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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2020
  • Medium
  • MCQ

Starting from the origin at time \(t=0,\) with initial velocity \(5 \hat{ j }\, ms ^{-1},\) a particle moves in the \(x-y\) plane with a constant acceleration of \((10 \hat{ i }+4 \hat{ j })\, ms ^{-2}\). At time \(t\), its coordinates are \(\left(20\, m , y _{0}\, m \right) .\) The values of \(t\) and \(y _{0},\) are respectively

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Answer & Solution

Correct Answer

(C) \(2\,s\) and \(18\, m\)

Step-by-step Solution

Detailed explanation

Given \(\quad \overrightarrow{ u }=5 \hat{ j } m / s , \overrightarrow{ a }=10 \hat{ i }+4 \hat{ j }, \quad\) final coordinate \(\left(20, y_{0}\right)\) in time \(t\)

\(S_{x}=4 u_x \times {t}+\frac{1}{2} a_{x} t^{2}\)

\(\frac{1}{2} \times 10 \times t^{2}\)

\(t=2 \sec\)

\(S_{y}=u_{y} \times t+\frac{1}{2} a_{y} t^{2}\)

\(y_{0}=5 \times 2+\frac{1}{2} 4 \times 2^{2}=18 m\)

\(2 \sec\) and \(18\, m\)

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