A force vec{F}=(40 hat{i}+10 hat{j}) , N acts | vedclass

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(B) Given: Force $\vec{F} = (40 \hat{i} + 10 \hat{j}) \, N$,mass $m = 5 \, kg$,initial velocity $\vec{u} = 0$,time $t = 10 \, s$.Using Newton's second

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$(400 \hat{i}+100 \hat{j}) \, m$

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(B) Given: Force $\vec{F} = (40 \hat{i} + 10 \hat{j}) \, N$,mass $m = 5 \, kg$,initial velocity $\vec{u} = 0$,time $t = 10 \, s$.Using Newton's second law,acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{40 \hat{i} + 10 \hat{j}}{5} = (8 \hat{i} + 2 \hat{j}) \, m/s^2$.Since the body starts from rest,the position vector $\vec{r}$ at time $t$ is given by the kinematic equation $\vec{r} = \vec{u}t + \frac{1}{2} \vec{a} t^2$.Substituting $\vec{u} = 0$,$\vec{a} = (8 \hat{i} + 2 \hat{j}) \, m/s^2$,and $t = 10 \, s$:$\vec{r} = 0 + \frac{1}{2} (8 \hat{i} + 2 \hat{j}) (10)^2$$\vec{r} = \frac{1}{2} (8 \hat{i} + 2 \hat{j}) (100)$$\vec{r} = (4 \hat{i} + 1 \hat{j}) (100) = (400 \hat{i} + 100 \hat{j}) \, m$.

$A$ force $\vec{F}=(40 \hat{i}+10 \hat{j}) \, N$ acts on a body of mass $5 \, kg$. If the body starts from rest,its position vector $\vec{r}$ at time $t=10 \, s$ will be -

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