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. $x=y^2-3y+2$
• D. $x=y^3-2y^2+2$

Answer 1

Answer: (B)

Velocity along $x$ -axis, $v_x=\frac{dx}{dt}=8t-2$
or $dx=(8t-2)dt$
Integrating it, we get $x=\frac{8t^2}{2}-2t+C$
$=4t^2-2t+C$
where $C$ is a constant of integration.
At $t=2,x=14;$ so $14=4\times 2^2-2\times 2+C$ or
$C=2$ $\therefore x=4t^2-2t+2\dots$(i)
Also, $v_y=\frac{dy}{dt}=2$ or $dy=2dt$
Integrating it, we get $y=2t+C^{\prime }.$
where $C^{\prime }$ is a constant of integration.
At $t=2,y=4;$ so $4=2\times 2+C^{\prime }$ or $C^{\prime }=0$
$\therefore y=2t\dots$(ii)
In order to find the equation of path of projectile we have to eliminate $t$ from (i) and (ii).
From (ii), we get $t=\frac{y}{2}$
Putting it in (i), we get
$x=\frac{4y^2}{4}-\frac{2y}{2}+2$
$=y^2-y+2$