The potential energy `U` in joule of a particle of mass `1 kg` moving in `x-y` plane obeys the law`U = 3x + 4y`, where `(x,y)` are the co-ordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t = 0` then :

Options

A. the particle has constant acceleration.
B. the particle has zero acceleration.
C. the speed of the particle when it crosses y-axis is 10 m/s.
D. co-ordinate of particle at `t=1` s is (4.5, 2).

Answer

A, C, D

Explanation

We know that, `F_(x) = (-d U)/( dx) =-3`
and `F_(y) = (-d U)/( dy) =-4 therefore overset(to) ( F) =-(3 overset(^)(i) + 4 overset(^)(j) ) N`
For particle to cross y-axis, x = 0
`x=u_(x) t + (1)/(2) a_(x) t^(2) rArr -6 =0 - (1)/(2) xx 3t^(2) rArr t=2s`
For resultant velocity
`overset(to)(v) = 0- (3 overset(^)(i) + 4 overset(^)(j) ) xx 2 rArr | overset(to)(v)| =10 m s^(-1)" "[ because | overset(to)(v) | = sqrt(3(2) + 4^(2) )=5 m s^(-1)]`
`therefore Delta x = 0 - (1)/(2) xx 3 xx 1^(2) =-1.5`
`Delta y=-0 - (1)/(2) xx 4 xx 1^(2) =-2`.
So, co-ordinate `=(6-1.5 , 4-2)=(4.5,2)`.