# 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?

Mar 18, 2018

$x = {y}^{2} - y + 2$

#### Explanation:

Given ${v}_{x} \left(t\right) = 8 t - 2$ and $v \left(t\right) = 2$, $x \left(2\right) = 14$, and $y \left(2\right) = 4$

Substitute ${v}_{x} \left(t\right) = \frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}}$

$\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} = 8 t - 2$

Integrating:

$x \left(t\right) = 4 {t}^{2} - 2 t + x \left(0\right) \text{ [1]}$

Substitute $x \left(2\right) = 14$

$14 = 4 {\left(2\right)}^{2} - 2 \left(2\right) + x \left(0\right)$

$x \left(0\right) = 2$

Substitute into equation [1]:

$x \left(t\right) = 4 {t}^{2} - 2 t + 2 \text{ [1.1]}$

Substitute ${v}_{y} \left(t\right) = \frac{\mathrm{dy} \left(t\right)}{\mathrm{dt}}$

$\frac{\mathrm{dy} \left(t\right)}{\mathrm{dt}} = 2$

Integrate:

$y \left(t\right) = 2 t + y \left(0\right) \text{ [2]}$

Substitute $y \left(2\right) = 4$

$4 = 2 \left(2\right) + y \left(0\right)$

$y \left(0\right) = 0$

Substitute into equation [2]:

$y \left(t\right) = 2 t \text{ [2.1]}$

Solve equation [2.1] for t:

$t = \frac{y}{2}$

Substitute into equation [1.1]:

$x = 4 {\left(\frac{y}{2}\right)}^{2} - 2 \left(\frac{y}{2}\right) + 2 \text{ [1.2]}$

Simplify:

$x = {y}^{2} - y + 2 \text{ [1.3]}$