A force vec{F} = a hat{i} + b hat{j} + c hat{ | vedclass

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(A) Given force $\vec{F} = a \hat{i} + b \hat{j} + c \hat{k}$.Using Newton's second law, the acceleration $\vec{a}_{acc} = \frac{\vec{F}}{m} = \frac{a

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$\frac{at^2}{2m}, \frac{bt^2}{2m}, \frac{ct^2}{2m}$

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(A) Given force $\vec{F} = a \hat{i} + b \hat{j} + c \hat{k}$.Using Newton's second law, the acceleration $\vec{a}_{acc} = \frac{\vec{F}}{m} = \frac{a}{m} \hat{i} + \frac{b}{m} \hat{j} + \frac{c}{m} \hat{k}$.Since the body starts from rest at the origin, the initial velocity $\vec{u} = 0$ and initial position $\vec{r}_0 = 0$.Using the kinematic equation $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}_{acc} t^2$, we get $\vec{r} = \frac{1}{2} \vec{a}_{acc} t^2$.Substituting the components:$x = \frac{1}{2} (\frac{a}{m}) t^2 = \frac{at^2}{2m}$$y = \frac{1}{2} (\frac{b}{m}) t^2 = \frac{bt^2}{2m}$$z = \frac{1}{2} (\frac{c}{m}) t^2 = \frac{ct^2}{2m}$Thus, the coordinates are $(\frac{at^2}{2m}, \frac{bt^2}{2m}, \frac{ct^2}{2m})$.

$A$ force $\vec{F} = a \hat{i} + b \hat{j} + c \hat{k}$ is acting on a body of mass $m$. The body was initially at rest at the origin. The coordinates of the body after time $t$ will be:

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