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(B) The relationship between acceleration $a$ and velocity $v$ is given by $a = v \frac{dv}{dx}$,which implies $\int_{u}^{v} v \ dv = \int_{0}^{x} a \

Vedclass · Accepted Answer

$1.2$

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(B) The relationship between acceleration $a$ and velocity $v$ is given by $a = v \frac{dv}{dx}$,which implies $\int_{u}^{v} v \ dv = \int_{0}^{x} a \ dx$.This means $\frac{v^2 - u^2}{2} = 	ext{Area under the } a-x 	ext{ graph}$.Given $u = 0.8 \ m/s$ at $x = 0$.The area under the $a-x$ graph from $x = 0$ to $x = 1.4$ consists of a rectangle and a trapezoid:Area $1$ (from $x=0$ to $x=0.4$): $0.4 	imes 0.4 = 0.16 \ m^2/s^2$.Area $2$ (from $x=0.4$ to $x=0.8$): $\frac{1}{2} 	imes (0.4 + 0.2) 	imes 0.4 = 0.12 \ m^2/s^2$.Area $3$ (from $x=0.8$ to $x=1.4$): $0.2 	imes (1.4 - 0.8) = 0.2 	imes 0.6 = 0.12 \ m^2/s^2$.Total Area $= 0.16 + 0.12 + 0.12 = 0.40 \ m^2/s^2$.Using the formula: $\frac{v^2 - (0.8)^2}{2} = 0.40$.$v^2 - 0.64 = 0.80$.$v^2 = 1.44$.$v = 1.2 \ m/s$.

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the graph. If the velocity of the particle is $0.8 \ m/s$ at $x = 0$,find the velocity of the particle at $x = 1.4 \ m$ (in $m/s$).

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