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 particle when it crosses the y-axis is 10 m/s
D. co-ordinates of particle at t = 1s are (4.5, 2)

Answer

A, C, D

Explanation

`vecF = (-dU)/(dx) hati + (-dU)/(dy) hatj implies veca = (vecF)/(m) = - 3 hati - hatj` is constant
At `t = 1s, vecs = vecu t + 1/2 veca t^2 = 0 +1/2 (-3 hati - 4 hatj) (1)^(2) = (-3)/2 hati - 2 hatj = hatr_(f) - hatr_(i)`
`implies vecr_(f) = (-(3 hati)/(2) - 2 hatj) + (6 hati + 4hatj) = 4.5 hati + 2 hatj`
`a_x = - 3m//s^2`, displacement along x - axis when it crosses y = axis = -6m
`implies -6 = 0 + 1/2(-3) t^2 implies t = 2 sec, " " {:(V_x = 0 + 2(-3) , = -6m//s),(V_y = 0+2(-4) , = -8 m//s):}} implies |vecV|= 10 m//s`.