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(A) The distance travelled is equal to the area under the velocity-time $(v-t)$ graph.First, we construct the $v-t$ graph from the given acceleration-

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$4$

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(A) The distance travelled is equal to the area under the velocity-time $(v-t)$ graph.First, we construct the $v-t$ graph from the given acceleration-time $(a-t)$ graph:$1$. For $0 \le t \le 1 \, s$, $a = 2 \, m/s^2$. Since $v = \int a \, dt$, $v = 2t$. At $t = 1 \, s$, $v = 2 \, m/s$.$2$. For $1 \le t \le 2 \, s$, $a = 0$, so velocity remains constant at $v = 2 \, m/s$.$3$. For $2 \le t \le 3 \, s$, $a = -2 \, m/s^2$. Velocity decreases from $2 \, m/s$ to $0 \, m/s$ at $t = 3 \, s$.The resulting $v-t$ graph is a trapezoid with parallel sides of length $1 \, s$ (from $t=1$ to $t=2$) and $3 \, s$ (from $t=0$ to $t=3$), and height $2 \, m/s$.Distance = Area of the trapezoid = $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$Distance = $\frac{1}{2} 	imes (1 + 3) 	imes 2 = 4 \, m$.

The acceleration-time graph for a particle is given in the figure. If it starts motion at $t=0$, the distance travelled in $3 \, s$ will be ........... $m$.

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