Question

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

Answer 1

`x = y^2-y+2`

Explanation:

Given `v_x(t) = 8t -2` and `v(t) = 2`, `x(2) = 14`, and `y(2) = 4`

Substitute `v_x(t) = (dx(t))/dt`

`(dx(t))/dt = 8t -2`

Integrating:

`x(t) = 4t^2-2t+x(0)" [1]"`

Substitute `x(2) = 14`

`14 = 4(2)^2-2(2)+x(0)`

`x(0) = 2`

Substitute into equation [1]:

`x(t) = 4t^2-2t+2" [1.1]"`

Substitute `v_y(t) = (dy(t))/dt`

`(dy(t))/dt = 2`

Integrate:

`y(t) = 2t+y(0)" [2]"`

Substitute `y(2) = 4`

`4 = 2(2) + y(0)`

`y(0) = 0`

Substitute into equation [2]:

`y(t) = 2t" [2.1]"`

Solve equation [2.1] for t:

`t = y/2`

Substitute into equation [1.1]:

`x = 4(y/2)^2-2(y/2)+2" [1.2]"`

Simplify:

`x = y^2-y+2" [1.3]"`