The potential energy (in SI units) of a particle of mass `2kg` in a conservative field is `U=6x-8y`. If the initial velocity of the particle is `vecu=-1.5hati+2hatj`, then find the total distance travelled by the particle in the first two seconds.

Answer

`15m`

Explanation

`vecF=-(delU)/(delx)hati-(delU)/(dely)hatj=-6hati+8hatj`
`veca=vecF//m-3hati+4hatj` has same direction as that of
`vecu=-1.5hati+2hatj`
`|veca|=5,|vecv|=5/2`
Since `vecu` and `veca` are in same direction, particle will move along a straight line, so the distance covered
`S=5/2xx2+1/2xx5xx2^2=15m`