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An object of mass $3 \mathrm{~kg}$ is at rest. Now a force $F=6 t^2 \hat{i}+4 \hat{j}$ is applied on the object then the velocity of the object at $t=3 \mathrm{~s}$ is:
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Verified Answer
The correct answer is:
$18 \hat{i}+6 \hat{j}$
From question $\vec{a}=\frac{\vec{F}}{m}=2 t^2 \hat{i}+\frac{4}{3} t \hat{j}$
$\therefore d \vec{V}=\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t$
Integrated both sides, We have,
$\vec{V}=2\left[\frac{t^3}{3}\right] \hat{i}+\frac{4}{3}\left[\frac{t^2}{2}\right] \hat{j}$
at $t=3 \mathrm{sec}, \vec{V}=\frac{2}{3}(3)^3 i+\frac{4}{6}(3)^2 j$
$=18 \hat{i}+6 \hat{j}$
$\therefore d \vec{V}=\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t$
Integrated both sides, We have,
$\vec{V}=2\left[\frac{t^3}{3}\right] \hat{i}+\frac{4}{3}\left[\frac{t^2}{2}\right] \hat{j}$
at $t=3 \mathrm{sec}, \vec{V}=\frac{2}{3}(3)^3 i+\frac{4}{6}(3)^2 j$
$=18 \hat{i}+6 \hat{j}$
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