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^(3) - y^(2) + 2`
B. `x = y^(2) - y + 2`
C. `x = y^(2) - 3y + 2`
D. `x = y^(3) - 2y^(2) + 2`

Answer

B

Explanation

Velocity along x-axis , `v_x = (dx) / (dt) = 8t - 2`
or `dx = (8t - 2) dt `
Integrating it , we get `x = (8t^2)/2 - 2t + C = 4 t^(2) - 2 t + C`
where C is a constant of integration .
At `t = 2, x = 14 , " so " 14 = 4 xx 2^(2) - 2 xx 2 + C " or " C = 2 `
`:. x = 4t ^(2) 0 2t + 2 " "...(i)`
Integrating it ,we get ` y = 2t + C.`
where C. is a constant of integration .
At `t = 2 , y = 4 , " so " 4 = 2 xx 2 + C. " or " C. = 0 `
`:. y = 2t " "...(ii)`
In order to find the equation of path of projectile we have to eliminate t from (i) and (ii) .
From (ii) , we get `t/2`
putting it in (i) , we get `x = (4y^2)/4 - (2y)/2 + 2 =y^(2) - y + 2`