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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2022
  • Medium
  • Numerical

A force on an object of mass \(100\, g\) is \((10 \hat{i}+5 \hat{j}) N\). The position of that object at \(t=\) \(2 s\) is \((a \hat{i}+b \hat{j}) m\) after starting from rest. The value of \(\frac{a}{b}\) will be..........

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Correct Answer

(A) \(2\)

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Detailed explanation

\(\overrightarrow{ F }=10 \hat{ i }+5 \hat{ j }\)

\(m =100 \,g =0.1 \,kg\)

\(\vec{a}=\frac{\overrightarrow{ F }}{ m }=100 \hat{ i }+50 \hat{ j }\)

\(\overrightarrow{ S }=\overrightarrow{ u t}+\frac{1}{2} \overrightarrow{ a }{ }^{2}=\frac{1}{2} \overrightarrow{ a } t ^{2}(\) as \(\overrightarrow{ u }=0)\)

\(=\frac{1}{2}(100 \hat{ i }+50 \hat{ j }) 2^{2}\)

\(=200 \hat{i}+100 \hat{j}\)

\(=a \hat{i}+b \hat{j}\)

\(a=200, b=100\)

\(\therefore \frac{a}{b}=2\)

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