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The position of a particle moving in the $xy-$ plane at any time $t$ is given by $x = (3t^2 -6t)\, metres$, $y = (t^2 -2t)\,metres$. Select the correct statement about the moving particle from the following
The acceleration of the particle is zero at $t = 0\, second$
The velocity of the particle is zero at $t = 0\, second$
The velocity of the particle is zero at $t = 1\, second$
The velocity and acceleration of the particle are never zero
Solution
$\mathrm{x}=3 \mathrm{t}^{2}-6 \mathrm{t} \quad \mathrm{y}=\mathrm{t}^{2}-2 \mathrm{t}$
$\mathrm{v}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}}=6 \mathrm{t}-6 \quad \mathrm{v}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-2$
at $t=1 \sec \rightarrow v_{x}=0$ and $v_{y}=0$
hence $\quad v=\sqrt{v_{x}^{2}+v_{y}^{2}}=0$
