A particle leaves the origin with an initial | vedclass

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(B) The initial velocity is $\vec{u} = 3 \hat{i} 	ext{ m s}^{-1}$ and acceleration is $\vec{a} = -1 \hat{i} - 0.5 \hat{j} 	ext{ m s}^{-2}$.Using the

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$\frac{9}{2}(\hat{i} - \frac{\hat{j}}{2}) 	ext{ m}$

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(B) The initial velocity is $\vec{u} = 3 \hat{i} 	ext{ m s}^{-1}$ and acceleration is $\vec{a} = -1 \hat{i} - 0.5 \hat{j} 	ext{ m s}^{-2}$.Using the equation of motion $\vec{v} = \vec{u} + \vec{a}t$, the velocity at time $t$ is:$\vec{v}(t) = (3 - t) \hat{i} - 0.5t \hat{j}$.For the maximum $x$-coordinate, the $x$-component of velocity must be zero:$v_x = 3 - t = 0 \implies t = 3 	ext{ s}$.Now, using the position equation $\vec{r} = \vec{u}t + \frac{1}{2}\vec{a}t^2$:$\vec{r}(3) = (3 \hat{i})(3) + \frac{1}{2}(-1 \hat{i} - 0.5 \hat{j})(3)^2$$\vec{r}(3) = 9 \hat{i} + \frac{1}{2}(-9 \hat{i} - 4.5 \hat{j})$$\vec{r}(3) = 9 \hat{i} - 4.5 \hat{i} - 2.25 \hat{j} = 4.5 \hat{i} - 2.25 \hat{j}$.Factoring out $\frac{9}{2}$:$\vec{r}(3) = \frac{9}{2} \hat{i} - \frac{9}{4} \hat{j} = \frac{9}{2} (\hat{i} - \frac{\hat{j}}{2}) 	ext{ m}$.

$A$ particle leaves the origin with an initial velocity $\vec{v} = (3 \hat{i}) \text{ m s}^{-1}$ and a constant acceleration $\vec{a} = (-1 \hat{i} - 0.5 \hat{j}) \text{ m s}^{-2}$. The position vector of the particle, when it reaches its maximum $x$-coordinate, is:

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