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(A) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.Given:Initial velocity, $u = -5 \, m/s$From th

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$+15$

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(A) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.Given:Initial velocity, $u = -5 \, m/s$From the graph:$1$. The area under the $a-t$ graph from $t = 0$ to $t = 6 \, s$ is a triangle with base $6 \, s$ and height $10 \, m/s^2$.Area$_1 = \frac{1}{2} 	imes 6 	imes 10 = 30 \, m/s$.$2$. The area under the $a-t$ graph from $t = 6 \, s$ to $t = 8 \, s$ is a triangle with base $(8 - 6) = 2 \, s$ and height $-10 \, m/s^2$.Area$_2 = \frac{1}{2} 	imes 2 	imes (-10) = -10 \, m/s$.Net area under the graph = Area$_1 +$ Area$_2 = 30 + (-10) = 20 \, m/s$.Since $\Delta v = v - u = 	ext{Net Area}$, we have:$v - (-5) = 20$$v + 5 = 20$$v = 15 \, m/s$.Therefore, the velocity at $t = 8 \, s$ is $15 \, m/s$.

The acceleration-time graph for a particle moving along the $x$-axis is shown in the figure. If the initial velocity of the particle is $-5 \, m/s$, the velocity at $t = 8 \, s$ is ....... $m/s$.

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