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(B) Given force $\overrightarrow{F} = (20 \hat{i} + 10 \hat{j}) \, N$ and mass $m = 2 \, kg$.Using Newton's second law,the acceleration is $\overright

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$500$

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(B) Given force $\overrightarrow{F} = (20 \hat{i} + 10 \hat{j}) \, N$ and mass $m = 2 \, kg$.Using Newton's second law,the acceleration is $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{20 \hat{i} + 10 \hat{j}}{2} = (10 \hat{i} + 5 \hat{j}) \, m/s^2$.The initial velocity $\overrightarrow{u} = 0$. The time $t = 10 \, s$.The displacement vector is given by $\overrightarrow{s} = \overrightarrow{u}t + \frac{1}{2} \overrightarrow{a} t^2$.$\overrightarrow{s} = 0 + \frac{1}{2} (10 \hat{i} + 5 \hat{j}) (10)^2$.$\overrightarrow{s} = \frac{1}{2} (10 \hat{i} + 5 \hat{j}) (100) = 50 (10 \hat{i} + 5 \hat{j}) = (500 \hat{i} + 250 \hat{j}) \, m$.The displacement along the $x$-axis is the $i$-component of the displacement vector,which is $500 \, m$.

$A$ boy pushes a box of mass $2 \, kg$ with a force $\overrightarrow{F} = (20 \hat{i} + 10 \hat{j}) \, N$ on a frictionless surface. If the box was initially at rest,then the displacement along the $x$-axis after $10 \, s$ is ........... $m$.

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