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Given the velocity vector $(2t-1, {2t }^{ 2 }-5t+2)$ determine the position of the particle when it is at rest if at time $t = 0$, its position is $(0, 0)$.

$ (0, 0)$

$ \left(-\dfrac{1}{4}, \dfrac{11}{24}\right)$

$ (-1, 2)$

$ \left(6, -\dfrac{2}{3}\right)$