with width $1\,s$ and height $3\,m/s$. Area = $1 	imes 3 = 3\,m$.Total position at $t = 5\,s$ = $2 + 4 + 3 = 9\,m$.

$A$ particle starts from the origin at time $t = 0$ and moves along the positive $x-$ axis. The velocity-time graph is shown in the figure. What is the position of the particle at time $t = 5\,s$ (in $,m$)?

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