The potential energy in joules of a particle of mass 1kg moving in a plane is given by $U = 3 x + 4 y,$ the position coordinates of the point being $x$ and $y,$ measured in metres. If the particle is at rest at (6, 4) then

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Its acceleration is of magnitude $5 m / s^{2}$

It moves in a straight line passing through the origin (0, 0)

Its velocity when it crosses the y-axis is $10 m / s$

It crosses the y-axis $(x = 0)$ at $y = 4$

answer is A, B.

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Detailed Solution

$U = 3 x + 4 y F_{x} = - \frac{d u}{d x} = - 3 & F_{y} = - \frac{d u}{d y} = - 4 \bar{F} = - 3 \hat{i} - 4 \hat{j} \bar{a} = \frac{\bar{F}}{M} = - 3 \hat{i} - 4 \hat{j} | a | = \sqrt{3^{2} + 4^{2}} = 5 m / s^{2}$

Particle is at rest (6, 4) then
$u_{y} = 0, a_{y} = - 4 m / s^{2}$
$u_{x} = 0, a_{x} = - 3 m / s^{2}$
When particle cross the y-axis disp. Along x-disp = 6m at ‘t’
then $x = u_{x} t + \frac{1}{2} a_{x} t^{2}$
$- 6 = 0 - \frac{1}{2} \times 3 \times t^{2}$
$t = 2 \sec$
$u_{x} = 0 - 3 \times 2 = - 6 m / s$
$u_{y} = - 4 \times 2 = - 8 m / s$
$u = \sqrt{u_{x}^{2} + u_{y}^{2}} = 10 m / s$
& $y = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$= 0 - \frac{1}{2} \times 4 \times 2$
Disp $= - 8$
Then value of $y = - 4 m$ when it corss y-axis