Question

A particle moves in the $x-y$ plane with velocity $v_{x}=8 t-2$ 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. $x=y^{2}-3 y+2$
• D. $x=y^{3}-2 y^{2}+2$

Answer 1

Answer: (B)

Velocity along $x$ -axis, $v_{x}=\frac{d x}{d t}=8 t-2$
or $d x=(8 t-2) d t$
Integrating it, we get $x=\frac{8 t^{2}}{2}-2 t+C$
$=4 t^{2}-2 t+C$
where $C$ is a constant of integration.
At $t=2, x=14 ;$ so $14=4 \times 2^{2}-2 \times 2+C$ or
$C=2$ $\therefore x=4 t^{2}-2 t+2 \dots$(i)
Also, $v_{y}=\frac{d y}{d t}=2$ or $d y=2 d t$
Integrating it, we get $y=2 t+C' .$
where $C'$ is a constant of integration.
At $t=2, y=4 ;$ so $4=2 \times 2+C'$ or $C'=0$
$\therefore y=2 t \dots$(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
$x=\frac{4 y^{2}}{4}-\frac{2 y}{2}+2$
$=y^{2}-y+2$