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At any instant the velocity of a particle of mass $500 \mathrm{~g}$ is $\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{ms}^{-1}$. If the force acting on the particle at $t=1 s$ is $(\hat{i}+x \hat{j}) N$. Then the value of $\mathrm{x}$ will be:
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Mass of particle, $\mathrm{m}=500 \mathrm{~g}=0.5 \mathrm{~kg}$
velocity of a particle, $\vec{v}=2 t \hat{i}+3 t^2 \hat{j}$
$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=2 \hat{\mathrm{i}}+6 \mathrm{t} \hat{\mathrm{j}}$
at $\mathrm{t}=1, \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$
Force acting on the particle,
$\overrightarrow{\mathrm{F}}=\mathrm{ma}=0.5(2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+x \hat{\mathrm{j}} \quad$ Hence $x=3$
velocity of a particle, $\vec{v}=2 t \hat{i}+3 t^2 \hat{j}$
$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=2 \hat{\mathrm{i}}+6 \mathrm{t} \hat{\mathrm{j}}$
at $\mathrm{t}=1, \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$
Force acting on the particle,
$\overrightarrow{\mathrm{F}}=\mathrm{ma}=0.5(2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+x \hat{\mathrm{j}} \quad$ Hence $x=3$
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