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A particle moves in the x-y plane with v...

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

A

`x= y^2 -y+2`

B

`x=2y^2+2y-3`

C

`x= 3y^2 +5`

D

can not be found above data

Text Solution

Verified by Experts

The correct Answer is:
A
`(dx)/(dt)=8t-2 rArr x=4t^(2)-2t+c_(1_), At t=2, x=14 rArr c_(1)=2, (dy)/(dt)=2 rArr y=2t +c_(2):`
at `t=2, y=4 rArr c_(2)=0 therefore x=4t^(2)-2t+2 & y=2t rArr x=y^(2)-y+2`
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