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(C) Given: mass $m = 2 \, kg$,force $\vec{F} = (2 \hat{i} + 3 \hat{j} + 5 \hat{k}) \, N$,initial velocity $\vec{u} = 0$,initial position $\vec{r}_0 =

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$12$

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(C) Given: mass $m = 2 \, kg$,force $\vec{F} = (2 \hat{i} + 3 \hat{j} + 5 \hat{k}) \, N$,initial velocity $\vec{u} = 0$,initial position $\vec{r}_0 = (0, 0, 0)$,time $t = 4 \, s$.Using Newton's second law,the acceleration $\vec{a}$ is:$\vec{a} = \frac{\vec{F}}{m} = \frac{2 \hat{i} + 3 \hat{j} + 5 \hat{k}}{2} = \hat{i} + 1.5 \hat{j} + 2.5 \hat{k} \, m/s^2$.Using the kinematic equation for position $\vec{r} = \vec{r}_0 + \vec{u}t + \frac{1}{2} \vec{a} t^2$:$\vec{r} = 0 + 0 + \frac{1}{2} (\hat{i} + 1.5 \hat{j} + 2.5 \hat{k}) (4)^2$$\vec{r} = \frac{1}{2} (\hat{i} + 1.5 \hat{j} + 2.5 \hat{k}) (16)$$\vec{r} = 8 \hat{i} + 12 \hat{j} + 20 \hat{k}$.Comparing this with the given coordinates $(8, b, 20)$,we find $b = 12$.

$A$ body of mass $2 \, kg$ moves under a force of $(2 \hat{i} + 3 \hat{j} + 5 \hat{k}) \, N$. It starts from rest and was at the origin initially. After $4 \, s$,its new coordinates are $(8, b, 20)$. The value of $b$ is ........ . (Round off to the Nearest Integer)

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