A particle moves in the xy -plane with veloci | vedclass

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Question

$\frac{\mathrm{dx}}{\mathrm{dt}}=8 \mathrm{t}-2$

$\mathrm{dx}=8 \mathrm{t} \mathrm{dt}-2 \mathrm{dt}$

$\Rightarrow \quad \int \mathrm{d} \mathrm

Vedclass · Accepted Answer

$x = y^2 -y + 2$

Vedclass · Answer

$\frac{\mathrm{dx}}{\mathrm{dt}}=8 \mathrm{t}-2$

$\mathrm{dx}=8 \mathrm{t} \mathrm{dt}-2 \mathrm{dt}$

$\Rightarrow \quad \int \mathrm{d} \mathrm{x}=8 \int \mathrm{tdt}-2 \int \mathrm{dt}$

$\Rightarrow \quad x=\frac{8 t^{2}}{2}-2 t+c$

$\Rightarrow \quad x=4 t^{2}-2 t+c$

at $t=2 \quad x=14$

$14=4 	imes(2)^{2}-2(2)+c$

$14=16-4+c$

$\mathrm{c}=2 	herefore \mathrm{x}=4 \mathrm{t}^{2}-2 \mathrm{t}+2 \ldots \ldots(1)$

$\operatorname{again} \frac{\mathrm{dy}}{\mathrm{dt}}=2$

$\Rightarrow \int \mathrm{d} \mathrm{y}=2 \int \mathrm{dt}$

$\mathrm{y}=2 \mathrm{t}+\mathrm{c}^{1}$

at $t=2, y=4$ 

$	herefore 4=2 	imes 2+c^{1}$

$\Rightarrow \mathrm{c}^{1}=0$

$	herefore \mathrm{y}=2 \mathrm{t} \Rightarrow \mathrm{t}=\frac{\mathrm{y}}{2}$

by equation $(1)$

$x=4\left(\frac{y}{2}ight)^{2}-2\left(\frac{y}{2}ight)+2$

$\Rightarrow x=y^{2}-y+2$

A particle moves in the $xy$ -plane with velocity $u_x = 8t -2$ and $u_y = 2$. If it passes through the point $(14, 4)$ at $t = 2\, s$, the equation of its path is

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