The potential energy `U`(in `J`) of a particle is given by `(ax + by)`, where `a` and `b` are constants. The mass of the particle is `1 kg` and `x` and `y` are the coordinates of the particle in metre. The particle is at rest at `(4a, 2b)` at time `t = 0`.
Find the speed of the particle when it crosses x-axis

Options

A. `2sqrt(a^(2) + b^(2))`
B. `sqrt(a^(2) + b^(2))`
C. `(1)/(2)sqrt(a^(2) + b^(2))`
D. `sqrt(((a^(2) + b^(2)))/(2))`

Answer

1

Explanation

`vecF = -(d)/(dr) = -a hati - bhatj`
`ac c. = (vecF)/(m)=- a hati - bhatj`
`(dv)/(dt) = - at hati - bhatj`
`vecV = - at hati - bthatj`
`vecS=(4a-(a)/(2)t^(2))^(2) hati + (2b-(bt^(2))/(2))hatj`
`vecS =(4a-(a)/(2)t^(2))hati +(2b-(bt)/(2))hatj`
When it cross x axis y=0
So t= 2
So `V = 2 sqrt((a^(2) + b^(2)))`