A) (a) The acceleration of the particle is zero at \[t=0\] second
B) The velocity of the particle is zero at \[t=0\] second
C) The velocity of the particle is zero at \[t=1\] second
D) The velocity and acceleration of the particle are never zero
Correct Answer: C
Solution :
\[{{v}_{x}}=\frac{dx}{dt}=\frac{d}{dt}(3{{t}^{2}}-6t)=6t-6\]. At \[t=1,\ {{v}_{x}}=0\] \[{{v}_{y}}=\frac{dy}{dt}=\frac{d}{dt}({{t}^{2}}-2t)=2t-2\]. At \[t=1,\ {{v}_{y}}=0\] Hence \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}=0\]
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