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=2 s, then the equation of the path is

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

Answer 1

Answer: (B)

$x=y^2-y+2$

As given $Vx=8t-2$ and $Vy=2$

integrating both we get

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

$x=\frac{8t^2}{2}-2t+C_1=4t^2-2t+C_1\to \left(i\right)$

Also, we have

$\int Vy=\int 2\int \frac{dy}{dt}=2t+C_2$

$y=2t+C_2$
As it is already given;

$x=14$and $y=4$at $t=2$

Putting the values in $\left(i\right)$

$x=4t^2-2t+C_{1}x=4(2{)}^2-2(2)+C_{1}14=16.x+C_1{C}_1-2$

$y=2t+C_{2}y=2\left(2\right)+C_{2}4=4+C_2{C}_2=0$

Now,

$y=2t$ and $x=4t^2-2t+2$

$t=\frac{y}{2}\to x=4t^2-2t+2$

Put the value of t we have

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

Option B