A particle moves in xy plane with a velocity given by `vecV = (8t-2)hati+2hatj.`. If it passes through the point `(14,4)` at `t=2 sec , then give eqution of the point

Options

A. `x + y - y^(2) = 2`
B. `y^(2)+x=4`
C. `x-y-y^(2)=2`
D. `y + x + y^(2)=2`

Answer

A

Explanation

`V = 8t -2`
`intdx = int(8t-2)dt`
`x = 4t^(2)-2t +c`
At `t =2,x = 14 implies " "14=4(4) -2(2)+c " implies "c = 2`
`x = 4t^(2)-2t+2,v_(y)=2`
`intdy int2dt`
`y = 2t + c`
At `t =2, y = 4 " implies "c = 0`
` y = 2t`
Equation `implies " " x = (2t)^(2)-2t +2`
`implies" " x=y^(2)-y+2`
`implies " "x + y-y^(2) =2`