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

Since potential energy of the particle is equal to ϕ, x-component of the force acting on the particle is equal toFx=−dϕdxand y-component of the force on the particle is equal to Fy=−dϕdyHence, force acting on the particle is F→=−(3i^+4j^) NIt means, force acting on the particle is constant. Hence, the particle moves with constant acceleration. So option (1) is correctAcceleration of the particle, a→=F→m2=−(3i^+4j^) ms−2Its magnitude is a=32+42=5 ms−2Since, the particle was initially at rest at (6, 4), position of the particle at time t is given byx=6+12axt2=6−32t2 mand y=4+12ayt2=4−2t2 mwhen the particle crosses the x-axis, y = 0, t1=2 sDisplacement of the particle during this time, s1=12at2=5 mHence, work done by the force, up to this instantFs1=5×5 J=25 JHence, Option (2) is also correct. The particle crosses y-axis when x = 0 Hence, 6−32t22=0 or t2=2sSpeed of the particle at this instant will be v=at2=5×2=10ms−1Hence, option (3) is also correct. At t=4s,x=6−32(4)2=−18mand y=4−2(4)2=−28mHence, option (4) is also correct.