A force of (2.6 hat{i} + 1.6 hat{j}) text{ N} | vedclass

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(D) Given:Force $\vec{F} = (2.6 \hat{i} + 1.6 \hat{j}) 	ext{ N}$Mass $m = 2 	ext{ kg}$Initial velocity $\vec{v}_0 = (3.6 \hat{i} - 4.8 \hat{j}) 	ex

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$6$

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(D) Given:Force $\vec{F} = (2.6 \hat{i} + 1.6 \hat{j}) 	ext{ N}$Mass $m = 2 	ext{ kg}$Initial velocity $\vec{v}_0 = (3.6 \hat{i} - 4.8 \hat{j}) 	ext{ ms}^{-1}$ at $t = 0$.Using Newton's second law, $\vec{a} = \frac{\vec{F}}{m} = \frac{(2.6 \hat{i} + 1.6 \hat{j})}{2} = (1.3 \hat{i} + 0.8 \hat{j}) 	ext{ ms}^{-2}$.The velocity at any time $t$ is given by $\vec{v}(t) = \vec{v}_0 + \vec{a}t$.$\vec{v}(t) = (3.6 \hat{i} - 4.8 \hat{j}) + (1.3 \hat{i} + 0.8 \hat{j})t$$\vec{v}(t) = (3.6 + 1.3t) \hat{i} + (-4.8 + 0.8t) \hat{j}$.For the velocity to be along the $x$-axis only, the $y$-component of the velocity must be zero:$-4.8 + 0.8t = 0$$0.8t = 4.8$$t = \frac{4.8}{0.8} = 6 	ext{ s}$.

$A$ force of $(2.6 \hat{i} + 1.6 \hat{j}) \text{ N}$ acts on a body of mass $2 \text{ kg}$. If the velocity of the body at time $t = 0$ is $(3.6 \hat{i} - 4.8 \hat{j}) \text{ ms}^{-1}$, the time at which the body will just have a velocity along the $x$-axis only is: (in $\text{ s}$)

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