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The potential energy 'U' in Joule of a particle of mass $1 kg$ , moving in $X - Y$ plane, obeys the law $U = 3 x + 4 y$ , where $(x, y)$ are the co-ordinates of the particle in meter. If the particle is at rest at $(6, 4)$ at $t = 0$ , then (Assuming there are no non-conservative forces acting on the partcile)

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the particle has constant acceleration

the work done by the external forces, from the positon of rest of the particle to the instant of crossing the X-axis is $50 J$

the speed of the particle when it crosses the Y-axis is $10 {ms}^{- 1}$

the co-ordinates of the particle at time $t = 4 s$ are $(- 18, - 28)$

answer is A, C, D.

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Detailed Solution

$U = 3 x + 4 y \vec{F} = (- 3) \overset{\land}{i} + (- 4) \overset{\land}{j} = m \vec{a} \Rightarrow \vec{a} = c o n s \tan t \vec{r} = (6 - \frac{3}{2} t^{2}) \overset{\land}{i} + (4 - 2 t^{2}) \overset{\land}{j} A n s . A C D$

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