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, the equation of the path is?

• A. $x=y^2-y+2$
• B. $x=y^2-2$
• C. $x=y^2+y-6$
• D. None of these

Answer 1

The correct option is A $x=y^2-y+2$

Velocity in y direction $v_y=2$

So, ${\int }_4^{y}dy={\int }_2^{t}2dt$

Or $y{|}_4^y=2\times t{|}_2^t$

$\therefore y-4=2(t-2)$

$⟹t=\frac{y}{2}$ ........(1)

Similarly, velocity in x direction $v_x=8t-2$

Or ${\int }_{14}^{x}dx={\int }_2^t(8t-2)dt$

Or $x{|}_{14}^x=(4t^2-2t){|}_2^t$

Or $x-14=4\times (t^2-4)-2(t-2)$

Or $x-14=4(\frac{y^2}{4}-4)-2(\frac{y}{2}-2)$

$⟹x=y^2-y+2$