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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2021
  • Easy
  • MCQ

The velocity of a particle is \(v = v _{0}+ gt + Ft ^{2}\). Its position is \(x=0\) at \(t=0\); then its displacement after time \(( t =1)\) is :

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Answer & Solution

Correct Answer

(B) \(v_{0}+\frac{g}{2}+\frac{F}{3}\)

Step-by-step Solution

Detailed explanation

\(v = v _{0}+ g t + Ft ^{2}\)

\(\frac{ ds }{ dt }= v _{0}+ gt + Ft ^{2}\)

\(\int ds =\int_{0}^{1}\left( v _{0}+ gt + Ft ^{2}\right) dt\)

\(s=\left[v_{0} t+\frac{g t^{2}}{2}+\frac{F t^{3}}{3}\right]_{0}^{1}\)

\(s=v_{0}+\frac{g}{2}+\frac{F}{3}\)

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