Question

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=2s, then the equation of the path is:

• A. $x=y^3-y^2+2$
• B. $x=y^2-y+2$
• C. $x=y^2-3y^2+2$
• D. $x=y^3-2y^2+2$

Answer 1

Answer: (B)

$x=y^2-y+2$

$\frac{dx}{dt}=8t-2$

Intergrating both sides and subtracting $x=14\&t=2$

We get

$x=4t^2-2t+2\to \left(1\right)\frac{dy}{dt}=2$

Integrating both sides and substituting $y=4\&t=2$

We get

$y=2t\to \left(2\right)$

Equating $\left(2\right)in\left(1\right)$

$x=4(\frac{y}{2}{)}^2-2\times \frac{y}{2}+2x=y^2-y+2$