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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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..........

  1. A \(2\)
  2. B \(99\)
  3. C \(20\)
  4. D \(60\)
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Answer & Solution

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…
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