At any instant velocity of a particle of mass 500,g is left(2 t hat{ i }+3 t ^2 hat{ j }right),ms ^{

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4-1.Newton's Laws of Motion

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At any instant the velocity of a particle of mass $500\,g$ is $\left(2 t \hat{ i }+3 t ^2 \hat{ j }\right)\,ms ^{-1}$. If the force acting on the particle at $t=1 s$ is $(\hat{i}+x \hat{j}) N$. Then the value of $x$ will be:

A

$3$

B

$4$

C

$6$

D

$2$

(JEE MAIN-2023)

Solution

$\overrightarrow{ v }=2 t \hat{ i }+3 t ^2 \hat{ j }$

$\overrightarrow{ a }=2 \hat{i}+6 t \hat{j}$

at $t=1, \vec{a}=2 \hat{i}+6 \hat{j}$

$\overrightarrow{ F }= ma =0.5(2 \hat{i}+6 \hat{j})=\hat{i}+3 \hat{j}$

$\overrightarrow{ F }=\hat{i}+x \hat{j} \quad \text { Hence } x=3$

Standard 11

Physics

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