A particle moves from the point (2.0hat i + 4 | vedclass

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(B) Given:Initial position $\vec{r}_0 = (2.0\hat i + 4.0\hat j) \, m$Initial velocity $\vec{u} = (5.0\hat i + 4.0\hat j) \, m/s$Acceleration $\vec{a}

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$20\sqrt{2} \, m$

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(B) Given:Initial position $\vec{r}_0 = (2.0\hat i + 4.0\hat j) \, m$Initial velocity $\vec{u} = (5.0\hat i + 4.0\hat j) \, m/s$Acceleration $\vec{a} = (4.0\hat i + 4.0\hat j) \, m/s^2$Time $t = 2 \, s$Using the equation of motion $\vec{r}(t) = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:$\vec{r}(2) = (2.0\hat i + 4.0\hat j) + (5.0\hat i + 4.0\hat j)(2) + \frac{1}{2}(4.0\hat i + 4.0\hat j)(2)^2$$\vec{r}(2) = (2.0\hat i + 4.0\hat j) + (10.0\hat i + 8.0\hat j) + \frac{1}{2}(4.0\hat i + 4.0\hat j)(4)$$\vec{r}(2) = (2.0\hat i + 4.0\hat j) + (10.0\hat i + 8.0\hat j) + (8.0\hat i + 8.0\hat j)$$\vec{r}(2) = (2.0 + 10.0 + 8.0)\hat i + (4.0 + 8.0 + 8.0)\hat j$$\vec{r}(2) = (20.0\hat i + 20.0\hat j) \, m$The distance from the origin is the magnitude of the position vector:$|\vec{r}(2)| = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, m$.

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