A particle moves in the x-y plane with velocity `v_x = 8t-2 and v_y = 2.` If it passes through the point `x =14 and y = 4 at t = 2 s,` the equation of the path is

Options

A. `x=y^2-y+2`
B. `x=y^2-2`
C. `x = y^2 + y-6`
D. None of these

Answer

A

Explanation

`v_x=(dx)/(dt)=(8t-2)`
`:. int_14^xdx=int_2^t(8t-2)dt`
`x-14=(4t^2-2t)-(12)`
`:. x=4t^2-2t+2…(i)`
`v_y=(dy)/(dt)=2`
`:. int_4^ydy=int_2^t 2dt`
`:. y-4=2t-4`
or `y=2t`
`:. t=y/2`
Substituting this value of t in Eq.(i) we have,
`x=y^2-y+2`