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(B) We know that $a = v \frac{dv}{dx}$, which implies $v dv = a dx$.Integrating both sides, we get $\int_{u}^{v} v dv = \int_{x_1}^{x_2} a dx$.$\frac{

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$1.2$

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(B) We know that $a = v \frac{dv}{dx}$, which implies $v dv = a dx$.Integrating both sides, we get $\int_{u}^{v} v dv = \int_{x_1}^{x_2} a dx$.$\frac{v^2 - u^2}{2} = 	ext{Area under the } a-x 	ext{ graph}$.$v^2 = u^2 + 2 	imes (	ext{Area under the } a-x 	ext{ graph})$.The area under the $a-x$ graph from $x=0$ to $x=1.4$ consists of a rectangle, a trapezoid, and another rectangle:Area $1$ (from $x=0$ to $0.4$): $0.4 	imes 0.4 = 0.16$.Area $2$ (from $x=0.4$ to $0.8$): $\frac{1}{2} 	imes (0.4 + 0.2) 	imes 0.4 = 0.12$.Area $3$ (from $x=0.8$ to $1.4$): $0.2 	imes 0.6 = 0.12$.Total Area $= 0.16 + 0.12 + 0.12 = 0.4$.Given $u = 0.8 \,ms^{-1}$, so $u^2 = 0.64$.$v^2 = 0.64 + 2 	imes (0.4) = 0.64 + 0.8 = 1.44$.$v = \sqrt{1.44} = 1.2 \,ms^{-1}$.

Column-$I$	Column-$II$
$(1)$ Acceleration is positive	$(a)$ Speed of the particle decreases
$(2)$ Acceleration is negative	$(b)$ Speed of the particle increases
	$(c)$ Speed of the particle keeps changing

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \,ms^{-1}$ at $x=0$, then its velocity at $x=1.4 \,m$ is (in $ms^{-1}$)

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Column-$I$ Column-$II$

$(1)$ Acceleration is positive $(a)$ Speed of the particle decreases

$(2)$ Acceleration is negative $(b)$ Speed of the particle increases

$(c)$ Speed of the particle keeps changing

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